The problem statement is as follows:
An ideal is a subring $\mathcal{J} \subset R$ with the property that $\forall r \in R\ \text{and}\ a \in \mathcal{J}$ we have $$ra,ar \in \mathcal{J}$$Prove that the kernel of a ring homomorphism is an ideal.
Preliminary Definitions
A ring homomorphism is a map $\phi: R \to S$ such that $\forall a,b \in R$: $$\phi(a + b) = \phi(a) + \phi(b)\ \text{and}\ \phi(ab) = \phi(a)\phi(b)$$
- Note that a ring homomorphism is a homomorphism of additive groups. The kernel of a ring homomorphism $ker(\phi) = \phi^{-1}(0)$ is thus an additive group.
The kernel of a homomorphism is a subset: $$ker(\phi) = \{r \in R \mid \phi(r) = e \}$$
Attempt at a proof
First we will show that $ker(\phi)$ is a subring of $R$. For that let $a,b \in ker(\phi)$. First since $\phi(e) = e$ we know that $ker({\phi})$ is nonempty. Then, since $\phi$ is a homomorphism we have: $$\phi(a+b) = \phi(a) + \phi(b) = e + e = e$$ By definition of additive identity. Thus $(a + b) \in ker({\phi})$. Similarly, $$\phi(ab) = \phi(a)\phi(b) = (e)(e) = e$$ By definition of multiplicative identity. Thus $ab \in ker({\phi})$ Finally, $$\phi(a^{-1}) = (\phi(a))^{-1} = (e)^{-1} = e.$$ Thus $a^{-1} \in ker({\phi})$. Consequently $ker({\phi})$ is a subring of $R$.
Next we will prove the condition that $\forall r \in R\ \text{and}\ a \in \mathcal{J}$ we have $ra,ar \in \mathcal{J}$. $$\phi(ar) = \phi(a)\phi(r) = e\phi(r) = \phi(r)...$$ $$\phi(ra) = \phi(r)\phi(a) = \phi(r)e = \phi(r)...$$
Now here is where I get stuck, I'm struggling to see how to show that the products $ar, ra \in ker(\phi)$. I know we want to show that $\phi$ sends them both to the identity $e \in R$, but I can't seem to find a way to either show that $\phi(r) = e$ or otherwise that the image under $\phi$ of a product of $a \in ker(\phi)$ and $r \in R$ is $e \in S$. Any hints?