8 Appendix: Polynomial Rings


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1 8 Appendix: Polynomial Rings Throughout we suppose, unless otherwise specified, that R is a commutative ring. 8.1 (Largely) a reminder about polynomials A polynomial in the indeterminate X with coefficients from the ring R is an expresssion of the form a n X n + a n 1 X n a 1 X + a 0 or, more briefly, n i=0 a ix i, where the a i are elements of R and where we adopt the convention that X 0 = 1. If a n 0 then this polynomial has degree n;: deg(f) = n. Typically we use notations such as f, g or, if we want to display the indeterminate, f(x), g(x), for polynomials. A polynomial of degree 0 can be regarded as simply an element of R  these are the constants and the term a 0 of a polynomial such as that above is called the constant term. The expressions such as a i X i appearing in the polynomial f above are called the terms of f and the term of highest degree is called the leading term and denoted lt(f). The element a i of R is the coefficient of the term a i X i ; the leading coefficient, denoted lc(f), is the coefficient of the leading term. For instance, the degree of f = 5X 3 + X 2 1 is 3, its leading term is 5X 3, its leading coefficient is 5, its constant term is 1. We let R[X] denote the set of all polynomials in the indeterminate X with coefficients from R. This becomes a ring when we equip it with the usual addition and multiplication of polynomials. We call it the polynomial ring in one indeterminate (X) over R (of course, taking, for example, T for the indeterminate rather than X gives the ring R[T] in place of R[X] but these rings are clearly isomorphic  so it s reasonable to talk about the polynomial ring in one indeterminate over R). Notice that there is another operation on polynomials which is not there in the ring structure of R[X], namely substitution: given polynomials f, g R[X] we can replace every occurrence of X in f(x) by the polynomial g(x) to get the result, f(g(x)), of substituting g(x) for X in f. For example if f(x) = 3X 2 X + 14 and if g(x) = X 1 then f(g(x)) = 3(X 1) 2 (X 1) + 1 = 3X 2 7X + 5. All this generalises to polynomials in more than one indeterminate. Given indeterminates X 1,...X t and a ring R, a monomial, in X 1,...X t with coefficient from R, is a term of the form ax n1 1...Xnt t with the n i nonnegative integers (possibly zero) and a R. The total degree of this monomial is n n t and the degree of X i in this monomial is n i. A polynomial in X 1,...X t with coefficients from R, is a sum of such monomials. The total degree of the polynomial and the degree of X i in the polynomial are defined to be the maximum of such over all monomials appearing (that is, with nonzero coefficient). We use deg(f) to denote the total degree of f which we also sometimes refer to just as the degree of f. A brief notation for f(x 1,...,X t ) is f(x)  so X denotes the tuple (X 1,...,X t ) of indeterminates. For example 3X1X 2 2 X 1 X is a polynomial in X 1,...X 5 with integer coefficients. Its (total) degree is 8, the degree of X 1 in it is 2, the degree of X 2 in it is 1, the degree of X 3 in it is 0, etc. 58
2 Notation (multiindices): we let the symbol X σ denote a term of the form X σ1 1...Xσt t where each σ i is a nonnegative integer, so then a polynomial in X 1,...X t with coefficients from R can be represented briefly as σ a σx σ with the a σ R. As with polynomials in one indeterminate, we can add and multiply such polynomials, obtaining the ring R[X 1,...,X t ] of polynomials in X 1,...X t with coefficients from R. When the number of indeterminates (also called variables) is small we use briefer notation such as R[X, Y, Z]. Any polynomial in X 1,...X t with coefficients in R (we also say over R) can we regarded as a polynomial in X t with coefficients from the polynomial ring R[X 1,...X t 1 ]. For instance: X 1 X 3 2 X 1 X 2 X 3 + 4X 3 + 2X may be regarded as a polynomial 2X (4 X 1 X 2 )X 3 + (X 1 X ) of degree 2 in X 3 with coefficients from Z[X 1, X 2 ]. Notice that, regarded as such, the constant term of this polynomial is X 1 X ! (but, of course, regarded as a polynomial in X 1, X 2, X 3, the constant term is 1). The point being made here is that we can regard the polynomial ring R[X 1,...X t ] as being obtained by adding one indeterminate at at time, that is, as R[X 1 ][X 2 ]...[X t ]. 