Chapter 8, Projective Dimension Video Solutions, A Course in Ring Theory

Problem 1

If $R$ is not a Wedderburn ring, prove that gl $\operatorname{dim} R=1+\sup \left\{\mathrm{pd}_{R} I\right\}$ where $I$ runs through all right ideals of $R$. For this, first observe that every cyclic $R$-module is of the form $R / I$.

Mohan Jain

Numerade Educator

06:22

Problem 2

If $N$ is a nilpotent ideal of $R$, show that gl $\operatorname{dim} R=\sup \left\{\mathrm{pd}_{R} V \mid\right.$ $V N=0\} .$

Chris Trentman

Numerade Educator

04:48

Problem 3

Let $D$ be a division ring, let $R$ be the ring of upper triangular $n \times n$ matrices over $D$ and let $N$ be the set of strictly upper triangular matrices. First show that $R$ is an Artinian ring with $N=\operatorname{Rad}(R)$ and that $R$ has precisely $n$ irreducible modules, namely $V_{i} \cong e_{i} R / e_{i} N$ for $i=1,2, \ldots, n$, where $e_{i}=e_{i, i} .$ Furthermore, prove that $e_{i} N \cong e_{i+1} R$ as right $R$-modules for $i<n$ and conclude that gl $\operatorname{dim} R \leq 1$

Anthony Ramos

Numerade Educator

04:33

Problem 4

Let $R \supseteq S$ be rings with $R_{S}$ a projective $S$-module. If $V$ is an $R$ module, prove that $\mathrm{pd}_{R} V \geq \mathrm{pd}_{S} V .$ If in addition $R=S+I$ with $I \triangleleft R$, prove that $\mathrm{gl} \operatorname{dim} R \geq \mathrm{gl} \operatorname{dim} S$. For the latter, observe that any $S$-module is an $R$-module by way of the homomorphism $R \rightarrow R / I \cong S$ and that this $R$-module restricts properly to $S$.

Let $A$ be an $R$-module. A projective resolution for $A$ is a long exact sequence
$$
\cdots \rightarrow P_{2} \stackrel{\alpha_{2}}{\longrightarrow} P_{1} \stackrel{\alpha_{1}}{\longrightarrow} P_{0} \stackrel{\alpha_{0}}{\longrightarrow} A \rightarrow 0
$$
with each $P_{i}$ a projective $R$-module. If all $P_{i}$ are free, then this sequence is a free resolution for $A$.

Lucía Guerrero

Numerade Educator

03:33

Problem 5

If
$$
\cdots \rightarrow Q_{2} \stackrel{\beta_{2}}{\longrightarrow} Q_{1} \stackrel{\beta_{1}}{\longrightarrow} Q_{0} \stackrel{\beta_{0}}{\longrightarrow} A \rightarrow 0
$$
is a second projective resolution for $A$, prove that
$$
\begin{aligned}
&\operatorname{Ker}\left(\alpha_{n}\right) \oplus Q_{n} \oplus P_{n-1} \oplus Q_{n-2} \oplus P_{n-3} \oplus \cdots \\
&\cong \operatorname{Ker}\left(\beta_{n}\right) \oplus P_{n} \oplus Q_{n-1} \oplus P_{n-2} \oplus Q_{n-3} \oplus \cdots
\end{aligned}
$$
for all $n \geq 0$. In particular, observe that $\operatorname{Ker}\left(\alpha_{n}\right) \sim \operatorname{Ker}\left(\beta_{n}\right)$. This is,

Gaurav Kalra

Numerade Educator

03:13

Problem 6

Prove that $A$ has a projective resolution and that if $R$ is Noetherian and $A$ is finitely generated, then $A$ has such a resolution with all $P_{i}$ finitely generated. Show that $\mathcal{K}^{n}[A]=\left[\operatorname{Ker}\left(\alpha_{n-1}\right)\right]$ for all $n \geq 1$ and then characterize pd $A$ in terms of the projective resolutions of $A$.

