# THROUGH THE FIELDS AND FAR AWAY

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1 THROUGH THE FIELDS AND FAR AWAY JONATHAN TAYLOR I d like to thank Prof. Stephen Donkin for helping me come up with the topic of my project and also guiding me through its various complications. Contents 1. Introduction 1 2. Basic Definitions and Theorems of Field Extensions 2 3. Algebraic Extensions 7 4. Integral Ring Extensions Transcendental Extensions Affine Varieties Algebraic Geometry Conclusion 41 References 42 Date: 12th May 2008.

2 THROUGH THE FIELDS AND FAR AWAY 1 1. Introduction In this project we will be examining the relationship between field extensions and algebraic geometry. Specifically, we will be looking at how the transcendence degree of a field extension can be used to define the notion of dimension. Field extensions, whose degrees are finite, are generally studied via Galois theory. However, when the degree of the extension is infinite, this tends to lie outside the scope of classical Galois theory. This second case is the one we will explore in greater depth during this project. The main bulk of this project will be the discussion of field extensions. An important step to reaching our main goal will be to prove a theorem known as Hilbert s Nullstellensatz (roughly translated, theorem of zeros ). In fact this result will be a corollary of a much more general theorem dealing with the embedding of one field into another. Our study of field extensions will be considered in three parts: algebraic extensions, integral ring extensions and transcendental extensions. By initially studying algebraic extensions we will be able to introduce some of the more basic concepts and fundamental results to the project. Then, we will concentrate for a substantial period of time on the discussion of integral ring extensions, which will benefit us later on in our study of algebraic geometry and affine varieties. Throughout each section covering extensions, we will always have the underlying goal of proving our main theorem. Therefore each section will finish with the formulation and proof of this theorem for each of the special cases. As we approach the end of section 5, we will encounter the corollary that will benefit us in our discussion of algebraic geometry. In section 6 we shall introduce the language of affine varieties, which we shall use to ground our study in section 7. In our final section we shall reformulate Hilbert s Nullstellensatz. There are, in fact, two forms of the Nullstellensatz in the language of affine varieties. They are referred to as the weak and strong form. It is the strong form of the Nullstellensatz that will be the restatement of our theorem from our discussion of field extensions. Finally in this section we shall show how the notion of dimension and transcendence degree are linked.

3 2 JONATHAN TAYLOR 2. Basic Definitions and Theorems of Field Extensions We note that all content from this section has been adapted from [1]. We start by introducing some of the basic definitions and facts that we will require for this project. Definition. Let E and F be fields, then if E F is a subfield of F we say F is an extension of E. This can be denoted by F/E but throughout this project we will use the notation E F. Example 2.1. We have the fields R C and hence the complex numbers are a field extension of the real numbers. We can view the field F as a vector space over the field E. We say that the extension is either finite or infinite, depending on whether the vector space has finite or infinite dimension. We denote the dimension of the vector space by [F : E] and call it the degree of the extension. This is a very coarse measure as to the size of the extension. Given any integral domain R we can define a ring R[X] called the polynomial ring of R. In this project we will be interested in the polynomial ring of a field E. We can define an element of the polynomial ring in the following way:- Consider a sum over all possible n-tuples ν = (ν 1,..., ν n ), then define f(x 1,..., x n ) = ν a (ν) x ν 1 1 x νn n for some a (ν) R. Hence R[X] is simply the set of all polynomials with coefficients in R. Definition. Let F be a field and p(x) F [X] such that deg(p) 1. We say that p(x) is reducible over F if there exists f(x), g(x) F [X], with deg(f), deg(g) 1, such that p(x) = f(x)g(x). If no such factorisation exists then we say p(x) is irreducible over F. We note that it is important to state which field the polynomial is irreducible over. Hardly any polynomials are universally irreducible over all fields. Example 2.2. We can see that x R[X] is not reducible over the field R. However, we have that x = x 2 ix + ix + 1 = (x + i)(x i). Thus it is reducible over C. Definition. Let E F be a field extension. Then any element α F is said to be algebraic over E if there exists a non-zero polynomial p(x) E[X] such that p(α) = 0. If no such polynomial exists then we say α is transcendental over E. Now that we have set up some of the fundamental definitions of our topic of discussion we move on to prove our first main theorem. However, before we do this we need to prove a small lemma about irreducible polynomials, (this has been taken from [2]). Lemma 2.1. Let F be a field and λ F where λ 0. We have that f is an irreducible polynomial over F λf is an irreducible polynomial over F.

4 THROUGH THE FIELDS AND FAR AWAY 3 Proof. Assume f(x) is reducible over F, then there exists g(x), h(x) F [X] such that f(x) = g(x)h(x). Hence f is reducible over F if and only if λf is reducible over F, because λf(x) = ( λg(x) ) h(x). Thus we have f irreducible over F if and only if λf is irreducible over F. The following theorem allows us to create a relationship between algebraic elements of a field and the polynomial rings in which they reside. This theorem and proof have been taken from [2]. Theorem 2.1. Let E F be a field extension and α F be algebraic over E. Then there exists a unique polynomial p(x) E[X] such that p is monic, irreducible over E and has α as a root. Proof. Let α F be algebraic over E, then there exists a polynomial f(x) E[X] such that f(α) = 0. Assume f(x) is the smallest polynomial in E[X] such that α is a root, (i.e. f has the least degree of all such polynomials in E[X]). We have that f must be irreducible over E. If not, then there exists a non-trivial factorisation of f, such as f(x) = g(x)h(x), where g(x), h(x) E[X] and deg(g), deg(h) < deg(f). However f(α) = 0 g(α) = 0 or h(α) = 0 but if this is true then this invalidates the choice of f as the smallest such polynomial. Hence f is irreducible over E. Let a n E be the coefficient of the highest term in f and define p(x) = 1 a n f(x). By Lemma 2.1, f irreducible over E implies p(x) is irreducible over E. Now we have shown that there exists a polynomial such that it is monic, irreducible over E and has α as a root. Assume that there exist two polynomials f 1, f 2 E[X] satisfying these conditions. Then we have that f 1 (α) = 0 and f 2 (α) = 0 (f 1 f 2 )(α) = 0. However, because f 1, f 2 are monic, we must have that the deg(f 1 f 2 ) is strictly less than either deg(f 1 ) or deg(f 2 ). This means the only possibility is that f 1 = f 2 p(x) is unique. We call the polynomial in Theorem 2.1 the minimum polynomial of α over E because it has the smallest degree of any such polynomial. We refer to a sequence of field extensions as a tower. For example, we may have E 1 E 2 E n. The idea of a tower will be important to us throughout our discussion of extensions and so we prove a small Lemma to do with the degree of a tower of extensions, (this has been taken from [2]). Lemma 2.2 (The Tower Law). Let E F L be a tower of field extensions. Then we have that [L : E] = [L : F ][F : E].

