Showing that an element generates the kernel of a homomorphism

Question

I have the ring homomorphism $\phi:\mathbb{Z}[\sqrt{-3}]\rightarrow \mathbb{Z}_{19}$ given by $\phi(a+b\sqrt{-3})=\overline{a+4b}$.

I am being asked to find an element $a+b\sqrt{-3}$ such that $\ker(\phi)=(a+b\sqrt{-3})$, and show that the element does indeed generate $\ker(\phi)$. I have shown that $\ker(\phi)=(4-\sqrt{-3})$. However I am not really sure what the second part is asking.

Does this mean that $\ker(\phi)$ cannot be written as any other ideal, i.e. cannot be written in the form $(a+b\sqrt{-3})$ for different $a$ or $b$, or $(a+b\sqrt{-3},c+d\sqrt{-3})$ for example?

Answer 1

If you have shown that $\operatorname{ker}\phi=(4-\sqrt{-3})$, then you have completed the task.

The other questions you asked are just that, other questions. For instance, if it turns out you have a unique factorization domain (UFD), or a principal ideal domain (PID), then we could probably answer both questions in the affirmative. But this would require some proof.

Actually, since $\Bbb Z_{19}$ is a field, the ideal $(4-\sqrt{-3})$ is maximal. I suspect that means that the generator, $4-\sqrt{-3}$, is prime.

Then the answers to the other questions appear to be yes, but I am going to have to defer to authority here, as ring theory is not my strong suit.

Answer 2

As Chris Custer says, if you show $\ker\phi = (a + b\sqrt{-3})$, then you have shown that this element generates the kernel. But it is not necessarily true that this is the only element generating the ideal, for instance, if $u$ is a unit in $\mathbb Z[\sqrt{-3}]$, then $u\cdot (a + b\sqrt{-3})$ will generate the same ideal.

By solving the equation $a^2 + 3 b^2 = 1,$ we see that the only units in the ring are $1$ and $-1$, so in a sense, $4 - \sqrt{-3}$ and $-4 + \sqrt{-3}$ are the only elements that generate the kernel as a principal ideal.

I'm thinking that the question might have been "find an element $a+b\sqrt{-3}$ such that $(a + b\sqrt{-3}) \subset \ker\phi$, and show that this element does indeed generate $\ker\phi$". Which would then turn our focus to proving $4 -\sqrt{-3}$ is irreducible.

To do so, we have a look at the norm function $N(a+b\sqrt{-3}) = a^2 + 3b^2$, which we have already used to find units ($a+b\sqrt{-3}$ is a unit if and only if it has norm $1$). This function satisfies $N(\beta\gamma) = N(\beta)N(\gamma)$. We have $N(4-\sqrt{-3}) = 27$, so factorisations of $4-\sqrt{-3}$ correspond roughly to factorisations of $27=1\cdot 27 = 3\cdot 9$.

Looking for an $\alpha$ such that $N(\alpha) = 3$, we find $\alpha = \pm\sqrt{-3}$. However, neither of these divide $4-\sqrt{-3}$: if $4-\sqrt{-3} = (c+d\sqrt{-3})(\pm\sqrt{-3}) = \mp 3d \pm c\sqrt{-3}$, which cannot happen since $3$ doesn't divide $4$. Hence, if $4 - \sqrt{-3} = \beta\gamma$, we must have $N(\beta\gamma) = N(\beta)N(\gamma) = 27$, with $N(\beta) = 1$ and $N(\gamma)=27$. This implies that $\beta$ is a unit, and so $4-\sqrt{-3}$ is irreducible.