Directly irreducible ring and central idempotents

Question

My question is mainly concerned with the characterisation of directly irreducible rings as defined on Wikipedia.

We say that a (unital, not necessarily commutative) ring $R$ is directly irreducible, if the following property holds: Suppose that we have a decomposition $R\cong R_1\times R_2$ with rings $R_1,R_2$. Then $R_1=0$ or $R_2=0$.

My goal is to prove the following:

Theorem: Let $R$ be a ring. Then the following are equivalent:

1. $R$ is directly irreducible.
2. The centre $Z(R) = \{r\in R \mid rs=sr \; \forall s\in R\}$ of $R$, which is a ring too, is directly irreducible.
3. The only idempotents in $Z(R)$ are the trivial ones, namely $0$ and $1$.

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I could show the following:

• $(2)\implies(3)$: Every idempotent $e\in Z(R)$ gives rise to a decomposition $Z(R)\cong Z(R)e\times Z(R)(1-e)$. Note that $Z(R)e$ and $Z(R)(1-e)$ are indeed rings with identity element $e$ and $1-e$, resp., since $Z(R)$ is commutative. The isomorphism is given by $r\mapsto (re,r(1-e))$, and its inverse is given by $(re,s(1-e))\mapsto re+s(1-e)$.

  By assumption, the only decompositions are $Z(R)\cong0\times Z(R)$ and $Z(R)\cong Z(R)\times0$, hence $0$ and $1$ are the only idempotents in $Z(R)$.

• $(3)\implies(2)$: Assume that $Z(R)\cong Z_1\times Z_2$ is a decomposition into rings with $Z_1\ne0\ne Z_2$. This would give at least four idempotents, namely $(0,0)$, $(1,1)$, $(0,1)$, $(1,0)$. But by assumption, $Z(R)$ has at most two idempotents, a contradiction.

• $(3)\implies(1)$: Let $R\cong R_1\times R_2$ be a decomposition into rings. Then $Z(R)\cong Z(R_1)\times Z(R_2)$ is a decomposition into rings. By the same argument as in '$(3)\implies(2)$', we must have $Z(R_1)=0$ or $Z(R_2)=0$. W.l.o.g. assume that $Z(R_1)=0$. Since $0_{R_1},1_{R_1}\in Z(R_1)$, we obtain $0_{R_1}=1_{R_1}$, whence $R_1=0$.

• $(2)\implies(1)$: This is similar to '$(3)\implies(1)$', but we skip the step '$(3)\implies(2)$'. Instead, apply the assumption directly.

So the real problem is how to deduce '$(1)\implies(2)$', or more precisely, how do we obtain a decomposition $R\cong R_1\times R_2$ from a given decomposition $Z(R)\cong Z_1\times Z_2$ such that $Z_i\subseteq R_i$ for $i=1,2$. The same problem arises if I attempt to show '$(1)\implies(3)$'.