5 4 JONATHAN TAYLOR Proof. Let {x 1,..., x n } be a basis for the vector space F over E and {y 1,..., y m } be a basis for the vector space L over F. Let a be any element of L then, because L is a vector space over E, we can express a as a = m α i y i, i=1 with the coefficients α i F. Now F forms a vector space over E, hence each element α i F can be expressed as α i = n β ij x j, for some β j E. Thus this gives us that a can be expressed as a = m i=1 j=1 n β ij x j y i. j=1 Hence we have that the field L is spanned by the set {x j y i }. Are these elements linearly independent? Let γ ij E, then we have n i=1 m γ ij x i y j = 0 j=1 n γ ij x i = 0 γ ij = 0 i=1 for all i, j because of the linear independence of {x 1,..., x n } and {y 1,..., y m }. Thus the set {x i y j } form a basis for the vector space L over E, proving our result. We carry on with our discussion of basic theorems by developing more theory behind the polynomial ring. Specifically we wish to develop some special results for when R is a field. We are aiming to show that if f is an irreducible polynomial over F then F [X]/ f is a field, where f is the principal ideal generated by f. The following Lemma and proof have been taken from [4]. Lemma 2.3. Let F be a field, then every ideal in F [X] is principal. Proof. Let I F [X] be an ideal. Take f(x) I to be the polynomial with smallest degree in I. Then given any other g(x) I we have by the division algorithm, (see Section 4, Chapter V of [1]), that there exists q(x), r(x) F [X], with deg(r) < deg(f) or r(x) = 0, such that g(x) = q(x)f(x) + r(x). If r(x) = 0 then we have g(x) = q(x)f(x) and hence any element of I can be expressed as a multiple of f, which implies I = f. If r(x) 0 then we have r(x) = g(x) q(x)f(x) I,

6 THROUGH THE FIELDS AND FAR AWAY 5 by the ideal properties. However deg(r) < deg(f) and this invalidates the choice of f as the polynomial in I with smallest degree. Thus I must be a principal ideal generated by the polynomial with smallest degree. Definition. Let R be a ring and I R an ideal of R. We say I is a maximal ideal if given any other ideal J of R such that I J R, then J = I or J = R. The following proof and Theorem have been taken from [4]. Theorem 2.2. Take any ring R and let I R be an ideal. Then the quotient ring R/I is a field if and only if I is a maximal ideal. Proof. We have that R/I is a field if and only if the only ideals are {0 + I} and R/I. By the correspondence theorem for ideals, (see Section of [6]), we have R/I has no proper ideals if and only if there does not exist an ideal J R with I J R. This is equivalent to saying I is a maximal ideal. Now we have proved our result in the more general case of a ring, the result we desire comes out as a small corollary. The proof of the following Corollary has been taken from [2]. Corollary 2.1. Given a field F and a polynomial f F [X], we have that the quotient ring F [X]/ f is a field f is irreducible over F. Consequently F [X]/ f is a field extension of F. Proof. We start by showing that f is a maximal ideal if and only if f is irreducible over F. Assume f is maximal then if there exists h(x) F [X] such that h f then we have that f h but by the maximality of f we must have h = f or h = F [X], i.e. h = f, or h is a constant. Assume f is irreducible over F and that there exists an ideal g, such that f g F [X]. We must have that g f but f is irreducible and so g = λ, a constant, or g = λf. However λf = f and λ = F [X] and so f is a maximal ideal. Hence by combining the above and Theorem 2.2 we get that F [X]/ f is a field f is irreducible over F. Now we notice that there exists an isomorphic copy of F in F [X]/ f, because any element a F can be expressed uniquely as a + f in F [X]/ f. Hence F is a subfield of F [X]/ f and so is a field extension. Example 2.3. Consider the polynomial ring R[X] and the irreducible polynomial x over R. By our theorem we have that R[X]/ x is a field. Any element of this field will be of the form g(x) + x 2 + 1, where g(x) = ax + b with a, b F. If g(x) has degree greater than or equal to 2, then use the division algorithm, (see Chapter 4, Section V of [1]), to divide through by x Thus given any non-zero (ax + b) + x we claim the the element, ( a a 2 + b 2 x + ) b + x R[X]/ x 2 + 1, a 2 + b 2 is its inverse. We check this using the standard multiplication in the quotient ring.

7 6 JONATHAN TAYLOR ( )( ax + b a a 2 + b 2 x + ) b + x = a 2 + b 2 = 1 a 2 + b 2 ( a2 x 2 + abx abx + b 2 ) + x 2 + 1, 1 a 2 + b 2 ( a2 x 2 a 2 + a 2 + b 2 ) + x 2 + 1, = a2 a 2 + b 2 (x2 + 1) + a2 + b 2 a 2 + b + 2 x2 + 1, = 1 + x So, every non-zero element has an inverse and hence R[X]/ x is a field. 1 Looking at the above, we notice that the elements ax+b and ( ax+b) bear a striking a 2 +b 2 1 resemblance to the complex number b + ai and its multiplicative inverse (b ai). In b+ai fact, this striking resemblance is not a coincidence. It is possible to prove that R/ x = C, which can be seen in any book on classical Galois theory.

