Chapter 10, Local Rings Video Solutions, A Course in Ring Theory

Problem 1

Prove Lemma 10.1(ii) directly from the quasi-regular property of $J=$ $\operatorname{Rad}(R) .$ To this end, suppose $\left\{v_{1}, v_{2}, \ldots, v_{n}\right\} \subseteq V$ is a minimal generating set for $V$ modulo $W$ and write $v_{n} \in W+V J=W+\sum_{i} v_{i} J$. Now solve for the generator $v_{n}$.

Chris Trentman

Numerade Educator

11:54

Problem 2

Let $\Gamma$ be the semigroup consisting of all symbols $x^{q}$, where $q$ is a nonnegative rational number, and let the multiplication in $\Gamma$ be given by $x^{a} \cdot x^{b}=x^{a+b} .$ Form the semigroup algebra $R=K[\Gamma]$ and let $\bar{R}=R / x^{1} R$. Show that $\bar{R}$ is a local ring with $\bar{J}=\operatorname{Rad}(\bar{R})$ a nil, but not nilpotent, ideal. Furthermore, show that $\bar{J}^{2}=\bar{J} \neq 0$, so that Nakayama's Lemma can indeed fail for infinitely generated modules.

Chris Trentman

Numerade Educator

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Problem 3

Now let $R$ be a local ring and suppose $C=\left(c_{i, j}\right) \in \mathrm{M}_{n}(R)$ with $c_{i, i} \notin \operatorname{Rad}(R)$ and $c_{i, j} \in \operatorname{Rad}(R)$ for all $i \neq j .$ Show that $C$ is both row and column equivalent to the identity matrix. Conclude from this that $C$ is invertible.

Nick Johnson

Numerade Educator

02:58

Problem 4

Suppose $T$ is a multiplicatively closed subset of the commutative domain $R$. If $\tilde{T}$ is the set of divisors of elements of $T$, prove that $\tilde{T}$ is multiplicatively closed and that $R T^{-1}=R \tilde{T}^{-1}$. In particular, if $R$ is a unique factorization domain, show that the rings $R T^{-1}$ correspond in a one-to-one manner to the sets of prime elements of $R$.

If $V$ is an $R$-module, then $\operatorname{Rad}(V)$ is defined to be the intersection of all maximal submodules of $V$. In particular, $\operatorname{Rad}\left(R_{R}\right)$ is the usual Jacobson radical of $R$.

Abigail Martyr

Numerade Educator

09:31

Problem 5

Let $W \subseteq V .$ If $M$ is a maximal submodule of $V$, show that either $M \supseteq W$ or that $W \cap M$ is a maximal submodule of $W$. Conclude that $\operatorname{Rad}(W) \subseteq W \cap \operatorname{Rad}(V)$ with equality if $W \mid V .$ Find an example of \mathbb-modules where equality fails.

Chris Trentman

Numerade Educator

04:33

Problem 6

Show that $\operatorname{Rad}(V) \supseteq V \cdot \operatorname{Rad}(R)$ with equality if $V$ is free and then if $V$ is projective. Show that equality fails when $R=\mathbb{Z}$ and $V=\mathbb{Q}_{\mathbb{Z}}$.
An $R$-module $\mathrm{V}$ has a projective cover $P \stackrel{\alpha}{\longrightarrow} V \rightarrow 0$ if $P$ is projective and if no proper submodule of $P$ maps onto $V$.

Lucía Guerrero

Numerade Educator

03:58

Problem 7

If $P \stackrel{\alpha}{\longrightarrow} V \rightarrow 0$ and $Q \stackrel{\beta}{\longrightarrow} V \rightarrow 0$ are projective covers for $V$, prove that $P \cong Q$. Furthermore, if $R$ is Artinian, use the argument of Theorem $5.9$ (iii) to show that every $R$-module has a projective cover.

Anthony Ramos

Numerade Educator

04:01

Problem 8

If $P \stackrel{\alpha}{\longrightarrow} V \rightarrow 0$ is a projective cover, show that $\operatorname{Ker}(\alpha) \subseteq \operatorname{Rad}(P)$ Conclude from Exercise 6 that if $\operatorname{Rad}(R)=0$, then $\alpha$ must be an isomorphism. In particular, if $\operatorname{Rad}(R)=0$, then only projective $R$ modules have projective covers.

Anthony Ramos

Numerade Educator

04:33

Problem 9

If $P$ is a cyclic projective $R$-module, show that $P \cong e R$ for some idempotent $e \in R$. Conclude that if $R$ has no idempotents other than 0 or 1 and if $V$ is an irreducible $R$-module with a projective cover, then $R$ is a local ring. Conversely, if $R$ is a local ring, show that all finitely generated $R$-modules have projective covers.

Lucía Guerrero

Numerade Educator

07:44

Problem 10

Let $F \stackrel{\alpha}{\longrightarrow} V \rightarrow 0$ be a projective cover for $V$ with $F$ a free $R$-module. Since $F \neq F \cdot \operatorname{Rad}(R)$, conclude that $V \neq V \cdot \operatorname{Rad}(R)$. Now use Exercise 2 to show that not every module over a local ring has a projective cover.

Anthony Ramos

Numerade Educator