How do you prove a ring homomorphism is onto using the range and kernel?

Question

The problem is:

Suppose $A$ is a commutative ring and $a \in A$. If $a^2 = a$, prove that the function $\pi_a(x) = ax$ is a homomorphism from $A$ into $A$. Show that the kernel of $\pi_a$ is $I_a$, the annihilator of $a$. Show that the range of $\pi_a $ is $\langle a\rangle$. Conclude by the Fundamental Homomorphism Theorem that $A/I_a\cong\langle a\rangle$.

I'm stuck on the last step. How has $\pi_a$ been shown to be onto/surjective by the above steps?

The question is CH19-D3 from Pinter, Charles C. "A book of abstract algebra. Reprint of the second (1990) edition." (2010).

Answer 1

Using your definition we can modify $\pi$ as follows $ \pi : A \to \langle a \rangle $ then $\pi$ becomes onto and then you can use your theorem about $A/\ker \pi$. Observe this shows that for any homomorphism $\phi : A \to B$ we have $$ A/\ker\phi \cong Im \phi $$

Answer 2

Note that the word onto is often used to mean surjective and should not be confused with the codomain / target of a function. Imo saying homomorphism from $A$ onto $B$ for a surjective homomorphism $A\rightarrow B$ is a very bad habit, as it may easily be confused with the strictly more general notion of homomorphism from $A$ to $B$.

To solve the exercise you need to observe / prove the following things:

• The subset $\langle a \rangle = \{ra \mid r\in R\} \subseteq R$ is a ring with unit $a$ (this uses $a^2=a$!). Note that it is not a subring (unless $a=1$).
• The function $\pi_a: R \rightarrow R, r \mapsto ra$ has the image $\langle a\rangle\subseteq R$. In fact, restricting the codomain to the image, it defines a surjective ring homomorphism $\overline{\pi_a}:R \rightarrow \langle a \rangle$.
• The kernel of $\overline{\pi_a}$ is the kernel of $\pi_a$, which is the annihilator of $a$.

The isomorphism theorem now tells you that the canonical map $A/\ker{\pi_a} \overset{\cong}{\longrightarrow} \langle a \rangle, [r] \mapsto ra$ is an isomorphism.