8.2 Homogeneous polynomials A polynomial f(x) is homogeneous, of degree d, if f is a sum of monomials each of degree d: f(x) = σ a σx σ where each term X σ has total degree d. Notice that a homogeneous polynomial of degree 0 is just a constant and that a homogeneous polynomial of degree 1 is a linear polynomial. (Homogeneous polynomials of degree d are sometimes called forms of degree d.) Lemma 8.1. Every polynomial f(x) of total degree d can be expressed as a sum f = f d +f d 1 + +f 1 +f 0 of homogeneous polynomials f i with f i either the zero polynomial or homogenous of degree i. This homogeneous decomposition of f is unique in the following sense: if also f = g d + g d g 1 + g 0 with each g i homogeneous of degree i or 0 then g i = f i for all i = d,...,0. Proof. A formal proof of this pretty obvious point can be given by inducting on the degree d. (To do it properly one also has to be quite precise about what a polynomial is and one has to define what it means for two polynomials to be equal. If you really want to pursue this I suggest that you regard a polynomial as a formal expression built up from certain symbols and that you define polynomials to be equal (I really mean equivalent ) if they can be transformed to identical expressions by using certain algebraic manipulations.) Lemma 8.2. Let K be a field and let f(x) K[X]. If f(x) is homogeneous of degree d then for every element λ K we have f(λx) = λ d f(x) where, if X = (X 1,...,X t ) then λx means (λx 1,...,λX t ). If the field K is infinite then the converse also is true. Proof. Suppose first that f is homogeneous of degree d. Then f is a sum of monomials of the form a σ X σ1 1...Xσt t where σ σ t = d. The corresponding monomial in f(λx) is a σ (λx 1 ) σ1...(λx t ) σt = a σ λ σ1 X σ1 1. X..λσt t σt = a σ λ σ1+ +σt X σ1 1...Xσt t = a σ λ d X σ1 1...Xσt t = λ d a σ X σ1 1...Xσt t. So, if f(x) = σ a σx σ then f(λx) = σ λd a σ X σ = λ d σ a σx σ = λ d f(x), as claimed. 59
3 For the converse suppose that f is a polynomial of degree n, say, such that f(λx) = λ d f(x) for every λ K. As in Lemma 8.1 write f = f n + + f 0 where each f i is homogeneous of degree i or is 0. Now notice that we can choose λ K so that λ d is different from each power λ i for 1 =,...,n, i d (this is possible since K is infinite  we just have to avoid the finitely many solutions to the equations T d T i = 0 (1 = 0,...,n, i d)). By the first part, we have f(λx) = λ n f n (X) + + λ i f i (X) + + f 0 (X) and by assumption this equals λ d f(x) = λ d f n (X) + + λ d f i (X) + + λ d f 0 (X). Comparing homogeneous components (so using the uniqueness part of 8.1) we obtain λ d f i (X) = λ i f i (X) for each i and so, since λ d λ i for i d, we conclude that f i (X) is the zero polynomial for each i d. Hence f(x) = f d (X), and f is homogeneous of degree d, as claimed. 8.3 Polynomial functions If R is any ring and f(x) R[X 1,...,X t ] then f defines the corresponding polynomial function from R t to R which takes any ttuple (r 1,...,r t ) R t to the result f(r 1,...,r t ) R of substituting r 1 for X 1, r 2 for X 2..., r t for X t in f(x). A point to note: when we say that two polynomials are equal we mean that they are equal as polynomials, not as polynomial functions. The latter is a weaker condition, as the next example shows. Example 8.3. Since 3 is a prime the ring, Z 3, of integers modulo 3 is a field. Consider the polynomials f(x) = X 3 + X + 1 and g(x) = 2X in Z 3 [X]. Then you can check that these define the same polynomial function from Z 3 to Z 3 yet they are different polynomials: f g. In fact, over infinite fields two polynomials are equal iff they define the same polynomial function but this is not so over finite fields. 8.4 The division algorithm The following is a result that you have seen already. Gröbner basis methods are exactly what allow us to extend it to polynomials in more than one indeterminate. Theorem 8.4. Division Theorem Let K be a field and let f, g K[X] be polynomials in one indeterminate with coefficients from K. Suppose that g 0. Then there are polynomials q, r K[X] such that f = qg + r and deg(r) < deg(g). The polynomials q and r are uniquely determined by these conditions. (We call q the quotient of f by g and r the remainder when f is divided by g.) Proof. We recall the proof (briefly) to see what, if anything, goes wrong if f, g are in more than one indeterminate. So consider the set A of all polynomials of the form f hg where h K[X]. We choose a member r A, say r=f qg, of lowest degree. So f = qg + r. Also deg(r) < deg(g)  otherwise we could subtract a multiple of g from r to get something of lower degree, contradicting minimality of the degree of r. You should write this part out carefully because you will see that it uses the assumption that K is a field (as opposed to just a 60