8 THROUGH THE FIELDS AND FAR AWAY 7 3. Algebraic Extensions In this section we will develop the theory of algebraic extensions in order to prepare us for the section concerning transcendental extensions. Primarily, we will be dealing with the notion of algebraic closure. Please note, all elements of this section have been adapted from [1], unless otherwise stated. We start by defining some general terminology. Definition. A field extension F of E is said to be algebraic if every element of the field F is algebraic over E. Definition. Let E F be a field extension and α an element of F. Then we write E(α) to represent the smallest field containing both E and α. We say that α is adjoined to E. The above definition is taken from [2]. It is important for us to consider how the property of being algebraic is maintained throughout a tower of extensions. To do this we consider the relationship between finite and algebraic extensions. Proposition: Let E F be a finite field extension. Then we must have that F is algebraic over E. Proof. We have that the dimension of F as a vector space over E is finite. Thus, the powers of β are linearly dependent, and hence there exists an equation a 0 + a 1 β + + a n β n = 0, such that a i E and not all a i = 0. Thus β is algebraic over E, and so F is algebraic over E. The following Lemma and proof have been taken from [2]. Lemma 3.1. Let α be algebraic over a field E. Then we have that E(α) = E[α] and E(α) is finite over E. The degree [E(α) : E] is equal to the degree of the minimum polynomial of α. Proof. Let p(x) E[X] be the minimum polynomial of α. Pick any polynomial f(x) E[X] such that f(α) 0 then we have p(x) f(x). Hence by the division algorithm we have that there exists polynomials g(x), h(x) E[X] such that g(x)f(x) + h(x)p(x) = 1. However because p(α) = 0 we get g(α)f(α) = 1 and hence f(α) has an inverse in E[α]. Thus E[α] is not only a ring but a field. We now want to show that E[α] is smallest among all such fields that contain E and α. Assume that E[α] is not the smallest field containing both E and α then we certainly have that E(α) E[α]. Now E(α) contains α and all elements of E and because E(α) is a field it must contain any polynomial f(α) with coefficients in E. This gives us that E[α] E(α) E(α) = E[α], as required. Let d = deg(p) then the powers of α 1, α, α 2,..., α d 1

9 8 JONATHAN TAYLOR are linearly independent over E. If this is not so then we will have f(α) = a 0 + a 1 α + + a d 1 α d 1 = 0 for some non-zero f(x) E[X]. Because p(x) is the minimum polynomial of α, we can see p(x) f(x) but deg(f) < deg(p), which is a contradiction. We claim that the set V = {1, α, α 2,..., α d 1 } forms a basis for E[α] as a vector space over E. Let g(x) E[X], then we have g(α) E[α]. By the division algorithm, there exist polynomials q(x), r(x) E[x], with deg(r) < d, such that g(x) = q(x)p(x) + r(x). However, this gives us that g(α) = r(α) and hence every element in E[α] can be expressed using only the elements of V, because deg(r) < d. Thus V forms a basis for E[α] = E(α) as a vector space over E and hence [E(α) : E] = d, as required. Lemma 3.2. Let F = E(a 1,..., a n ) be a finitely generated extension of a field E. Assume that each a j (j = 1,..., n) is algebraic over E, then we must have that F is algebraic over E. Proof. Consider the extension E F as a tower of extensions E E(a 1 ) E(a 1, a 2 ) E(a 1,..., a n ) = F. Each step in this tower is generated by adjoining one extra element to the field before it. By Lemma 2.2 we have that the degree [F : E] is [F : E] = [F : E(a 1,..., a n 1 )][E(a 1,..., a d 1 ) : E(a 1,..., a d 2 )]... [E(a 1 ) : E]. By Lemma 3.1 above we have that each one of these degrees is equal to the minimum polynomial of some a i over the field E(a 1,..., a i 1 ), i.e. each degree is finite. Then this gives us that [F : E] is finite and by the very first proposition of this section we have F is algebraic over E. Theorem 3.1. Given a tower of field extensions E F K we have that K is algebraic over E F is algebraic over E and K is algebraic over F. Proof. Start by assuming that K is algebraic over E, then given any element α K there exists a polynomial p(x) E[X], such that p(α) = 0. Now F K and hence F is algebraic over E. Also E F, hence p(x) E[X] p(x) F [X] and so K is algebraic over F. Now assume that K is algebraic over F and F is algebraic over E. We want to show that K is algebraic over E. Let α K then 0 f(x) F [X] exists, such that f(α) = a n α n + + a 1 α + a 0 = 0. Consider the field F = F (a n,..., a 0 ). Then F is a finitely generated extension of E and each a j is algebraic over E. Thus by Lemma 3.2 we have F is algebraic over E. We have a tower of extensions

10 THROUGH THE FIELDS AND FAR AWAY 9 E F F (α), where F (α) is a finite extension of E and hence is algebraic over E. Hence α is algebraic over E and so any element of K is algebraic over E, as required. We note that we refer to an embedding as a ring homomorphism which induces an isomorphism on its image. Any injective homomorphism will do so, and hence when we refer to an embedding we think of an injective homomorphism. In this section we will primarily be developing theorems about algebraic closure by looking at embeddings from one field to another. The major result that we are leading up to is to show that any field is in fact a subfield of an algebraically closed field. We say a field E is algebraically closed when given any polynomial f(x) E[X] with deg(f) 1 then f(x) has a root in E. On the road to this theorem we start with a small proposition, which has been taken from [2]. Proposition: Let E be a field and f(x) a polynomial in E[X] with deg(f) 1. Then there exists a field extension F of E such that f has a root in F. Proof. Assume that f is an irreducible polynomial. If f is not an irreducible polynomial then factorise f as f = gh, where g is irreducible, and continue with g. Consider the field E[X]/ f, i.e. the field of left cosets of the ideal f in E[X]. Elements of this field take the form g(x) + f, where deg(g) < deg(f). Now let us assume f(x) has the form f(x) = a 0 + a 1 X + a 2 X a n X n, where each a i E. Now f(x) has an isomorphic copy of itself in the field E[X]/ f and it takes the form f(x) = (a 0 + f ) + (a 1 + f )X + (a 2 + f )X (a n + f )X n. We want to show that f(x) has a root in the field E[X]/ f, which is an extension of the field E. Consider X + f as a possible root of f. This gives us f(x + f ) = (a 0 + f ) + (a 1 + f )(X + f ) + + (a n + f )(X + f ) n, = (a 0 + f ) + (a 1 X + f ) + + (a n X n + f ), = (a 0 + a 1 X + + a n X n ) + f, = f + f, = f. We note that in the field F [X]/ f the zero element is f and hence f has a root in F [X]/ f. Corollary 3.1. Let E be a field and f 1 (X),..., f n (X) be polynomials in E[X] with deg(f i ) 1. Then there exists a field extension E F such that every f i has a root in F.

