Let $R$ be an indecomposable ring with $(R,+)\cong(\Bbb{Z}_2\oplus \Bbb{Z}_4,+)$. Suppose that $R$ is noncommutative and has no multiplicative identity. If $R$ has an nonzero idempotent element $e$, show that $2e\neq 0$.
The original question come from the paper "Rings of order $p^3$", writed by V. G. Antipkin and V. P. Elizarov. Which paper classify the rings of order $p^3$. http://link.springer.com/article/10.1007%2FBF00968650
The author separate the problem into many cases, one of the cases is
- $R$ is indecompoable
- $(R,+)\cong (\Bbb{Z}_p\oplus \Bbb{Z}_{p^2},+)$
- $R$ is noncommutative
- $R$ has no unity
- $R$ has a nonzero idempotent element $e$
The author prove that the additive order of $e$ is $p^2$ by contradiction. He suppose that the additive order of $e$ is $p$. There are three cases, $Re=R$, $|Re|=p^2$ and $|Re|=p$. My question occur in the case $|Re|=p^2$. The author says that:
If $|Re|=p^2$, then $char Re=char R(1-e)=p$ and we find again that $char R=p$, a contradiction. (Because $char R=p^2$.)
I doubt the statement $char Re=char R(1-e)$ is wrong. I find out a counterexample when $p=2$. Since $(R,+)\cong (\Bbb{Z}_2\oplus \Bbb{Z}_4,+)$, suppose that $(R,+)\cong \langle b\rangle +\langle e\rangle$, where $e$ is the idempotent and the additive order of $e$ is $2$ and the additive order of $b$ is 4. Then $R=\{0,b,2b,3b,e,b+e,2b+e,3b+e\}$, $Re=\{0,be,e,be+e\}$, $R(1-e)=R-Re=\{0,2b,3b-be,b-be\}$. $$ 2=char Re\neq char R(1-e)=4. $$
The complete table of the rings of order 8 http://bfhaha.blogspot.tw/p/rings-of-order-8.html
Lemma 1: Suppose $R$ is a ring. If $A$ and $B$ are two ideals of $R$ and $R=A+B, A\cap B=\{0\}$, then $R\cong A\oplus B$. See the Hungerford's book: Algebra, p.130, theorem 2.24.
Lemma 2: Let $R$ be a ring not necessary has a multiplicative identity. Suppose that $e$ is a nonzero idempotent element and for all $r\in R$, $re=er$. That is, $e\in Z(R)$, the center of the ring $R$. Let $A=eR=\{ er\mid r\in R\}$, $B=\{r-er\mid r\in R\}$. Then $A$ and $B$ are two ideals of $R$ and $R\cong A\oplus B$.
Lemma 3: Let $R$ be a ring not necessary has a multiplicative identity. If $e$ is an idempotent element in $R$ and $eR=Re$, then $e\in Z(R)$. See T. Y. Lam's book: Exercises in classical ring theory 2e, p.316, exercise 22.3 A.
The proof of these lemma see the end of this answer.
We state the theorem again, the hypothesis noncommutative, $\nexists$, has no unity are redundant.
Let $R$ be an indecomposable ring with $(R,+)\cong(\Bbb{Z}_2\oplus \Bbb{Z}_4,+)$. If $R$ has a nonzero idempotent element $e$, then $ord(e)=4$, where $ord(x)$ denote the additive order of the element $x$.
The proof of the theorem.
Since $|(R,+)|=4$, by Lagrange's Theorem, $ord(e)\in\{1, 2, 4\}$. Since $e\neq 0$, so $ord(e)\neq 1$. If $ord(e)=2$, then we select $b\in R$, where $ord(b)=4$, suct that $R=\langle e\rangle +\langle b\rangle=\{0,b,2b,3b,e,b+e,2b+e,3b+e\}$. It is easy to know that $Re=\{0,be,e,be+e\}$ and $eR=\{0,eb,e,eb+e\}$. There are three cases, $|Re|=4$, $|eR|=4$ and $|Re|=|eR|=2$. We show that whatever which case, there is a contradiction. Therefore, $ord(e)=4$.
If $|Re|=4$, then $be\neq e$ and $be\neq 0$. Since $2(be)=b(2e)=0$ and there are three elements $2b, e$ and $2b+e$ in $R$ whose additive order is 2. Thus, $be\in \{2b, 2b+e\}$. If $be=2b$, then $be=be^2=(be)e=2be=b(2e)=0$, a contradiction. Similarly, $be=2b+e$ implies that $be=be^2=(be)e=(2b+e)e=2be+e^2=e$, a contradiction. Therefore, $|Re|\neq 4$. By the similar argument, we have $|eR|\neq 4$.
The remain case is $|eR|=|Re|=2$. In this case, since $\{0,e\}\subseteq Re,eR$, hence $Re=eR$. By Lemma 3, $e\in Z(R)$. Furthermore, by Lemma 2, $R\cong eR\oplus (R-eR)$. Since $eR\neq R$, contrary to the hypothesis $R$ is indecomposable.
The proof of Lemma 2: $A$ and $B$ are ideals of $R$ could be directly verified, we only verify some axioms of the ideal. $$ \forall er_1, er_2\in eR, (er_1)(er_2) \stackrel{e\in Z(R)}{=} e^2 r_1 r_2 \stackrel{e^2=e}{=} e r_1 r_2\in eR, $$ $$ \forall er\in eR, r'\in R, r'er \stackrel{e\in Z(R)}{=} err'\in eR. $$ For all $r\in R$, $r=er+(r-er)\in A+B$, then $R=A+B$. If $x=er_1=r_2-er_2\in A\cap B$, then multiplying $e$ to both side of this identity, we get $$ ex=e^2r_1=er_2-e^2 r_2=er_2-er_2=0. $$ Then $$ x=er_1=e^2 r_1=e(er_1)=ex=0. $$ That is, $A\cap B=\{0\}$. Then by Lemma 1, $R\cong A\oplus B$.
The proof of Lemma 3: For all $r\in R$. \begin{eqnarray*} & & er-ere=e(r-re)\in eR=Re\\ & \Rightarrow & er-ere=r_1 e...(*)\\ & \stackrel{\mbox{right multiplying }e}{\Rightarrow} & ere-ere^2=r_1 e^2=r_1 e\\ & \Rightarrow & r_1 e=0\\ & \stackrel{(*)}{\Rightarrow} & er-ere=r_1 e=0\\ & \Rightarrow & er=ere. \end{eqnarray*}
Similarly, since $re-ere=(r-er)e\in Re=eR$, we can get $re=ere$. Therefore, $er=ere=re$.