If $(A,+,\cdot)$ and $(B,+,\cdot)$ are rings and there exists a homomorphism such that $\phi:A\rightarrow B$ then $\phi(1_A)=1_B$? The idea of my proof was that if we have a neutral element in a ring than it is unique.
I have showed that 
$$\phi(a)=\phi(a)\cdot \phi(1_A)=\phi(1_A)\cdot \phi(a).$$
For all elements that are in the imageset of $\phi$, $\phi(1_A)$ acts as a neutral element, how can I argue that because of that $\phi(1_A)$ must be a neutral element of all elements in $B$?
 A: No, you cannot argue like that. Just consider$$\begin{array}{rccc}\phi\colon&\mathbb R&\longrightarrow&\mathbb R^{2\times2}\\&x&\mapsto&\begin{bmatrix}x&0\\0&0\end{bmatrix},\end{array}$$which is a ring homomorphism. However,$$\phi(1)=\begin{bmatrix}1&0\\0&0\end{bmatrix}\neq\begin{bmatrix}1&0\\0&1\end{bmatrix}=\operatorname{Id}.$$Note, however, that it is usually part of the definition of ring homomorphism that, when we are dealing with rings with unity, $\phi$ must be such that $\phi(1_Q)=1_B$.
A: Well, this is not the case. Take the zero mapping $\phi:a\mapsto 0_B$ which is a homomorphism.
In general, if $\phi:G\rightarrow H$ is a homomorphism between groups, then $\phi(1_G)=1_H$.
But if $\phi:M\rightarrow N$ is a homomorphism between monoids (similarly to in your case of rings), then not necessarily $\phi(1_M)=1_N$.
A: This won't work; we take for instance $\phi:\mathbb{Z}\to \mathbb{Z}$ by $\phi(x)=0$ for all $x$. $\phi(x+y)=\phi(x)+\phi(y)$ and $\phi(xy)=\phi(x)\phi(y)$. 
The issue is that here $\phi(1)=0$, and indeed $\phi(1)\phi(x)=\phi(x)$, but this is just because $\phi(x)=0$ for all $x\in \mathbb{Z}$.
A: There is no argument for that. 
(Observe e.g. that the function that sends every element of $A$ to $0_B$).
That is why it is actually a part of the definition of (unital) ringhomomorphism concerning rings with identity.

Sidenote:
The fact that in the definition of grouphomomorphism it is not demanded that they send identity to identity must be looked at as exceptional.
It is a consequence of the fact that based on $\phi(ab)=\phi(a)\phi(b)$ it can be shown that $\phi$ respects idempotent elements (i.e. if $a^2=a$ then $\phi(a)^2=\phi(a^2)=\phi(a)$). Further there is only one idempotent element in a group which is the identity. This guarantees the $\phi$ will send identity to identity. But again: this is exceptional. 
