How to prove, that $(p, 1+\sqrt{-5})$ is a prime ideals in $\mathbb Z[\sqrt{-5}]$ for prime $p\in\mathbb Z$?

Question

I am wondering if $I:=(p,1+\sqrt{-5})$ is a prime ideal of $R:=\mathbb Z[\sqrt{-5}]$. I just tried to use the definition of the index and used to show that this is $2$. Since $R$ is a Dedekind domain, this shows the primality.

I can't find any argument for proving $[R:I]=2$. Could someone help me out, please?

Thanks

Answer 1

This is not entirely true. For most primes $p$, the ideal $I$ is the unit ideal. If $(1)$ is a prime ideal depends on your taste I guess.

Anyways, here's one way to compute in this case: note that $\mathbb Z[\sqrt{-5}] = \mathbb Z[x]/(x^2+5)$. Here $x$ correspond to $\sqrt{-5}$. Dividing out by the ideal $I$ is the same computing $$ \mathbb Z[x]/(x^2+5,p,1+x). $$

Here I use that dividing out by $I$ first then by $J$ is the same as dividing out by $I+J$. Explicitly, any ideal $J \subset R/I$ lifts to an ideal $\overline J$ containing $I$. So we can write $\overline J = I+ J'$ (just let overline $J'$ be generated by the elements of $\overline J$ not in $I$). Then $R/(I+J') \simeq (R/I)/J$.

But dividing out by $1+x$ is the same as putting $x=-1$ in the quotient ring. Do you see the rest?