# Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman

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1 Lecture Notes Math 371: Algebra (Fall 2006) by Nathanael Leedom Ackerman October 17, 2006

2 TALK SLOWLY AND WRITE NEATLY!! Integral Domains and Fraction Fields Theorems Now what we are going to do is to consider ways in which a ring R can be embedded in a field F. Now we saw previously that if a R is a zero divisor then we can t adjoin an inverse without killing some elements. And so we can t embed a ring with zero divisors into a field. This turns out to be the only obstacle though. Integral Domain Definition A ring R is an integral domain if it has no zero divisors. I.e. 0 1 and if ab = 0 then a = 0 or b = 0. Cancellation Law

3 Theorem If R is an integral domain then it satisfies the cancellation law: ( a, b, c)(a 0) (ab = ac) (b = c) Proof. We have ab ac = a(b c) = 0 so b c = 0 as a 0. Integral Domain of Polynomials Theorem Let R be an integral domain. Then R[x] is an integral domain. Proof. ****************** Finite Integral Domain is a Field Theorem An integral domain with finitely many elements is a field. Proof. ****************** Field of Fractions 2

4 Theorem Let R be an integral domain. Then there exists an embedding φ : R F into a a field F Proof. The way we are going to show this is to mimic how the rational numbers are created from the integers. 3 Definition of Field of Fractions Definition A Fraction will be a pair a/b where a, b R and b 0. Two fractions a 1 /b 1, a 2 /b 2 are called equivalent a 1 /b 1 a 2 /b 2 if a 1 b 2 = a 2 b 1. First we want to check that is an equivalence relations. Reflexivity and Symmetry are obvious so all that is left is transitivity. Suppose a 1 /b 1 a 2 /b 2 and a 2 /b 2 a 3 /b 3. We then have a 1 b 2 = a 2 b 1 and a 2 b 3 = a 3 b 2. Hence a 1 b 2 b 3 =

5 a 2 b 1 b 3 = a 3 b 2 b 1 by the assumptions. But because we are in an integral domain we can cancel the b 1 and we get 4 a 2 b 3 = a 3 b 2 and hence is an equivalence relation. The Field of Fractions of R is defined to be the set of equivalence classes of fractions of R. We define (a/b)(c/d) = ac/bd, a/b + c/d = ad + bc bd We have to check that the axioms of a field are satisfied and that replacing a/b, c/d by equivalent elements doesn t change the answer, but this is easy and we won t do it here. As an example consider the the case of the ring K[x] where K is an integral domain. Then in this case the field of fractions is K(x) = {[f/g] : f, g are polynomials with g 0 and [f/g] is the equ

6 Extending a map onto field of fractions 5 Theorem Let R be an integral domain with field of fractions F and let ϕ : R K be any injective homomorphism from R into a field K. Then the rule Φ(a/b) = φ(a)φ(b) 1 defines the unique extension of ϕ to a homomorphism from F R. Proof. First we need to check that this extension is well defined. Since the denominator in a/b is not allowed to be zero and since ϕ is injective we have ϕ(b) 0 for all a/b. Hence ϕ(b) is invertible in K. and hence ϕ(a)ϕ(b) 1 is an element of K. Next we need to check that equivalent fractions have the same image. But a 1 /b 1 a 2 /b 2 a 1 b 2 = a 2 b 1

7 and hence ϕ(a 1 )ϕ(b 2 ) = ϕ(a 2 )ϕ(b 1 ) and Φ(a 1 /b 1 ) = ϕ(a 1 )ϕ(b 1 ) 1 = ϕ(a 2 )ϕ(b 2 ) 1 = Φ(a 2 /b 2 ) as required. 6 Hence Φ is a homomorphism. And showing that Φ is unique is easy. 0.2 Maximal Ideals Definitions Now lets look at surjective maps from a ring R to a field F. If ϕ : R F is such a map then we know by the first isomorphism theorem that F = R/ker(ϕ). So in particular we can recover F, ϕ from R, ker(ϕ). So we must look at ideals M such that R/M is a field. By the Correspondence Theorem (??4.3??) the ideals of R/M correspond to the ideals of R which contain M. So R/M is a field if R has exactly two ideals containing M.

8 7 I.e. M, R. Maximal Ideal Definition An ideal M is maximal if M R but M is not contained in any other ideals than M, R. Quotient by Maximal Ideal Lemma An ideal M is maximal if and only if R/M is a field. And the zero ideal is maximal if and only if R is a field. Proof. This is immediate from the definition of maximal ideal Examples Maximal Ideals in Integers Theorem The maximal ideals of the ring Z of integers are the principle ideals generated by the prime integers.

