Chapter 6, Hereditary Rings Video Solutions, A Course in Ring Theory

Problem 1

In the following two problems, let $\mathbb{Z}$ be the ring of integers, $\mathbb{Q}$ the field of rationals, and let $R$ be the subring of $\mathrm{M}_{2}(\mathbb{Q})$ given by $R=$ $\left(\begin{array}{ll}\mathbb{Z} & Q \\ 0 & Q\end{array}\right) .$ Set $N=\left(\begin{array}{ll}0 & Q \\ 0 & 0\end{array}\right)$ and let $V=\left(\begin{array}{ll}0 & 0 \\ 0 & Q\end{array}\right) .$ We show that $R$ is hereditary but not left hereditary.
1. First prove that $N$ and $V$ are minimal right ideals of $R$, that $V_{R}$ is projective and that $N_{R} \cong V_{R}$. Next show that any right ideal of $R$ properly containing $N+V$ has the form $\left(\begin{array}{cc}z \mathbb{Z} & \mathbb{Q} \\ 0 & Q\end{array}\right)$, where $0 \neq$ $z \in \mathbb{Z}$, and that, as a right $R$-module, this ideal is isomorphic to $R_{R}$ Conclude that $R$ is hereditary. To this end, observe, for example, that if $I$ is a right ideal of $R$ with $I \nsupseteq N$, then $I \cap N=0 .$ Thus $I+N$ is a right ideal containing $N$ and it suffices to show that $I+N$ is projective.

Chris Trentman

Numerade Educator

04:33

Problem 2

Note that both $N$ and $A=N+V$ are two-sided ideals of $R$ and that $\bigcap_{n=1}^{\infty} n R=A=A R$. It follows that if $F$ is a free left $R$-module, then $\bigcap_{n=1}^{\infty} n F=A F .$ Now show that $_{R} N$ is not projective. Indeed, if $F$ is a free left $R$-module with $F=N+W$, then
$$
N=\bigcap_{n=1}^{\infty} n N \subseteq \bigcap_{n=1}^{\infty} n F=A F=A N+A W=A W
$$
Thus $R$ is not left hereditary.
Let $R$ be a ring and let $n$ be a positive integer. Then $R$ is said to be an $n-f i r$ if all $n$-generator right ideals of $R$ are free of unique rank.

Lucía Guerrero

Numerade Educator

03:07

Problem 3

. Prove that $R$ is a 1 -fir if and only if it is a domain, that is a ring without zero divisors. In one direction, suppose $R$ is a 1-fir, let $0 \neq a \in R$ and consider the short exact sequence $0 \rightarrow I \rightarrow R \rightarrow a R \rightarrow 0$, where the epimorphism is left multiplication by $a .$ Since $a R$ is free, $R \cong I \oplus a R$ and hence $I$ is also a 1-generator right ideal of $R$. The other direction follows easily from Lemma $2.5 .$

Kevin Harmer

Numerade Educator

07:08

Problem 4

Let $V$ be an $R$-module with a composition series and let $\theta \in \operatorname{End}_{R}(V)$. Observe that the ascending chain $\left\{\operatorname{Ker}\left(\theta^{i}\right) \mid i=1,2, \ldots\right\}$ and the descending chain $\left\{\operatorname{Im}\left(\theta^{i}\right) \mid i=1,2, \ldots\right\}$ must both terminate and choose integer $n \geq 1$ so that $\operatorname{Ker}\left(\theta^{n}\right)=\operatorname{Ker}\left(\theta^{2 n}\right)$ and $\operatorname{Im}\left(\theta^{n}\right)=\operatorname{Im}\left(\theta^{2 n}\right)$ Prove that $V=\operatorname{Ker}\left(\theta^{n}\right)+\operatorname{Im}\left(\theta^{n}\right)$, that $\theta$ is a nilpotent endomorphism on the first summand and that $\theta$ is an automorphism of the second. In
particular, if $V$ is indecomposable, conclude that $\theta$ is either nilpotent or an automorphism of $V$. This is Fitting's Lemma.

