Is it possible to construct a ring morphism $A \times A \to B$ from a ring morphism $A \to B$? Let $A$ and $B$ be two rings and $\varphi : A \to B$ a ring morphism.
We equip $A \times A$ with the induced ring structure from $A$.
I was wondering if it is possible to construct a ring morphism $\tilde{\varphi} : A \times A \to B$ from $\varphi$ different from the composition with the natural projections (i.e. different from $(x, y) \mapsto \varphi(x)$ and $(x,y) \mapsto \varphi(y)$) ?
Thanks in advance for any hint or help.
K. Y.
 A: It can be constructed if $B$ isn't directly reducible. Then there exists a non-trivial central idempotent element in $B$.
Suppose $\overline\phi(1,0)=a,\overline \phi(0,1)=b$, then we have $ab=0$, $a+b=1$. So $a(1-a)=0$ or $a^2=a$, so existence of idempotent element is a necessary condition.
It's sufficient if we also requires $a$ to be in the center, then we can take $\overline \phi:(m,n)\mapsto a\cdot \phi(m)+b\cdot\phi(n)$.
Addition:
\begin{align}
\overline\phi((m_1,n_1))+\overline \phi((m_2,n_2))&=a\cdot\phi(m_1)+b\cdot\phi(n_1)+a\cdot\phi(m_2)+b\cdot\phi(n_2)\\
&=a\cdot\phi(m_1+m_2)+b\cdot\phi(n_1+n_2)\\
&=\overline\phi((m_1+m_2,n_1+n_2))
\end{align}
Multiplication, we need
$$b^2=(1-a)^2=1-2a+a^2=1-a=b$$
Then:
\begin{align}
\overline\phi((m_1,n_1))\cdot\overline\phi((m_2,n_2))&=(a\cdot\phi(m_1)+b\cdot\phi(n_1))(a\cdot \phi(m_2)+b\cdot\phi(n_2))\\
&=a^2\cdot\phi(m_1m_2)+b^2\cdot\phi(n_1n_2)\\
&=a\cdot\phi(m_1m_2)+b\cdot\phi(n_1n_2)\\
&=\overline\phi((m_1m_2,n_1n_2))
\end{align}
In this construction, we only used that $a$ commutes with $\mathrm{im}\, \phi $, which I'm not sure is it a necessary condition.
One example that can be constructed is: $\mathbb Z \times \mathbb Z\to \mathbb Z/6\mathbb Z$, sending $(m,n)\mapsto 3m-2n$.
