Fulton 2 - Lemmas

Lemma 1. If $N$ is a submodule of a finitely generated $R$ -module $M,$ then $M/N$ is finitely generated.

Let $\{s_1,\ldots,s_n\}$ be a generating set for $M.$ Then $\{s_1+N,\ldots,s_n+N\}$ generates $M/N.$

Lemma 2. Suppose $N$ is a submodule of an $R$ -module $M$ and $N,M/N$ are finitely generated over $R.$ Then $M$ is finitely generated over $R.$

Let $N=\sum Ra_i,$ $M/N=\sum R(b_i+N).$ We will show that $M=\sum Ra_i+\sum Rb_i.$ If $x\in M$ then clearly $x+N\in M/N.$ Therefore $x+N=\sum_{i=1}^mr_ib_i+N$ for some $r_i\in R.$ Therefore $x-\sum_{i=1}^mr_ib_i=y\in N.$ Moreover, $y=\sum_{i=1}^ns_ia_i$ for some $s_i\in R.$ Therefore $x=\sum_{i=1}^mr_ib_i+\sum_{i=1}^ns_ia_i.$

Definition 3. A module is Noetherian if every submodule is finitely generated.

Lemma 4. Suppose $N$ is a submodule of an $R$ -module $M.$ Then $M$ is Noetherian if and only if $N,M/N$ are Noetherian. Equivalently if $N,M,P$ are $R$ -modules and $0\longrightarrow N\longrightarrow M\longrightarrow P\longrightarrow 0$ is exact, then $M$ is Noetherian if and only if $N,P$ are Noetherian.

Suppose $M$ is Noetherian. Then $N$ is clearly Noetherian since submodules of $N$ are submodules of $M.$ Moreover, any submodule $L$ of $M/N$ is $R$ -module isomorphic to $L&39;/N,$ where $L&39;$ is some submodule of $M$ containing $N.$ Since $L&39;$ is finitely generated, $L$ is finitely generated by Lemma 1. Hence $M/N$ is Noetherian as well.

Suppose conversely that $N,M/N$ are Noetherian. Let $L$ be a submodule of $M.$ Then $L\cap N$ is a submodule of $N$ and hence is finitely generated. Moreover, $L/(L\cap N)$ is isomorphic to $(L+N)/N.$ Since $(L+N)/N$ is a submodule of $M/N,$ $(L+N)/N$ is finitely generated. Thus $L\cap N$ and $L/(L\cap N)$ are both finitely generated, and hence $L$ is finitely generated by Lemma 2.

The result may be reinterpreted in terms of the exact sequence above by noting that $P$ is $R$ -module isomorphic to $M/N.$

Lemma 5. If $R$ is a Noetherian ring, then $R^n$ is Noetherian as an $R$ -module for all $n\ge 1.$

If $n=1$ then $R$ is Noetherian as an $R$ -modules since it is a Noetherian ring (the $R$ -submodules of $R$ are precisely the ideals of $R$ ).

Suppose that $R^n$ is a Noetherian $R$ -module for some $n\ge 1.$ Observe that $R^{n+1}/R$ is $R$ -module isomorphic to $R^n,$ where $R$ is identified with the $(n+1)$ th copy of $R$ in $R^{n+1}.$ Since $R^n,$ $R$ are each Noetherian $R$ -modules, $R^{n+1}$ is Noetherian by Lemma 4.

Lemma 6. If $R$ is a Noetherian ring and $M$ is an $R$ -module, $M$ is Noetherian if and only if it is finitely generated over $R.$

If $M$ is Noetherian, then it is finitely generated by definition. Suppose $M$ is generated by the finite set $\{s_1,\ldots,s_n\}.$ Note that there exists a natural surjective $R$ -module homomorphism $\varphi:R^n\to M$ mapping $1$ in the $k$ th copy of $R$ to $s_k.$ Thus $M$ is $R$ -module isomorphic to $R^n/\ker\varphi.$ By Lemma 5, $R^n$ is Noetherian. Hence $M$ is Noetherian by Lemma 4.