11 10 JONATHAN TAYLOR Proof. We prove this by induction. We have that for one polynomial f 1 (X) that the result is true by the above proposition. Assume that there exists a field extension E F such that the polynomials f 1 (X),..., f r (X) have a root in F. Consider a polynomial f r+1 (X) E[X], such that f r+1 {f 1,..., f r } and deg(f r+1 ) 1. Then by the proof of the above proposition, we can see that f r+1 will have a root in the field F [X]/ f r+1. Clearly E F F [X]/ f r+1 is a field extension of E, and so F [X]/ f r+1 contains a root for each f i, (with 1 i r + 1). Hence our result is proved. Theorem 3.2. Given any field E there exists a field extension E F, such that F is algebraically closed. Proof. We wish to construct a field which is algebraically closed, i.e. given any f(x) F [X] there exists an element α F such that f(α) = 0. Now consider the polynomial ring E[X] and the ideal generated by all polynomials f(x) E[X] with deg(f) 1. Call this ideal I. We can see that I E[X]. If I = E[X] then there exists a finite number of elements f 1,..., f n I and g 1,..., g n E[X] such that g 1 f g n f n = 1, which is not possible. Hence there exists a maximal ideal h such that I h and E[X]/ h is a field extension of E. We can see that for any polynomial f E[X], with deg(f) 1, we will have f h. Hence, by the same reasoning as in our previous proposition, f will have a root in E[X]/ h. By repeating this argument we can inductively create a sequence of fields E 1 E 2 E n, in which every polynomial f(x) E n [X], with deg(f) 1, has a root in E n+1. Now let F = i=1e i, which we can show produces a field. Take x, y F, then for some n sufficiently large we have x, y E n. Performing the operations xy and x + y in E n then gives us xy F and x + y F. Given any polynomial f(x) F we must have, for some sufficiently large n, that f(x) E n [X], and hence f(x) has a root in E n+1, (which implies f(x) has a root in F ). Therefore, F is algebraically closed, as required. Corollary 3.2. Let E be a field, then there exists a field extension E E, which is algebraically closed and algebraic over E. Proof. Start by constructing an extension E F which is algebraically closed. consider all extensions of E contained in F E F i F, such that F i is algebraic over E. Now define E = i=1f i, (i.e. the union of all algebraic subextensions of F ). Let α E then we have that α is algebraic over E, (because α resides in some F i ), and hence E is algebraic over E. We now want to show that E is algebraically closed. Let f(x) E[X] F [X], then f(x) has a root α F because F is algebraically closed. We have a tower of extensions E E F where α is algebraic over E and, by Theorem 3.1, α is algebraic over E. Now we must have α F i for some i and so α E. Hence E is algebraically closed and algebraic over E. Now

12 THROUGH THE FIELDS AND FAR AWAY 11 Example 3.1. If we consider the field of real numbers, then the complex numbers are an extension of R that contain a root for each p(x) R[X], by the Fundamental Theorem of Algebra. Also, given any complex number xi + y we can find a polynomial p(x) R[X], such that p(xi+y) = 0. Hence R C is an algebraically closed extension which is algebraic over R. On reflection, we can appreciate that the above proofs required little input from fields and are, in fact, more set-theoretic in nature. All the results, that were used above had been proven in small Lemmas and Propositions earlier in the section. In connection with this we now move on to proving a theorem about extending embeddings σ : E L to algebraic extensions F of E. We will look to prove a similar result when considering transcendental extensions of a field. Theorem 3.3. Let E be a field and F an algebraic extension of E. Consider an embedding σ : E L into some algebraically closed field L. Then there exists an extension of σ into an embedding F in L. If F is algebraically closed and L is algebraic over σ(e), then any such extension of σ is an isomorphism of E into L. Proof. We wish to create a set S of totally ordered pairs. Start by considering all subfields of F which contain E, i.e. the tower of extensions E F i F. With each of these subfields assign a corresponding embedding τ i : F i L. We now take all such pairs (F i, τ i ) and form a totally ordered set S = { (F i, τ i ) }. We say that (F i, τ i ) (F j, τ j ) if F i F j and τ j Fi = τ i. We can see that S because by the initial statement of the theorem we have (E, σ) S. Define F = i=1f i and τ such that for any F j we have τ Fj = τ j. Now clearly (F, τ ) is an upper bound of our set and so we have a chain of ordered pairs (E, σ) (F 1, τ 1 ) (F 2, τ 2 ) (F, τ ). By Zorn s Lemma there must exist a maximal element (K, λ) of S. We claim K = F and hence there exists λ : F L, an extension of σ as required. If K F then there exists an element α F such that α K. We consider the field K(α) = K[α]. Now F is algebraic over E and hence there exists a minimum polynomial p(x) E[X] such that p(α) = 0. We know L is algebraically closed and λ(p) L[X] hence there exists a root β L of λ(p) = p λ. Note that we define p λ (X) in the following way p λ (X) = λ ( p(x) ) = λ(a 0 + a 1 X + + a n X n ) = λ(a 0 ) + λ(a 1 )X + + λ(a n )X n. Let f(α) be an element of K(α) and define an extension of λ by mapping f(α) f λ (β). We check that this is well defined. Let g(α) K(α) then if g(α) = f(α) we must have g(α) f(α) = 0 p(x) ( g(x) f(x) ). This gives us that

13 12 JONATHAN TAYLOR p λ (X) (g f) λ (X) g λ (β) f λ (β) = 0 g λ (β) = f λ (β). Hence the extension of λ is well defined. We can see that the restriction of our extension to K is λ. However this pair (K(α), λ ) is an element of S and so invalidates the maximality of (K, λ). Thus we must have K = F and λ is an extension of σ to F. Let F be algebraically closed and L be algebraic over σ(e), then σ(f ) is algebraically closed and L is algebraic over σ(f ). Hence L = σ(f ). As a consequence of this theorem we have that any two algebraically closed and algebraic extensions of E are isomorphic, (see Section 2, Chapter VII of [1]). Thus, we consider the E in Corollary 3.2 to be unique up to isomorphism and refer to it as the algebraic closure of E.

