I have seen this as an answer to Prove that $p$ is prime in $\mathbb{Z}[\sqrt{-3}]$ if and only if $x^2+3$ is irreducible in $\mathbb{F}_p[x]$.

If you assume that $x^2+3$ is irreducible, then you have that $$\mathbb{Z}[\sqrt{-3}]/(p)\cong \mathbb{Z}[x]/(p,x^2+3)\cong \mathbb{F}_p[x]/(x^2+3)$$

However, I am having a very difficult time proving this.


Suppose $x^2+3$ is irreducible in $\mathbb{Z}_p[x]$. Then we have that $\frac{\mathbb{Z}_p[x]}{(x^2+3)}$ is a field. Let $\phi: \mathbb{Z}[x] \to \frac{\mathbb{Z}_p[x]}{(x^2+3)}$ be a homomorphism such that $p(x) \to q(x) \text{mod} (x^2+3)$ where $q(x) \in \mathbb{Z}_p[x]$. Let $K=(p,x^2+3)=\{ph(x) + (x^2+3)r(x)\}$. Then $\phi(K)=0$, thus $K$ is the kernel. Then by the first isomorphism theorem we have that $\frac{\mathbb{Z}[x]}{(p,x^2+3)} \cong \frac{\mathbb{Z}_p[x]}{(x^2+3)}$.

I am sure there are some errors in my attempt, furthermore I am having a difficult time defining these ring homomorphism correctly. Also, I am completely stuck as to how to prove that $$\mathbb{Z}[\sqrt{-3}]/(p)\cong \mathbb{Z}[x]/(p,x^2+3)$$


There is a cannonical homomorphism $\mathbb{Z}[x]\longrightarrow\mathbb{Z}(\sqrt{-3})$, which extends the identity map on $\mathbb{Z}$ and $x\mapsto\sqrt{-3}$. Compose this with the quotient map to get a map $\phi:\mathbb{Z}[x]\longrightarrow \mathbb{Z}(\sqrt{-3})/(p)$. Claim that the kernel is $I=(p,x^2+3)$. Clearly $I\subset \ker\ \phi$. Take $f(x)\in \ker\ \phi$. Then $f(\sqrt{-3})+(p)=\bar{0}$. That is $f(\sqrt{-3})\in (p)$. From this can you prove that $f(x)\in(x^2+3,p)$? Hint : Use Euclidean algorithm and write $f(x)=(x^2+3)q(x)+r(x)$ with appropriate conditions on $r$.


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