Lemma 7. Let $K$ be a field and $R\supset K$ an integral domain. Let $L$ be the field of fractions of $R.$ Then $R$ is module finite over $K$ if and only if $L$ is a finite field extension of $K.$

Suppose $R$ is module finite over $K$ and $\{v_1,\ldots,v_n\}$ a basis for $R.$ Note that since $R$ is module finite over $K,$ $v_1,\ldots,v_n$ are each algebraic over $K.$ So $K(v_1,\ldots,v_n)$ is a finite extension of $K.$ Since $R\subset K(v_1,\ldots,v_n),$ $L\subset K(v_1,\ldots,v_n).$ Thus $L$ is a finite extension of $K.$

Suppose $L$ is a finite extension of $K.$ Since fields are Noetherian, $L$ is Noetherian as a $K$ -module by Lemma 6. Thus since $K\subset R\subset L,$ $R$ is a finitely generated $K$ -module.

Lemma 8. Let $F\in k[X_1,\ldots,X_n]$ be a polynomial of degree 1. Then there exists an affine change of coordinates $T:\mathbb{A}^n\to\mathbb{A}^n$ such that $F^T=X_m.$

Suppose that $F=\sum_{i=1}^n a_iX_i+a_0$ with $a_1,\ldots,a_n\in k\setminus\{0\}.$ We will define an affine change of coordinates $T=T&39;\circ T&39;&39;$ with $T&39;$ a linear map and $T&39;&39;$ a translation. Let $T_i&39;&39;=X_i$ for $i\ne n$ and $T_n&39;&39;=X_n-a_0.$ Define $T&39;$ by setting $T_i&39;=X_i$ for $i<d,$ $T_d&39;=a_d^{-1}X_d,$ and $T_i&39;=a_i^{-1}X_i-a_i^{-1}X_{i-1}$ for $i>d.$ Observe that $T&39;$ is defined by the following matrix in $k^{n\times n},$ $T&39;= \begin{bmatrix} 1 & 0 & 0 & \ldots & \ldots & 0\\ 0 & a_1^{-1} & 0 & \ldots & \ldots & 0\\ 0 & -a_{2}^{-1} & a_{2}^{-1} & \ldots & \ldots & 0\\ \vdots & \ddots & \ddots & \ddots & \ddots & \vdots\\ 0 & \ldots & \ldots & -a_{n-1}^{-1} & a_{n-1}^{-1} & 0\\ 0 & \ldots & \ldots & 0 & -a_n^{-1} & a_n^{-1} \end{bmatrix}$ Note that $T&39;$ is upper triangular, and therefore invertible. So $T=T&39;\circ T&39;&39;$ is an affine change of coordinates. Observe that $\begin{aligned}[t] F^T &= \left(\sum_{i=1}^n a_i(T_i)\right)+a_0\\ &= \left(\sum_{i=1}^{n-1} a_i(T_i&39;(X_i)\right)+a_n(T_n&39;(X_n-a_0)+a_0\\ &= a_1(a_1^{-1}X_1)+\left(\sum_{i=2}^{n-1} a_i(a_i^{-1}X_i-a_i^{-1}X_{i-1})\right)+a_n(a_n^{-1}(X_n-a_0)-a_n^{-1}X_{n-1})+a_0\\ &= X_n. \end{aligned}$

Lemma 9. Let $T$ be an affine change of coordinates on $\mathbb{A}^n$ and $S\subset k[X_1,\ldots,X_n].$ If $I(S)\ne I(S\setminus\{F_i\})$ for some $i\in\{1,\ldots,r\},$ then $I(S)^T\ne I(S\setminus\{F_i\})^T.$

Suppose $G\in I(S).$ Observe that if $\widetilde{T}(G)\in I(S\setminus\{F_i\})^T,$ then $\widetilde{T}^{-1}(\widetilde{T}(G))=G\in I(S\setminus\{F_i\}).$