14 THROUGH THE FIELDS AND FAR AWAY Integral Ring Extensions All rings in this section will be commutative and have a unit element 1. Please note, all elements of this section have been adapted from [1], unless otherwise stated. Having spent time introducing algebraic field extensions, we now need to focus on a more general idea. By looking at the concept of integral ring extensions we can aim to generalise the result of Theorem 3.3 from the above section. Integral elements of rings are very similar to algebraic elements in field. We need this study of ring extensions to help prove our main result for transcendental field extensions and start by introducing the definition of an integral element. Definition. Let B be a ring and A a subring of B. We say an element α B is integral over A if there exists a polynomial p(x) = X n + a n 1 X n a 0 A[X], such that p(α) = 0. The way that this definition differs from the notion of an algebraic element is that the polynomial p(x) must be monic, i.e. the leading coefficient is 1. We note that there are two equivalent statements of the definition of integrality. We say α B is integral over A if: the subring A[α] is a finitely generated A-module, there exists a finitely generated A-module M such that, for any a A[α], am = 0 a = 0. (We will not prove the equality of these statements here but they can be found in Chapter 9, Section 1 of [1].) If we let A be a subring of a ring B, then we say B is integral over A if every element of B is integral over A. By an integral domain we refer to a ring which has no zero divisors. So, if A is an integral domain and a, b A such that ab = 0 then a = 0 or b = 0. Proposition: Let A be an integral domain and K its associated quotient field. If α A is algebraic over E then there exists a non-zero element c A, such that cα is integral over A. Proof. We have that α is algebraic over K and hence there exists some p(x) K[X] such that p(α) = k n α n + k n 1 α n k 0 = 0 and k n 0. We have that K is a field and thus k n has an inverse element kn 1. Multiply p(α) through by kn n 1 to get k n 1 n (k n α n ) + kn n 1 (k n 1 α n 1 ) + + kn n 1 k 0 = 0, (k n α) n + k n 1 (k n α) n k n 1 n k 0 = 0.

15 14 JONATHAN TAYLOR Each k i K is of the form a i b i for some a i A and non-zero b i A. Consider the product b = b n b 0, which is a non-zero element of A. Multiplying the above equation through by b n gives us (bk n α) n + bk n 1 (bk n α) n b n kn n 1 k 0 = 0. Hence each coefficient of the above polynomial is in A and c = bk n is a non-zero element of A. So cα is integral over A as required. Lemma 4.1. Let A be a subring of a ring B such that B is integral over A. If B is finitely generated as an A-algebra then B is finitely generated as an A-module. Proof. We note that given any subring A of a ring B we will always have that B is an A- algebra. So, if B is a finitely generated A-algebra then let w 1,..., w n be the generators of B. Now as {w 1,..., w n } is a subset of B we have that each w i is integral over A. Consider the following tower A A[w 1 ] A[w 1, w 2 ] A[w 1,..., w n ] = B. We start by showing that if w 1 is integral over A then we have A[w 1 ] is a finitely generated A-module. Let w 1 be integral over A then w 1 solves some monic polynomial p(x) A[X], with deg(p) 1. Let p be such a polynomial with least degree in A[X]. Now consider any polynomial f(x) A[X] with deg(f) 1. We have by the factor theorem that f(x) = g(x)p(x) + r(x), with deg(r) < deg(p) or r(x) = 0. If r(x) = 0 then f(w 1 ) = 0 f(x) = p(x), otherwise f(w 1 ) = r(w 1 ). If we have deg(p) = n then the elements 1, w 1,..., w1 n 1 clearly generate A[w 1 ]. Hence A[w 1 ] is a finitely generated A module. Now we can see from the tower above that B will be a finitely generated A-module. This is because each step of the tower is finitely generated. Proposition: Let A, B be subrings of a ring C such that A is a subring of B. Now we have that C is integral over A if and only if C is integral over B and B is integral over A. Proof. Assume that C is integral over A. Now α C α B and hence B is clearly integral over A. Also p(x) A[X] p(x) B[X] and so C is clearly integral over B. Now assume C is integral over B and B is integral over A. Given an element β C we have that there exists a monic polynomial in B[X] such that β n + b n 1 β n b 0 = 0. Consider the ring B = A[b n 1,..., b 0 ], then we have A is a subring of B and hence B forms a finitely generated A-algebra. Also each element of B is integral over A and hence by Lemma 4.1 we have that B is a finitely generated A-module. It s clear that given any element a B [β] then ab = 0 only if a = 0. Now βb B and so we must have that β is integral over A by the alternate definition of integrality. Thus C is integral over A, as required.

16 THROUGH THE FIELDS AND FAR AWAY 15 We now consider ring-homomorphisms and the notion of integral elements. As homomorphisms are structure preserving functions we would like to assume that they would also preserve integral elements. Lemma 4.2. Let A be a subring of a ring B such that B is integral over A. If σ : A B is a ring homomorphism then σ(b) is integral over σ(a). Proof. Let α B then we have that there exists some monic polynomial with coefficients a i A such that α n + a n 1 α n a 0 = 0. Now applying the homomorphism to the above equation we have that σ(α n + a n 1 α n a 0 ) = 0, σ(α n ) + σ(a n 1 )σ(α n 1 ) + + σ(a 0 ) = 0, σ(α) n + σ(a n 1 )σ(α) n σ(a 0 ) = 0. Thus σ(α) σ(b) is integral over σ(a) as each σ(a i ) σ(a). So, ring homomorphisms preserve the property of being integral. In the light of ring homomorphisms, it is going to be useful for us to adapt our idea of being integral. If f : A B is a ring homomorphism then we can consider B as an A-module by defining the map a b = f(a)b. Now we can see that this will allow us to satisfy the axioms of an A-module, which we detail below. Let a, b A and x, y B, then we have a (x + y) = f(a)(x + y) = f(a)x + f(a)y = a x + a y, (a + b) x = f(a + b)x = ( f(a) + f(b) ) x = f(a)x + f(b)x = a x + b x, (ab) x = f(ab)x = f(a) ( f(b)x ) = a (b x), 1 x = x (trivially). In fact B will form an A-algebra if f(a) lies in the centre of B, i.e. f(a)b = bf(a) for all a A and b B. However, all rings in this section are commutative and hence f(a) does lie in the centre of B. This gives us that for a A and x, y B we have x(a y) = x ( f(a)y ) = f(a)(xy) = a (xy), (a x)y = ( f(a)x ) y = f(a)(xy) = a (xy). Hence given any two rings, A and B, together with a ring homomorphism f : A B, we see that B forms an A-algebra. Now we extend the definition of being integral by saying that f is integral if B is integral over f(a). From a previous proposition we know that a tower of ring extensions is integral if and only if every step in the tower is integral. Now assume A B C is a tower of ring extensions.