Gideon Idumah

Numerade Educator

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Problem 7

Let
$$
\cdots \rightarrow C_{2} \stackrel{\gamma_{2}}{\longrightarrow} C_{1} \stackrel{\gamma_{1}}{\longrightarrow} C_{0} \stackrel{\gamma_{0}}{\longrightarrow} C \rightarrow 0
$$
be a complex of $R$-modules and $R$-homomorphisms. Suppose that there exist abelian group homomorphisms $\sigma: C \rightarrow C_{0}$ and $\sigma_{i}: C_{i} \rightarrow$ $C_{i+1}$ such that $\gamma_{0} \sigma=1_{C}, \sigma \gamma_{0}+\gamma_{1} \sigma_{0}=1_{C_{0}}$ and $\sigma_{n-1} \gamma_{n}+\gamma_{n+1} \sigma_{n}=$ $1_{C_{n}}$ for all $n \geq 1$. Prove that this sequence is exact.

Nick Johnson

Numerade Educator

08:25

Problem 8

Now assume that the preceding $C$-sequence is exact and that $f: A \rightarrow C$ is an $R$-module homomorphism. Prove that there exist homomorphisms $f_{i}: P_{i} \rightarrow C_{i}$ such that the diagram
Let $S$ be a ring and let $\Gamma$ be a multiplicative semigroup. Then the semigroup ring $R=S[\Gamma]$ is the set of all formal finite sums $\sum_{x \in \Gamma} s_{x} x$ with $s_{x} \in S$. Addition in $S[\Gamma]$ is componentwise and multiplication is determined distributively by $a x \cdot b y=(a b)(x y)$ for all $a, b \in S$ and $x, y \in \Gamma .$ For example, the polynomial ring $R=S\left[x_{14} x_{2}, \ldots, x_{n}\right]$ is a semigroup ring with $\Gamma=\left\langle x_{1}, x_{2}, \ldots, x_{n}\right\rangle .$ Furthermore, if $\Gamma=G$ is a multiplicative group, then $S[G]$ is, of course, the group ring.

Ely Crowder

Numerade Educator

01:25

Problem 9

If $R=S[\Gamma]$, use Problem 4 to prove that $\mathrm{gl} \operatorname{dim} R \geq \mathrm{gldim} S$

Chai Santi

Numerade Educator

14:09

Problem 10

. Suppose $R=S[G]$ with $G$ a multiplicative group and, for each $n \geq 0$, let $F_{n}$ be the free right $S$-module with basis consisting of all $(n+1)$ tuples $\left[g_{0}, g_{1}, \ldots, g_{n}\right]$ with $g_{i} \in G .$ If $x \in G$, dèfine $\left[g_{0}, g_{1}, \ldots, g_{n}\right] x=$ $\left[g_{0} x, g_{1} x, \ldots, g_{n} x\right]$ and, for all $n \geq 1$, let the map $\alpha_{n}: F_{n} \rightarrow F_{n-1}$ be determined by
$$
\left[g_{0}, g_{1}, \ldots, g_{n}\right] s \mapsto \sum_{k=0}^{n}(-)^{k}\left[g_{0}, \ldots, \hat{g}_{k}, \ldots, g_{n}\right] s
$$
where $\hat{g}_{k}$ indicates that $g_{k}$ is omitted. If $\alpha_{0}: F_{0} \rightarrow S$ is given by $\left[g_{0}\right] s \mapsto s$, show that
$$
\cdots \rightarrow F_{2} \stackrel{\alpha_{2}}{\longrightarrow} F_{1} \stackrel{\alpha_{1}}{\longrightarrow} F_{0} \stackrel{\alpha_{0}}{\longrightarrow} S \rightarrow 0
$$
is a free resolution for the $R$-module $S_{R}$. For exactness, observe that the maps $\sigma: S \rightarrow F_{0}$ and $\sigma_{n}: F_{n} \rightarrow F_{n+1}$ given by $\sigma: s \mapsto[1] s$ and
$$
\sigma_{n}:\left[g_{0}, g_{1}, \ldots, g_{n}\right] s \mapsto\left[1, g_{0}, g_{1}, \ldots, g_{n}\right] s
$$
satisfy the assumptions of Problem $7 .$

Anthony Ramos

Numerade Educator