Algebra 4 – April 3rd – Local rings and Nakayama’s Lemma

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.

Nakayama’s Lemma: Let M be a finitely generated module over A and let I be an ideal of A contained in the Jacobson radical J. If IM=M then M=0.

Proof: By the previous Lemma, there is an i in I such that im=m for all m. Thus (1-i)m=0 for all m, but 1-i is invertible, hence m=0 for all m.

References: Atiyah-Macdonald chap 1 & 2, link, link

Algebra 4 – March 27th – Modules over PID

Review of Rings

We finish proving that if A is a PID and M is an A-module free of finite rank s, then all submodules are finitely generated and are actually free of rank smaller than s. Then we prove that (still over a PID) all finitely generated modules are direct sum of cyclic modules. The primary decomposition theorem. Vector spaces over k as k[x]-modules. Application to rational normal form and Jordan form of a matrix.

References: Notes by Prof. Giudetti, notes by Prof. De Stefano, notes by Prof. Valla

Algebra 4 – March 19th and 20th- Review of Rings and Modules

Review of Rings

We finish the proof that $\mathbb{Z}[\frac{1+\sqrt{-19}}{2}]$ is a PID (we proved it’s not an euclidean ring last time). We recall the definitions and basic properties of modules, bases, generators. We start proving that if A is a PID and M is an A-module free of finite rank s, then all submodules are finitely generated and are actually free of rank smaller than s.

References: Chap 1 of Atiyah-Macdonald, PDF, notes by Prof. Giudetti, notes by Prof. De Stefano, notes by Prof. Valla

Algebra 4 – March 13th – Review of Rings

Review of Rings

We recall the main notions of ring theory: definitions of rings, ideals, quotients, R/I field iff I maximal and R/I domain iff I prime. Elements which are prime, irreducible, zero divisors, nilpotent. Euclidean rings, PID, UFD and relations with examples and counter examples. The set of nilpotent elements is an ideal and is the intersection of all prime ideals. The radical of an ideal. The Jacobson radical. An element x is in the Jacobson radical iff 1-xy is a unit for every y. Ring homomorphisms and ideals. Extensions and contractions. Definition of normal rings. UFD implies normal. Definition of local ring. The rings $\mathbb{Z}[\sqrt{-d}]$ for $d\geq 0$ , which are UFD and which not. Short discussion about the rings $\mathbb{Z}[\sqrt{d}]$ for $d\geq 0$ and $\mathbb{Z}[\frac{1+\sqrt{d}}{2}]$ . We show that $\mathbb{Z}[\frac{1+\sqrt{-19}}{2}]$ is a PID non an euclidean ring.

References: Chap 1 of Atiyah-Macdonald, PDF