17 16 JONATHAN TAYLOR Let f : A B and g : B C be integral ring homomorphisms then g f : A C is also integral. We note however that if g f is integral then this does not imply that f is integral. Given any ring A and a submonoid, under multiplication, of A, (say S), we can form the quotient ring of A by S. We do this by defining an equivalence relation for pairs (a, s) A S. We say that if (a, s) (a, s ) then there exists an element t S, such that (1) t(as a s) = 0. We denote the equivalence class containing a pair (a, s) by a and the quotient ring itself by s S 1 A. We define multiplication and addition in the quotient ring by the standard notation of fractions a s a = aa s ss, a s + a = as + sa. s ss By defining 0 S 1 A = 0 1 and 1 S 1 A = 1 1 we can see that the above operations will give S 1 A a ring structure. It can be shown using the relation above that these are, in fact, well-defined relationships. If A is an integral domain and S is the set of all non-zero elements in A, then we have that S 1 A is a field, which we refer to as the quotient field of A. For any given non-zero element a s S 1 A then a s s a = as sa = 1. We can see that this satisfies (1) because by letting t = 1 we have as 1 sa 1 = 0. Also s a S 1 A because a 0 and so S 1 A is a field. Example 4.1. Given our classic example of a ring Z, we can clearly see that Q is its associated quotient field. Our next step is to consider the integral relationship between two rings and their quotient rings. Let f : A B be an integral ring homomorphism and S a multiplicative submonoid of A. We can define a homomorphism between S 1 A and S 1 B, say S 1 f : S 1 A S 1 B by letting ( a ) (S 1 f) = f(a) s f(s) for all (a, s) A S. When we refer to S in S 1 B we strictly mean the image of S under f, i.e. ( f(s) ) 1 B. It is clear that the homomorphic property of f will endow S 1 f with the same property. Thus we have a commutative diagram

18 THROUGH THE FIELDS AND FAR AWAY 17 B f τ S 1 B S 1 f A τ S 1 A where τ is the canonical transformation τ(x) = x for x A or B. We now wish to show 1 that the property of being integral is preserved by S 1 f. Lemma 4.3. Let f : A B be an integral ring extension and S a multiplicative submonoid of A. Then S 1 f : S 1 A S 1 B is also integral. Proof. We have f is integral and so given some α B we see that α solves a monic polynomial in f(a)[x], say α n + f(a n 1 )α n f(a 0 ) = 0. If we apply the map τ to the above equation we find that ( ) n α + f(a ( ) n 1 n 1) α + + f(a 0) = 0, f(1) 1 f(1) 1 noting that f(1) = 1 because f is a ring homomorphism. Each coefficient of the above polynomial lies in S 1 A and hence α is integral over 1 S 1 A. Letting s be any element of S 1 then we multiply the above equation by to get f(s) n ( ) n α + f(a ( ) n 1 n 1) α + + f(a 0) f(s) f(s) f(s) f(s) = 0. n Each coefficient in the above polynomial lies in (S 1 f)(s 1 A) and so α f(s) S 1 B is integral over (S 1 f)(s 1 A). Thus, S 1 f is an integral ring homomorphism as required. Lemma 4.4 (Nakayama s Lemma). Let A be a ring and I an ideal of A which is contained in all maximal ideals of A. If M is a finitely generated A-module and IM = M then M = 0. Proof. We assume that M is finitely generated and IM = M. Let {w 1,..., w n } be a set of generators for M. Now as IM = M it follows that each of the generators, say w 1, has the following expression w 1 = α 1 w α n w n, for some α j I. Rearranging gives us that (2) (1 α 1 )w 1 = α 2 w α n w n and we argue that 1 α 1 must be a unit. If 1 α 1 is not a unit then it is contained in some maximal ideal J of A. As I is contained in all maximal ideals of A we have that

19 18 JONATHAN TAYLOR α 1 J 1 J by the ideal property of J. This is a contradiction and so 1 α 1 is a unit, thus we can multiply (2) by the inverse of 1 α 1. This implies that M can be generated by n 1 elements and so M = 0. Throughout this section, our aim has been to generalise Theorem 3.3 from the previous section to integral ring extensions. We need to look now at prime ideals and see how they affect integral rings. By looking at prime ideals of rings, we can consider local rings and then reduce some of our problems. We say an ideal p A is a prime ideal of a ring A if A/p is an integral domain. Equivalently we say p A is a prime ideal of A if ab p implies a p or b p. We can see this is true, as A/p is an integral domain (a + p)(b + p) = p, (a + p) = p or (b + p) = p, a p or b p. Now let A be a ring and p a prime ideal of A. Define S to be the set complement of p in A, i.e. S = A \ p. This is clearly a multiplicative submonoid of A. Take s, s S then if ss p we must have s p or s p but this is a contradiction and so S is closed under multiplication. We note that 1 p otherwise p is trivially the whole ring and so 1 S. Example 4.2. Let 3Z Z be an ideal of Z. It is easy to see that 3Z is prime. Given ab 3Z then ab = 3z for some z Z 3 3z and 3 ab. As 3 is a prime number we must have 3 a or 3 b and hence a 3Z or b 3Z. Taking the set complement of 3Z in Z we have S = Z \ 3Z = {z Z 3 z}. Clearly 3 0 and so 0 S, which means S forms a multiplicative submonoid of Z. Let p = 3Z, then we have which is a set of equivalence classes. { z } Z p = z Z and s S, s Let f : A B be a ring homomorphism, (i.e. B is an A-algebra), and consider the sets S 1 A and S 1 B. We denote these sets by A p and B p respectively and can consider B p as an A p -module. We make the following definition. Definition. Let A be a subring of a ring B and p, q be prime ideals of A and B respectively. We say that q lies above p if q A = p. Now let A be a subring of a ring B and p, q be prime ideals of A and B respectively such that q lies above p. If λ : A B is an injective homomorphism between the rings then we can induce an injective homomorphism λ : A/p B/q between the factor rings. Explicitly, this map will be λ(a + p) = λ(a) + q for all a A. This gives us a commutative diagram of injective homomorphisms.

