Fulton 1.1

Posted on November 15, 2017

1.1.* Let R be a domain.

(a) If F,G are forms of degree r,s respectively in R[X_1,\ldots, X_n], show that FG is a form of degree r+s.

Since F,G are forms F=\sum_{i=1}^m a_i X^{(i)} and G=\sum_{j=1}^l b_j X^{(j)} with a_i,b_j\in k and X^{(i)},X^{(j)} monomials of degree r,s for each (i),(j), respectively. Observe          FG=\sum_{i=1}^m\sum_{j=1}^l a_ib_jX^{(i)}X^{(j)}. Note each term is a monomial of degree r+s. Thus FG is a form.

(b) Show any factor of a form in R[X_1,\ldots,X_n] is also a form.

Suppose F is a form of degree n and G,H\in R[X_1,\ldots,X_n] such that GH=F. Then G=G_0+G_1+\ldots+G_r and H=H_0+H_1+\ldots H_s with r+s=n, G_i,H_j forms of degree i,j for each i\in\{0,\ldots,r\}, j\in\{0,\ldots, s\}, and G_r,H_s\ne 0.

Let i,j be the least integers such that G_i,H_j\ne 0. Then G_iH_j is a term of F with degree i+j. Since F is a form it must be that i+j=n and hence i=r, j=s.

1.2.* Let R be a UFD, K the quotient field of R. Show that every element z of K may be written z=a/b, where a,b\in R have no common factors; this representative is unique up to units of R.

Let z=a/b\in K. Since R is a UFD we have a=p_1\ldots p_n, b=q_1\ldots q_m with p_1,\ldots p_n,q_1,\ldots,q_m\in R irreducible. Suppose q_i\mid p_j for some i,j. Then p_j=uq_i for some unit u\in R. So          z=\frac{a}{b}=\frac{p_1\ldots p_j\ldots p_n}{q_1\ldots q_j\ldots q_m}                 =\frac{p_1\ldots uq_i \ldots p_n}{q_1\ldots q_i \ldots q_m}                 =\frac{up_1\ldots p_{j-1}p_{j+1}\ldots p_n}{q_1\ldots q_{i-1}q_{i+1}                         \ldots q_m}. By induction we may reduce a,b such that \gcd(a,b)=1. That is, if u\in R such that u\mid a and u\mid b, then u is a unit.

Now suppose z=a/b=c/d\in K such that \gcd(a,b)=\gcd(c,d)=1. So ad=bc. Since a\mid bc and \gcd(a,b)=1, it must be that a\mid c. Thus c=ra for some r\in R. Moreover, d=rb. Since \gcd(c,d)=1, r is a unit. Thus the representation is unique up to units of R.

1.3.* Let R be a PID. Let P be a nonzero, proper, prime ideal in R.

(a) Show that P is generated by an irreducible element.

Since R is a PID, P=(a) for some a\in P. Suppose a=rs for some r,s\in R. Since P is prime we may assume, without loss of generality, r=ac, c\in R. Thus a=acs. So cs=1 and s is a unit. Thus a is irreducible.

(b) Show that P is maximal.

Suppose Q=(b) is an ideal of R such that P\subset Q. Then a=br, r\in R. Since a is irreducible by part (a), one of b or r must be a unit. If r is a unit then P=(a)=(b)=Q. If b is a unit then bb^{-1}=1\in Q, so Q=(1)=R. Thus P is maximal.

1.4.* Let k be an infinite field, F\in k[X_1,\ldots,X_n]. Suppose F(a_1,\ldots,a_n)=0 for all a_1,\ldots,a_n\in k. Show that F=0.

Suppose n=1, that is, F\in k[X]. Then since F has infinitely many roots, F=0.

Suppose n>1 and the statement holds for k[X_1,\ldots,X_{n-1}]. Observe we may write      F=\sum_{i=0}^{\deg(F)}F_iX_n^i where F_i\in k[X_1,\ldots,X_{n-1}] for each i. Let a_1,\ldots,a_{n-1}\in k. If F_i(a_1,\ldots, a_{n-1})\ne 0 for any i then F(a_1,\ldots,a_{n-1},X_n) has finitely many roots. So F_i(a_1,\ldots,a_{n-1})=0 for each i. Thus by the inductive hypothesis, F_i=0 for each i and F=0.

1.5.* Let k be any field. Show that there are an infinite number of irreducible monic polynomials in k[X].

Suppose there exist only finitely many irreducible monic polynomials F_1, \ldots, F_n. Let F_{n+1}=F_1\ldots F_n+1. Let G be an irreducible polynomial dividing F_{n+1}. Since G\ne F_i for any i, we have a contradiction.

1.6* Show that any algebraically closed field is infinite.

Let k be an algebraically closed field. By Problem 1.5 there are infinitely many irreducible polynomials in k[X]. But since k is algebraically closed the irreducible polynomials are precisely those of the form (X-a), a\in k. Thus k is infinite.

1.7.* Let k be a field, F\in k[X_1,\ldots,X_n], a_1,\ldots,a_n\in k.

(a) Show that          F=\sum\lambda_{(i)}(X_1-a_1)^{i_1}\ldots(X_n-a_n)^{i_n},                 \ \lambda_{(i)}\in k.

Observe since K[X_1] is a Euclidean domain we may repeatedly divide by (X_1-a_1) a,b\in k and write F=\sum_{i=0}^{\deg(F)}\lambda_i(X_1-a_1)^i, \lambda_i\in k.

Suppose n>1 and the statement holds for k[X_1,\ldots,X_{n-1}]. Then we may write          F=\sum\mu_{(i)}(X_1-a_1)^{i_1}\ldots(X_{n-1}-a_{n-1})^{i_{n-1}}G_{(i)},                 \ \mu_{(i)}\in k, G_{(i)}\in k[X_n]. Moreover, for each (i) we may write G_{(i)}=\sum_{j=0}^{\deg(G_{(i)})}\mu_j (X_n-a_n)^j, each \mu_j\in k. By expanding and grouping by multidegree (i) we therefore may write          F=\sum\lambda_{(i)}(X_1-a_1)^{i_1}\ldots(X_n-a_n)^{i_n},                 \ \lambda_{(i)}\in k.

(b) If F(a_1,\ldots,a_n)=0, show that F=\sum_{i=1}^n(X_i-a_i)G_i for some (not unique) G_i in k[X_1,\ldots,X_n].

By part (a) we may write          F=\sum\lambda_{(i)}(X_1-a_1)^{i_1}\ldots(X_n-a_n)^{i_n},                 \ \lambda_{(i)}\in k. Since (a_1,\ldots,a_n) is a root of F the above sum has no constant term, that is, for each (i) we have i_j\ge 1 for some j\in\{1, \ldots,n\}. Hence we may regroup the sum in the form          F=\sum_{i=1}^n(X_i-a_i)G_i.