Local Rings and Nakayama’s Lemma

Definition of local ring and basic properties. Definition of Jacobson radical with elementary criterion. Exercise 1.13. Exercise 4, 5, 6, 7, 9,10,11,12. Nakayama’s lemma and corollaries.

Lemma: Let M be a finitely generated module over A and let I be an ideal of A. If IM=M then there exists an i in I such that im=m for all m in M.

Proof: Let $\{x_1,\ldots,x_n\}$ be a set of generators. Since $IM=M$ we can write each one as a linear combination of the others $x_i=\sum_j a_{ij}x_j$ . View this as matrices, where the generators are in a column X, we write $A=(a_{ij})$ , and thus X=AX, hence 1X=AX, so $(1-A)X=0$ . Multiply by the adjoin matrix, and get that the determinant of 1-A annihilates every generator. But the determinant of 1-A is of the form 1-i, where i is an element of the ideal I. So $(1-i)x_i=0$ for all i, thus $(1-i)m=0$ for all m, hence the thesis.

Corollary: (link) Let M be a finitely generated A-module. Every surjective endomorphism is bijective.

Proof: Let $f:M\to M$ be the endomorphism and consider M as an $A[X]$ -module via f, i.e., $pm= p(x)m=p(f)(m)$ . Let $I=(x)\subseteq A[X]$ so we have $IM=M$ and we know there is a $p\in I$ such that $p m=m$ for all m. Now suppose $m\in Ker(f)$ . Then
$m=pm=p'xm=p'f(m)=p'0=0$
so $m=0$ and the morphisms is injective, therefore bijective.