Fulton 1.1

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 $F=\sum_{i=1}^n(X_i-a_i)G_i.$