Definition 10. Let $R$ be a ring, $A\subset R,$ and $I=I(A).$ We say $A$ is a minimal generating set for $I$ if for all $B\subset I$ such that $I=I(B),$ $|B|\ge|A|.$

Lemma 11. Suppose $V=V(F_1,\ldots,F_r)$ is a linear subvariety of $\mathbb{A}^n.$ If $\{F_1,\ldots,F_r\}$ is a minimal generating set for $I(V),$ then $V$ has dimension $n-r.$

By Problem 2.14 there exists an affine change of coordinates $T$ on $\mathbb{A}^n$ such that $V^T=V(X_{m+1},\ldots,X_n).$ Also, $V^T=V(I(V)^T)=V(F_1^T,\ldots,F_r^T).$ Since $\{F_1,\ldots,F_r\}$ is a minamal generating set for $I(V),$ by Lemma 2.9, $\{F_1^T,\ldots,F_r^T\}$ is a minimal generating set for $I(V)^T.$ Therefore $n-(m+1)+1=r$ and $m=n-r.$

Lemma 12. Let $V$ be an algebraic set in $\mathbb{A}^n,$ and $L$ a linear subvariety of dimension $1.$ Then either $L\subset V$ or $V\cap L$ is finite.

Suppose $L=V(X_2,\ldots,X_n).$ Let $F\in I(V).$ Note that $V\subset V(F).$ Observe that the set $V(F)\cap L$ is the set of solutions to $F(X_1,0,\ldots,0),$ a polynomial of one variable. Thus either $F(X_1,0,\ldots,0)=0$ and $F\in I(L),$ or $V(F)\cap L$ is finite.

Suppose $L$ is any linear subvariety of dimension $1.$ By 2.14(b), there exists an affine change of coordinates $T$ such that $L^T=V(X_2,\ldots,X_n).$ So by the result above, either $L^T\subset V^T$ or $V^T\cap L^T$ is finite. Thus $L\subset V$ or $V\cap L$ is finite.

Note. The lemma above also has a proof using Corollary 4 of the Nullstellensatz, by showing that $k[X_1,\ldots,X_n]/I(V\cap L)$ is a finite dimensional vector space over $k.$

Lemma 13. Suppose $W$ is a subvariety of a variety $V,$ and $P$ is a point in $W.$ Then the natural homomorphism $\pi:\Gamma(V)\to\Gamma(W)$ extends uniquely to a homomorphism $\mathcal{O}_{P}(V)\to\mathcal{O}_{P}(W).$

Note that $\pi:\Gamma(V)\to\Gamma(W)$ is the homomorphism induced by the inclusion polynomial map $W\hookrightarrow V.$ Thus by Problem 2.21, $\pi$ extends uniquely to a homomorphism $\mathcal{O}_{P}(V)\to\mathcal{O}_{P}(W).$

Lemma 14. Let $V\subset\mathbb{A}^n$ be an algebraic set. Then $\dim_k(k[X_1,\ldots,X_n]/I(V))$ is finite if and only if $V$ is finite.

Let us decompose $V(I)$ into irreducible algebraic sets $V(I)=\bigcup_{i=1}^mV_i.$

Suppose $V$ is finite, that is, each $V_i$ is a point. Therefore the ideals $I(V_1),\ldots,I(V_n)$ are distinct maximal ideals. In particular, they are pairwise comaximal. Thus, by the Chinese Remainder Theorem, $k[X_1,\ldots,X_n]/I(V)\cong k[X_1,\ldots,X_n]/I(V_1)\times \ldots\times k[X_1,\ldots,X_n]/I(V_m)\cong k^m$ Conversely, suppose $\dim_k(k[X_1,\ldots,X_n]/I(V))=d<\infty.$ Since $I(V_i)\supset I(V)$ for each $i,$ $\dim_k(k[X_1,\ldots,X_n]/I(V_i))$ is finite. Thus, by Problem 2.4, $k[X_1,\ldots,X_n]/I(V_i)\cong k$ for each $i$ and the result follows as above. Moreover, $d=m.$