20 THROUGH THE FIELDS AND FAR AWAY 19 B λ B/q λ A A/p The horizontal mappings in the diagram are just the canonical homomorphisms. For a simple example of λ we can consider λ to be the inclusion map. Now if B is integral over A then we have, by Lemma 4.2, that B/q is integral over A/p. Lemma 4.5. Let A be a subring of a ring B such that B is integral over A. Let p be a prime ideal of A then we have that pb B and there exists a prime ideal of B, say q, lying above p. Proof. We start by proving the first part of the statement, namely that pb B. We have that B is integral over A and so B p is integral over A p by Lemma 4.3. The ideal p is prime and so A p is a local ring of A with unique maximal ideal m p = S 1 p, for S = A \ p, (see Section 3, Chapter II of [1] for details). We notice that, by the ideal property of p, we have pb p = pa p B p = m p B p. From this we can see that it will be enough to show that pb B when A is a local ring. By the existence of a prime ideal in A we note that we must have 1 0. It is also apparent that pb = B if and only if 1 pb. Now assume for a contradiction that pb = B, then 1 pb and so 1 = α 1 b α n b n, for some α i p and b i B. Consider the ring B = A[b 1,..., b n ] we note that A is a subring of B and so B is a finitely generated A-algebra. By Lemma 4.1 we have that B is a finitely generated A-module. Given any q(b 1,..., b n ) B and p p it s clear that pq(b 1,..., b n ) B because the coefficients will still lie in A p. Thus pb = B and so by Nakayama s Lemma B = 0, however this is a contradiction, so pb B. Let f represent the homomorphism S 1 f : S 1 A S 1 B and consider the following commutative diagram: B A B p f A p All other homomorphisms in the above diagram are the canonical homomorphisms. We have shown in the first part of this proof that m p B p B p and so m p is contained in some maximal ideal M of B. Consider the pre-image of M in A p f 1 (M) = {a A p f(a) M}.

21 20 JONATHAN TAYLOR Now we have that f 1 (M) is an ideal in A p and in fact will be a maximal ideal containing m p. We know that m p is a maximal ideal in A p and so f 1 (M) = m p, i.e. M A p = m p. Consider the canonical mapping τ : B B p and the pre-image of M under this map τ 1 (M) = {b B τ(b) M}. Now this will be an ideal in B and in fact will be a prime ideal. Let bc τ 1 (M) then we have that τ(bc) M τ(b)τ(c) M. Any maximal ideal is also a prime ideal and so we must have either τ(b) M or τ(c) M, which gives us b τ 1 (M) or c τ 1 (M). Hence τ 1 (M) is a prime ideal of B, say q. We note that the pre-image of m p in A is just p. Now consider the pre-image of M going both directions in the commutative diagram. We can see that B q = p and so q lies above p, as required. We now progress to the final and main theorem of this section, the generalisation of Theorem 3.3. We want to look at extending a homomorphism from an integral ring into an algebraically closed field. Theorem 4.1. Let A be a subring of a ring B such that B is integral over A. Consider L to be an algebraically closed field and ϕ : A L to be a homomorphism. Then we can extend ϕ into a homomorphism from B to L. Proof. We start by reducing this problem to one of local rings. Let p be the kernel of ϕ, then p is a prime ideal of A. This is because if bc Ker(ϕ) then ϕ(bc) = 0 ϕ(b)ϕ(c) = 0. Now ϕ(b), ϕ(c) L and L is a field, which is an integral domain and so ϕ(b) = 0 or ϕ(c) = 0 b Ker(ϕ) or c Ker(ϕ). Consider the set complement S = A \ p then we can describe a commutative diagram B S 1 B A S 1 A as we have done before. In fact S 1 A is just the local ring A p. Now we have that B p is integral over A p by Lemma 4.3. We extend the definition of ϕ such that for any a A s p ( ) a ϕ = ϕ(a) s ϕ(s). Hence we can factor ϕ into a mapping A A p L. We now wish to show that this can be extended to a homomorphism B p L. We consider ψ : A p L such that if τ : A A p is the canonical homomorphism, then ϕ = ψ τ. Let m be the unique maximal ideal of A p and assume Ker(ψ) = m. By Lemma 4.5 we know that there exists a maximal ideal, (because m is maximal - see Section 1, Chapter IX of [1]), say M, of B p that lies above m. Clearly B p /M and A p /m are fields and in fact A p /m B p /M is an algebraic extension. By the first isomorphism theorem, (see Section of [6]), we have A p /m = ψ(a p ).

22 THROUGH THE FIELDS AND FAR AWAY 21 It is possible to find an isomorphism from A p /m to ψ(a p ) such that the factored homomorphism A p A p /m L is the same as ψ. Now we choose an embedding of B p /M into L such that the following diagram is commutative. B p B p \ M A p A p \ m This has allowed us to extend our homomorphism ψ : A p L to a homomorphism from B p L. Thus by composing this with the canonical homomorphism from B B p we have extended our original homomorphism ϕ and so our theorem is proved. L

23 22 JONATHAN TAYLOR 5. Transcendental Extensions Please note, all elements of this section have been adapted from [1], unless otherwise stated. Now we have developed some of the theory of field extensions, we can move on to the discussion of transcendental extensions. We currently have a very coarse measure of size in a field extension, called the degree of the extension. To improve on this we introduce the idea of dimension in a field extension, much like dimension in a vector space. To create a basis in a vector space we look at the linearly independent elements of the space. We do a similar thing for a field extension by looking at the algebraically independent elements of the extension. We start by defining what is meant for elements to be algebraically independent. Definition. Let E F be a field extension and S = {x i } a subset of F, with S = n. We say that S is algebraically independent over E if, summing over all possible n-tuples ν = (ν 1,..., ν n ), we have 0 = a (ν) x ν i i ν i with a (ν) E a (ν) = 0. Another way of putting this is that there exists no non-zero polynomial q(x) E[X] such that q(x 1,..., x n ) = 0. Example 5.1. We have that i and 2 are algebraically dependent over R, because given the polynomial q(x 1, x 2 ) = 2x x 2 R[X], it s clear to see that q(i, 2) = = 0. Hence the set {i, 2} is not algebraically independent over R. Example 5.2. Using well known results from number theory we know that e and π are transcendental over Q, and hence neither satisfy a polynomial equation in Q[X]. Thus we have that {e} and {π} are algebraically independent subsets of R over Q. It s important to note, however, that this set is not algebraically independent over R. It is currently unknown whether the subset {e, π} of R is algebraically independent over Q, but it has been proved that the set { π, e π, Γ( 1 4 )} is algebraically independent over Q. These algebraically independent subsets can be ordered by inclusion, with F an upper bound of the inclusion. Clearly by Zorn s Lemma these sets have maximal elements, i.e. sets with largest cardinality. Let E F be a field extension and M be a subset of F such that M is algebraically independent over E. If M has maximum cardinality with respect to such subsets, then we call the cardinality of M the transcendence degree or dimension of F over E. A subset M of F, which is algebraically independent over E and maximal amongst all such subsets, shall be referred to as a transcendence base of F over E. We can see that if M = {m 1,..., m i } is a transcendence base, then F is algebraic over E(M). This is