Tim Strang

Numerade Educator

07:08

Problem 5

Let $S$ be a ring with the property that every element is either nilpotent or invertible. If $\alpha, \beta, \gamma \in S$ with $\alpha$ and $\beta$ nilpotent, show that $\alpha \gamma, \gamma \alpha$, and $\alpha+\beta$ are nilpotent. For the latter, first observe that $\alpha+\beta$ cannot equal 1. Conclude that $\operatorname{Nil}(S)$ is the set of all nilpotent elements of S. In view of the preceding problem, this applies to $S=\operatorname{End}_{R}(V)$ if $V$ is an indecomposable $R$-module with a composition series.

In the next two exercises let $V$ be an $R$-module with a composition series and write
$$
V=V_{1}+V_{2}+\cdots+V_{n}=V_{1}^{\prime} \dot{+} V_{2}^{\prime}+\cdots+V_{m}^{\prime}
$$
with all $V_{i}$ and $V_{j}^{\prime}$ indecomposable modules. Let $\pi_{i}: V \rightarrow V_{i}, \eta_{i}: V_{i} \rightarrow$ $V, \pi_{j}^{\prime}: V \rightarrow V_{j}^{\prime}$ and $\eta_{j}^{\prime}: V_{j}^{\prime} \rightarrow V$ be the natural projections and injections.

Tim Strang

Numerade Educator

02:04

Problem 6

For each $i=1,2, \ldots, n$ define $\alpha_{i}=\pi_{1}^{\prime} \eta_{i} \in \operatorname{Hom}_{R}\left(V_{i}, V_{1}^{\prime}\right)$ and $\beta_{i}=$ $\pi_{i} \eta_{1}^{\prime} \in \operatorname{Hom}_{R}\left(V_{1}^{\prime}, V_{i}\right) .$ Then $\alpha_{i} \beta_{i} \in \operatorname{End}_{R}\left(V_{1}^{\prime}\right)$ and note that $1=$ $\sum_{i=1}^{n} \alpha_{i} \beta_{i}$. Since $V_{1}^{\prime}$ is indecomposable, deduce from the preceding problem that some $\alpha_{i} \beta_{i}$ is not nilpotent, say $\alpha_{1} \beta_{1}$. Now show that $\beta_{1} \alpha_{1}$ is a nonnilpotent endomorphism of $V_{1}$ and conclude that both $\alpha_{1} \beta_{1}$ and $\beta_{1} \alpha_{1}$ are automorphisms.

Nick Johnson

Numerade Educator

01:05

Problem 7

Continuing with the preceding notation, show that $\alpha_{1}$ is an isomorphism with inverse $\left(\beta_{1} \alpha_{1}\right)^{-1} \beta_{1}=\beta_{1}\left(\alpha_{1} \beta_{1}\right)^{-1}$ and conclude that $V_{1} \cong$ $\therefore V_{1}^{\prime} .$ Next, prove that $V=V_{1}+V_{2}^{\prime}+\cdots+V_{m}^{\prime} .$ To this end, note that $\alpha_{1}$ maps $V_{1} \cap\left(V_{2}^{\prime}+\cdots+V_{m}^{\prime}\right)$ to zero and that $\pi_{1}^{\prime}$ maps $V_{1}+V_{2}^{\prime} \dot{+} \cdots+V_{m}^{\prime}$ onto $V_{1}^{\prime}$. Finally, show by induction on the total number of summands that $n=m$ and that, by suitably relabeling the submodules, we have $V_{i} \cong V_{i}^{\prime}$ for all $i$. This is the Krull-Schmidt Theorem.

Anthony Ramos

Numerade Educator

01:59

Problem 8

Show that any primitive ring is prime. Conversely, if $R$ is a prime ring with a minimal right ideal, prove that $R$ is primitive. Give an example of a prime ring that is not primitive.

James Chok

Numerade Educator

02:16

Problem 9

Prove that a prime Artinian ring is simple. Give an example of a primitive ring that is not simple. For this, let $V$ be an infinite dimensional $K$-vector space and let $R$ be a suitable subring of End $_{K}(V)$.

Anthony Ramos

Numerade Educator

03:38

Problem 10

Use the argument of the Hopkins-Levitzki Theorem to show that a ring is Artinian if and only if it is semiprimary and Noetherian.

Prathan Jarupoonphol

Numerade Educator