HW6

A division ring is a (possibly non-commutative) ring where each nonzero element admits an inverse. (each problem is 10 pts )

All rings have identity.

In this HW, some rings are possibly non-commutative. (in later HW, we will focus on commutative rings only)

(1) $R$ : a possibly non-commutative ring with more than one element. For all nonzero $a\in R$, exists a unique $b$ such that $aba=a$. Show that $R$ is a division ring.

\begin{proof} Assume that there exists $a,c\ne 0$ such that $ac=0$ and $b$ satisfies $aba=a$. Then $a(b+c)a=aba=a$ yields that $b+c=b$ and so $c=0$, which is a contradiction. Therefore, for any nonzero $a$, $a$ does not have non-zero zero divisor. It deduces that $a(ba-1)=(ab-1)a=0$ and so $b=a^{-1}$. Hence, $R$ is a division ring. \end{proof}

(2) $R$ : commutative ring and $\mathrm{char}R=p$ a prime number. If $a,b\in R$, then $(a\pm b{)}^{p^m}=$ $a^{p^m}\pm b^{p^m}$ for all $m\ge 1$.

\begin{proof} Note that $(a\pm b{)}^{p^m}=a^{p^m}\pm b^{p^m}+{\sum }_{k=1}^{p^m-1}(\genfrac{}{}{0ex}{}{p^m}{k})a^k(\pm b{)}^{p^m-k}$. Since $\mathrm{char}R=p$, $\left(\genfrac{}{}{0ex}{}{p^m}{k}\right)=0$ for all $k\in \{1,\cdots ,p^m-1\}$. Therefore, $(a\pm b{)}^{p^m}=$ $a^{p^m}\pm b^{p^m}$ for all $m\ge 1$. \end{proof}

(3) $R$ : ring. $a\in R$ is called nilpotent if exists some $n\ge 0$ (which depends on $a$) such that $a^n=0$. Show that: If $R$ is commutative, $a,b\in R$ both nilpotent, then so is $a+b$.

\begin{proof} Assume that $m,n$ are integers such that $a^n=b^m=0$. Since $R$ is commutative, we have that

$$(a+b{)}^{m+n}=\sum _{k=1}^{m+n}(\genfrac{}{}{0ex}{}{m+n}{k})a^k{b}^{m+n-k}.$$

If $a^k\ne 0$, then $k<n$, $m+n-k>m$ and $b^{m+n-k}=0$. Therefore, at least one of $a^k$ and $b^{m+n-k}$ equals $0$. Therefore, $(a+b{)}^{m+n}=0$ and so $a+b$ is nilpotent. \end{proof}

(4) Let $R={\prod }_{i=1}^{\infty }{R}_i$ where $R_i=\mathbb{Z}$ for all $i$. Find a ideal $I$ of $R$ such that $I$ cannot be written as a product of ideals $I_i\subset R_i$.

\begin{proof} Let $I=\{(a_1,a_2,a_3,\dots ):a_i=0\text{mbox}forallbutfinitelymanyi\}$. For any $r\in R$, it is easy to verify that $rI\subseteq I$ and so $I$ is an ideal. For each $i\in {\mathbb{N}}_{+}$, define $J_i=\{x\in R_i:(a_1,\cdots ,a_{i-1},x_i,a_{i+1},\cdots ,a_n)\in I\}$ and $J_i=\mathbb{Z}$. If $I={\prod }_{i=1}^{\infty }{I}_i$, then $I_i\supseteq J_i$ and so $I\supseteq {\prod }_{i=1}^{\infty }{J}_i=R$, which is impossible. Therefore, $I$ cannot be written as a product of ideals $I_i\subset R_i$. \end{proof}

(5) Suppose $R$ commutative. A $R$-mod $M$ is called simple if its only submodules are $0$ and $M$. Show that

• (a) Every simple mod is cyclic (i.e., generated by one element)
• (b) If $M$ is simple, then every $R$-mod endomorphism of $M$ is either the zero map or an isomorphism.

\begin{proof} i) For a nonzero element $m\in M$, $L:=\{rm:r\in R\}$ is a submodule of $M$ and so $L=M$. It follows that $M$ is generated by $m$ and so $M$ is cyclic.

ii) For any endomorphism $\phi :M\to M$, define $K:=\ker \phi$. Note that $\phi (rm)=\phi (r)\phi (m)$ for all $r\in M$ and $m\in K$, then $K$ is a submodule of $R$. Since $M$ is simple, by i) either $K=0$ or $K=M$. It deduces that $\phi$ is either an isomorphism or the zero map. \end{proof}

(6) Suppose $R$ commutative. $f:M\to M$ is $R$-mod homomorphism such that $f\circ f=f$. Then

$$M\simeq \mathrm{Kerf}\oplus \mathrm{Imf}$$

\begin{proof} We have proved $\ker f$ is a submodule in Exercise (5), and it is easy to verify $\mathrm{Im}f$ is a vector space. For any $\ell \in \mathrm{Im}f$, there exists $k$ such that $\ell =f(k)$. For any $r\in R$, $r\ell =rf(k)=f(rk)\in \mathrm{Im}f$ and so $\mathrm{Im}f$ is a submodule. Therefore, both $\ker f$ and $\mathrm{Im}f$ are submodules of $M$. As $\ker f\cap \mathrm{Im}f=0_M$, $\ker f\oplus \mathrm{Im}f\subseteq M$.

For any $m\in M$, $m=m-f(m)+f(m)$. Since

$$f(m-f(m))=f(m)-f(f(m))=f(m)-f(m)=0$$

and $f(f(m))=f(m)$, we have that $m-f(m)\in \ker f$ and $f(m)\in \mathrm{Im}f$. It follows that $M\subseteq \ker f\oplus \mathrm{Im}f$ and so $M\simeq \ker f\oplus \mathrm{Im}f$. \end{proof}