9 Proof. If (a) is an ideal and (a) (b) an ideal, then a divides b. Maximal Ideals in Polynomial rings over C 8 Theorem The maximal ideals of the polynomial ring C[x] are the principle ideals generated by the linear polynomials x a. The ideal generated by x a is the kernel of the substitution homomorphism s a : C[x] C sending f(x) f(a). So there is a bijective correspondence between maximal ideals in C[x] and C. Proof. First notice that if M is a maximal ideal then we know that M is a principle ideal generated by a monic polynomial f of least degree (Because C[x] is a PID). But since every complex polynomial has a root, f is divisible by some x a. But then f is in the principle ideal (x a)

10 and hence M (x a) so in fact M = (x a) as M is maximal. 9 Next we show that the kernel of the substitution homomorphism s a is generated by (x a). To say that a polynomial f is in the kernel of s a is to say that a is a root of g or that x a divides g. Thus (x a) generates s a. Since the image of s a is a field this shows that (x a) is a maximal ideal. Hilbert s Nullstellensatz Theorem (Hilbert s Nullstellensatz). The maximal ideals of the polynomial ring C[x 1,..., x n ] are in a bijective correspondence with points of complex n-dimensional space. A point a = a 1,..., a n in C n corresponds to the kernel of the substitution map s a : C[x 1,..., x n ] C which sends f(x) f(a). The

11 Kernel M a of this map is the ideal generated by the linear polynomials 10 x 1 a 1,..., x n a n Proof. Let a C n and let M a be the kernel of the substitution map s a. Since s a is surjective and C is a field M a is a maximal ideal. Next let us verify that M a is generated by the linear polynomials as asserted. To do so we expand f(x) in powers of x 1 a 1,..., x n a n writing f(x) = f(a)+ i c i (x i a i )+ i,j c ij (x i a i )(x j a j )+ The existence of such an expansion can be gotten by letting x = a + u and substituting it into f(x) and then replacing every u with x a. Now notice that every termon the right except f(a) is divisible by at least 1 of (x i a i ). So if f is in the kernel of s a, i.e. f(a) = 0 then f(x) is in

12 the ideal which is generated by (x 1 a 1 ),..., (x n a n ). Hence ({x i a i : i n}) generate M a. 11 Next we have to prove that every maximal ideal is of the form ({x i a i : i n}) for some point a C n. To do so lets let M be a maximal ideal and let K denote the field C[x 1,..., x n ]/M. We consider the restrictions of the canonical map π : C[x 1,..., x n ] K to the subring π i : C[x i ] K Lemma The kernel of π i is either 0 or else it is a maximal ideal. Proof. Assume the kernel isn t 0 and 0 f ker(π i ). Now since K isn t the zero ring ker(π i ) isn t the whole ring. So f is non-constant and hence divisible by a linear polynomial. Say f = (x i a i )g.

13 12 Then π(f) = π(x i a i )π(g) = 0 in K. a field either π(x i a i ) = 0 or π(g) = 0. Since K is So either (x i a i ) ker(π i ) or g(x i ) ker(π i ). So by induction on the degree of f ker(π i ) contains a linear polynomial. We are now going to show that ker(π i ) isn t the zero ideal and hence M contains a linear polynomial of the form (x i a i ). Because i was arbitrary this will then show that M is contained in the kernel of a substitution map and hence must be equal to the kernel as it is a maximal idea. Suppose ker(π i ) = 0. Then π i maps C[x i ] isomorphically to its image which is a subring of K. Hence π i can be extended to the field of fractions of C[x i ]. Hence K contains a subfield isomorphic to C(x), the field of ratio-

14 13 nal functions. Now the monomials x i = x i 1 1 x i n n C[x 1,... x n ] as a vector space of C. form a basis for So in particular C[x 1,..., x n ] has a countable basis as a vector space over C. And, as K is a quotient of C[x 1,..., x n ] there is a countable family which spans K (namely the residue of the x i ). However we will show that there are uncountably many linearly independent elements of C(x). It will follow that C(x) can t be isomorphic to a subfield of K. We need the assumption that C is uncountable and then the following two lemmas.

15 Lemma The uncountably many rational functions (x α) 1, α C are linearly independent. Proof. A rational function f/g defines an actual function by evaluation at points of the complex plane where g 0. The rational function (x α) 1 has a pole at α, which means that it takes on arbitrarily large values near α. It is also bounded near any other point. 14 Now consider the following linear combination. n i=1 c i x α i where α 1,..., α n are distinct complex numbers and say c 1 0. Then the first term is unbounded near α while the other functions are all bounded near it. Hence the linear combination doesn t define the zero function.

16 Lemma Let V be a vector space which is spanned by a countable family {v 1,..., v n,... } of vectors. Then every set L of linearly independent vectors in V is finite or countably infinite. Proof. Let L be a linearly independent subset of V and let V n be the span of the first n vectors and let L n = L V n. So L n is a linearly independent subset of a finite dimensional vector space and hence L n is finite. Moreover, 15 L = n L n. And, the union of countably many finite sets is finite or countable. 0.3 TODO Come up with A BUNCH of examples (more than I can use) so that I don t run out of time.

17 Go through Lang s book on the same topics. 16

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