Chapter 13, Graded Rings and Modules Video Solutions, A Course in Ring Theory

Problem 1

The graded ring $S=S_{0}+S_{1}+\cdots$ is said to be graded Noetherian if the set of graded right ideals of $S$ satisfies the ascending chain condition. Prove that $S$ is Noetherian if and only if it is graded Noetherian. Note that if $I$ is any right ideal of $S$ and if $I_{n}$ denotes the set of degree $n$ components of elements of $I \cap\left(S_{0}+S_{1}+\cdots+S_{n}\right)$, then $I_{0}+I_{1}+\cdots$ is a graded right ideal of $S$. With this, one can proceed as in the proof of the Hilbert Basis Theorem.

Bobby Barnes

University of North Texas

03:38

Problem 2

Show that $S_{0}$ Noetherian does not imply $S$ Noetherian in general.

Prathan Jarupoonphol

Numerade Educator

01:26

Problem 3

Let $S=S_{0} \dot{+} S_{1}+\cdots$ be a graded ring and assume that every projective $S_{0}$-module is free. Prove that every graded-projective $S$-module $P=$ $P_{0}+P_{1}+\cdots$ is graded-free. To this end, observe that $P S_{+}=N$ is a graded submodule of $P$ and that $P / P S_{+}=P / N=\cdot \sum_{i} P_{i} / N_{i}$ is a projective $S_{0}$-module by Lemma 9.11(i). Thus each $P_{i} / N_{i}$ is a projective and hence free $S_{0}$-module. Now use the argument of Lemma $10.2$ along with the Graded Nakayama's Lemma.

Ankit Gupta

Numerade Educator

08:25

Problem 4

Find all finitely generated graded $K[x]$-modules up to abstract isomorphism. Remember, $K[x]$ is a principal ideal domain.
Let $R$ be a ring and let $G$ be a multiplicative group. Assume that $G$ acts as automorphisms on $R$, so we have a group homomorphism $G \rightarrow \operatorname{Aut}(R)$. As usual, the fixed ring $R^{G}$ of $R$ is defined by
$$
R^{G}=\left\{r \in R \mid r^{g}=r \text { for all } g \in G\right\}
$$
where the map $r \mapsto r^{g}$ is the automorphism corresponding to $g \in G$. Furthermore, we define the skew group ring $R * G$ to be the set of all formal finite sums $\sum_{g \in G} r_{g} g$ with coefficients $r_{g} \in R$. Addition in $R * G$ is componentwise and multiplication is determined distributively by the formula
$$
a g \cdot b h=a b^{g^{-1}}(g h)
$$
for all $a, b, \in R$ and $g, h \in G .$ As with ordinary group rings, we identify $r$ with $r 1$ and $g$ with $1 g$ so $R, G \subseteq R * G .$ In Problems 6 and 7 we assume that $|G|=n<\infty$.

Ely Crowder

Numerade Educator

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Problem 5

Verify that $R^{G}$ is a subring of $R$ and that $R * G$ is an associative ring with 1 .

Nick Johnson

Numerade Educator

02:57

Problem 6

If $1 / n \in R$, show that $e=(1 / n) \sum_{g \in G} g$ is an idempotent of $R * G$ with $e(R * G) e=e R^{G} \cong R^{G}$.

Mohan Jain

Numerade Educator

01:51

Problem 7

Let $W \subseteq V$ be $R * G$-modules and assume that $V_{R}=W_{R} \dot{+} X_{R}$ for some $R$-submodule $X$ of $V .$ If $1 / n \in R$, show that $V=W+Y$ for some $R * G$-submodule $Y$ of $V$. To this end, let $\pi: V \rightarrow W$ be the $R$-module projection map determined by $V=W+X$ and define $\pi^{\prime}: V \rightarrow W$ by $\pi^{\prime}(v)=(1 / n) \sum_{g \in G} \pi\left(v g^{-1}\right) g .$ Prove that $\pi^{\prime}$ is an $R * G$-module projection map from $V$ to $W$. Explain why this generalizes Maschke's Theorem (Proposition 4.9).

Again let $G$ be a multiplicative group. A ring $S$ is said to be $G$ graded if $S=\cdot \sum_{g \in G} S_{g}$ is a direct sum of the additive subgroups $S_{g}$ and if $S_{g} S_{h} \subseteq S_{g h}$ for all $g, h \in G .$ Furthermore, $S$ is strongly G-graded if $S_{g} S_{h}=S_{g h}$ for all $g, h \in G .$ Observe that $S=R * G$ is strongly $G$-graded with $S_{g}=R g$.

Varsha Aggarwal

Numerade Educator

06:51

Problem 8

Let $S$ be a $G$-graded ring with 1 . Show that $1 \in S_{1}$ and that $S$ is strongly $G$-graded if and only if $1 \in S_{g} S_{g^{-1}}$ for all $g \in G .$ Furthermore, if $u \in S_{g}$ is a unit of $S$, prove that $u^{-1} \in S_{g^{-1}}$.

Linda Hand

Numerade Educator

01:07

Problem 9

Let $S=\mathrm{M}_{3}(K)$ and let $G=\{1, g\}$ be a group of order 2 . Define
$$
S_{1}=\left(\begin{array}{ccc}
K & K & 0 \\
K & K & 0 \\
0 & 0 & K
\end{array}\right) \quad \text { and } \quad S_{g}=\left(\begin{array}{ccc}
0 & 0 & K \\
0 & 0 & K \\
K & K & 0
\end{array}\right)
$$

Varsha Aggarwal

Numerade Educator

08:25

Problem 10

If $|G|=n$, let $\mathrm{M}_{G}(R)$ be the ring of $n \times n$ matrices over $R$ with rows and columns indexed by the elements of $G .$ Show that $S=\mathrm{M}_{G}(R)$ becomes $G$-graded by assigning a grade of $g^{-1} h$ to the entries in the $(g, h)$-position. Thus $\mathrm{M}_{n}(R)$ is a $G$-graded ring for any group $G$ of order $n$.

Ely Crowder

Numerade Educator