Consider the ring isomorphism $\phi: A \to B$
I have to prove that $a\in A $ is irreducible if and only if $\phi(a)$ is irreducible.
By definition, $a$ is irreducible in A if and only if:
1) $a$ is no cero
2) $a$ a is not a unit
3) If $a=bc$ where $c,b\in A$, then $b$ or $c$ must be a unit.
To prove $\to$ this direction I have seen that 1) $\phi(a)$ is no cero (by contradiction), and to prove that $\phi(a)$ a is not a unit I have explained the following:
By contradiction, let's suppose that $\phi(a)$ is a unit. So, it exists $b\in A$ where $\phi(a)\phi(b)=1$. As $\phi$ is an homomorpism, $\phi(a)\phi(b)=\phi(ab)=1$ and $\phi(1)=1$. Therefore, as $\phi$ is inyective,$ab=1$
We get a contradiction as a is not a unit.
Is my proof correct?
Could you help me to prove the 3rd condition, please? 3) If $\phi(a)=\phi(b)\phi(c)$, then $\phi(b)$ or $\phi(c)$ is a unit.
Thank you for your time and dedication.
Let us write $\phi(a)=\phi(b)\phi(c)=\phi(bc)$. This implies $a=bc$. We know that either $b$ or $c$ is a unit, as $a$ is irreducible. Thus, either $\phi(b)$ or $\phi(c)$ is a unit, as $B^\times=\phi(A^\times)$, where $R^\times$ denotes the group of units in the ring $R$. Actually, if you do not want to prove $B^\times=\phi(A^\times)$, you just need $\phi(A^\times)\subset B^\times$, and it is clear that a unit $u\in A^\times$ is sent to a unit (by any ring homomorphism): $$1_A=uu'\Rightarrow 1_B=\phi(1_A)=\phi(uu')=\phi(u)\phi(u').$$