4 ring) and, crucially, it completely breaks down if we re dealing with polynomials in more than one indeterminate. Exercise: exactly where does the proof break down? Associated with this theorem is the division algorithm  an algorithm which, given f and g, produces q and r  an algorithm with which you should be familiar. Also associated with the theorem are various results concerning greatest common divisors and relatively prime polynomials (just as with the division algorithm for the ring Z of integers). We will recall these as we need them. 8.5 Polynomial rings over domains Lemma 8.5. Let R be a commutative ring. Then the polynomial ring R[X] in one indeterminate over R is a commutative ring. Proof. First consider two monomials a i X i and b j X j. We have (a i X i )(b j X j ) = a i b j X i X j = b j a i X j X i = (b j X j )(a i X i )  so monomials commute. But every polynomial is a sum of monomials: say f = i a ix i, g = j b jx j and so fg = i a ix i. j b jx j = i j b jx j i a ix i = gf, as required. j (a ix i )(b j X j ) = j i (b jx j )(a i X i ) = Corollary 8.6. Let R be a commutative ring. Then the polynomial ring R[X 1,...,X t ] is a commutative ring. Proof. This is by induction using the previous corollary, since R[X 1,...,X t ] = (R[X 1,...,X t 1 ])[X t ] and R[X 1,...,X t 1 ] is commutative by induction. Lemma 8.7. Let R be a commutative domain. Then the polynomial ring R[X] in one indeterminate over R is a commutative domain. Proof. Commutativity is by the above. Suppose that f = n i=0 a ix i, g = m j=0 b jx j are nonzero polynomials with a n 0, b m 0. Then fg = n+m k=0 c kx k where c k = i+j=k a ib j. In particular, the term of highest degree is a n b m X n+m. Since R is a domain and since a n 0, b m 0 we have a n b m 0 and hence fg 0. Corollary 8.8. Let R be a commutative domain. Then the polynomial ring R[X 1,...,X t ] is a commutative domain. Proof. This is by induction on t, as for commutativity. Lemma 8.9. Suppose that R is a commutative domain and let f, g R[X 1,...X t ] be nonzero. Then deg(f g) = deg(f) + deg(g) and, if g is not a constant polynomial, then deg(f g) > deg(g). Proof. The proof of 8.7 shows that deg(fg) = deg(f)+deg(g) in the case where f, g are polynomials in one indeterminate. But the general case is not so obvious because, at first sight, it seems that cancellation between terms could lower the degree. We prove it by induction on deg(f) + deg(g). Suppose deg(f) = n, deg(g) = m. Write f = f + f where f is a homogeneous polynomial of degree n and deg(f ) < n. Write g = g +g where g is homogeneous of degree m and deg(g ) < m. Then fg = f g +(f g +f g +f g ). 61
5 Notice that (by induction or just because it s pretty obvious that we always have deg(fg) deg(f) + deg(g)) we have deg(f g + f g + f g ) < n + m. So we just have to show that deg(f g ) = n + m. Choose an indeterminate which actually appears in f  say X t (if it does so). Write f, g as polynomials in X t with coefficients from R[X 1,...,X t 1 ]: f = k i=0 a ixt, i g = l j=0 b jx j t, so a i, b j R[X 1,...,X t 1 ] and a k, b l 0 and, by assumption, k 1. Since f is homogeneous of degree n, the degree of a k as a polynomial in X 1,...,X t 1 is n k and, similarly, the degree of b l is m l. The leading monomial of the product f g has the form a k b l Xt k+l. By induction, deg(a k b l ) = (n k) + (m l) and hence deg(a k b l Xt k+l ) = (n k) + (m l) + (k + l) = n + m. But since no other term of f g can cancel with a k b l Xt k+l (every other term has smaller degree in X t ), we have deg(f g ) = n+m and hence deg(f g) = n + m, as required. The last part is left as an (easy) exercise. Theorem Let K be a field. Every ideal in the polynomial ring K[X] is principal. Proof. Let I be an ideal of K[X]. Since the zero ideal is blatantly principal, we may suppose that I 0. Choose an element g I \ {0} of least degree. We claim that g = I. Since g I, certainly g I (think about it!). Suppose, for the converse, that f I. Apply the Division Theorem (8.4) (noting that g 0) to obtain q, r K[X] with f = qg + r and deg(r) < deg(g). Since g,f Iso is r = f qg I. By minimality of deg(g) among nonzero elements of I, it must be that r = 0. But then f = qg and so f is, indeed, divisible by g and hence f g. Thus we have shown that I = g is principal, as required. 62
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