24 THROUGH THE FIELDS AND FAR AWAY 23 because if α F then we must have that α is algebraically dependent upon the set M over E because M is maximal. Therefore there exist some non-zero f(x) E[X] where f(α, m 1,..., m i ) = 0. If we define g(x) = f(x, m 1,..., m i ) then g(x) E(M)[X] and g(α) = 0, then we can see F is algebraic over E(M). These transcendence bases shall be the focus of our discussion in this section. To start, we will show that the cardinality of a transcendence base is unique. Theorem 5.1. Let E F be a field extension. If X and Y are two transcendence bases for F over E then we must have that X = Y. Proof. We start by dealing with the case when the transcendence base is finite. Let X be a finite transcendence base of F over E such that X = {x 1,..., x m }, m 1. We want to show that any other transcendence base must also have m elements. Let Y = {y 1,..., y n } be another transcendence base of F over E with n 1. By the definition of the transcendence base X is the largest algebraically independent subset of F over E. Following in our previous logic there exists a non-zero f(x) E[X], in m + 1 variables, where f(y 1, x 1,..., x m ) = 0. We have that X and Y are maximal algebraically independent subsets of F. Hence we must have at least one x i which is algebraically dependent on (y 1, x 1,..., x i 1, x i+1,..., x m ). After a suitable renumbering of the elements say this element is x 1. Now x 1 and y 1 must both occur in f and thus x 1 is algebraic over E(y 1, x 2,..., x m ). Assume for an inductive argument that after a finite application of the above process we can find {y 1,..., y l } (l < n), such that F is algebraic over E(y 1,..., y l, x l+1,..., x m ). Then there must exist a non-zero polynomial g(x) E[X], in m + 1 variables, where g(y l+1, y 1,..., y l, x l+1,..., x m ) = 0 and y l+1 occurs in g(x). Now the {y 1,..., y l+1 } and {x l+1,..., x m } are two algebraically independent sets over E. Due to the algebraic independence of the y i, some x j must also occur in g. After a suitable renumbering of {x l+1,..., x m }, if we assume that x l+1 occurs in g, then we have that x l+1 is algebraic over E(y 1,..., y l+1, x l+2,..., x m ). From the above arguments we have developed a tower of extensions, E E(x 1,..., x m ) E(y 1, x 2,..., x m ) E(y 1,..., y l+1, x l+2,..., x m ) F. By Theorem 3.1, a tower of algebraic extensions is algebraic. Hence we must have F is algebraic over E(y 1,..., y l+1, x l+2,..., x m ). If n m, then we can replace all the x j s by y i s and obtain F is algebraic over E(y 1,..., y m ). If n m then we can start this procedure with the basis Y and replace each of the y i s by x j s. This would give us F is algebraic over E(x 1,..., x m ), and hence we must have that n = m, i.e. X = Y. Now consider the case when X is infinite. If Y is finite then we can use a finite application of the above argument to replace each y i by an x j. However, this new set of

25 24 JONATHAN TAYLOR x j s will not be maximal as it will be strictly contained in X. Hence Y cannot form a transcendence base for the extension, which gives us that Y must be infinite. Theorem 5.2. Let E F be a field extension and Γ a set of generators of the extension, such that F = E(Γ). If S Γ is an algebraically independent subset, then S can be extended to a transcendence base M. Proof. If S is maximal with respect to the inclusion ordering of algebraically independent subsets, then it is a transcendence base already and we are done. If we assume S is not maximal then S = {s 1,..., s k } resides in some transcendence base X = {x 1,..., x n } of E F. Now the elements of S are algebraically independent and because S X there must exist some x i s j for all j = 1,..., k. Now after a sufficient renumbering of the elements of X, we can say this element is x k+1. Then we have that x k+1 is not algebraic over E(s 1,..., s k ) and so M = {s 1,..., s k, x k+1 } is algebraically independent over E. Repeat this argument until M = n and hence M is a transcendence base of E F. Lemma 5.1. Let E F L be a tower of field extensions. Then if n is the transcendence degree of E F and m is the transcendence degree of F L we have n + m is the transcendence degree of E L. Proof. Let X = {x 1,..., x n } be a transcendence base for the extension E F and Y = {y 1,..., y m } be a transcendence base for the extension F L. We want to show that the set M = {x 1,..., x n, y 1,..., y m } is a transcendence base for the extension E L. Now given any element α L it must be algebraically dependent upon the transcendence base Y. Hence there exists a non-zero polynomial p(x) F [X], such that p(y 1,..., y m, α) = 0. Each coefficient of p(x) is an element of F, which is algebraic over E(x 1,..., x n ). Hence we can replace each coefficient of p(x) with a non-zero polynomial of E[X] dependent upon x 1,..., x n. We can then consider p(x) to be the non-zero polynomial q(x) E[X], such that q(x 1,..., x n, y 1,..., y m, α) = 0. Renaming this polynomial g(x) = q(x 1,..., x n, y 1,..., y m, x) we have g(x) E(M)[X] and g(α) = 0. Thus α is algebraic over E(M) L is algebraic over E(M). To show that M is a transcendence base we have to show that it is maximal. If M is not maximal then there exists some M that contains M, which is algebraically independent over E. This suggests that there is some β M where {x 1,..., x n, y 1,..., y m, β} is an algebraically independent set. However, this gives us that {y 1,..., y m, β} is an algebraically independent subset of L over F, which invalidates the maximality of Y. Hence M is maximal and forms a transcendence base for E L with M = n + m.

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