Proving that an ideal is prime - is it correct?

Question

I need to prove that although $X^2 + 3X +1 \in \mathbb{Z} [X]$ is irreducible, the ideals $(5,X^2 + 3X +1 )$ and $(11, X^2 + 3X +1)$ are not prime.

I know that an ideal $I$ is prime iff ${\mathbb{Z}[X]}/I$ has no zero divisors.

So when it comes to the first ideal, we have: ${\mathbb{Z}[X]}/(5,X^2 + 3X +1)={\mathbb{Z_5}[X]}/{(X^2 + 3X +1 )} = {\mathbb{Z_5}[X]} / {((X-1)^2 )}$.

That means that $X-1 + ((X-1)^2)$ is a zero divisor in the quotient ring so the ideal is not prime. Is that correct?

Similarly, for $(11, X^2 + 3X +1)$ we have that $X^2 + 3X +1 = X^2 -8X + 12 = (X-2)(X-6)$ in $\mathbb{Z}_{11}[X]$.

For what $p \in \mathbb{Z}$ prime is the ideal $(p,X^2 + 3X +1 )$ prime?

Could you help me a bit?

Answer 1

If $p=2$, then $X^2+3X+1$ is irreducible modulo $p$.
For $p\ge 3$ one can write $X^2+3X+1=(x+3/2)^2-5/4=[(2X+3)^2-5]/4$ and your question reduces to the following: Determine all primes $p$ for which $5$ is a quadratic residue modulo $p$. (Maybe I had to say that $\mathbb Z[X]/(p,X^2 + 3X +1)\simeq(\mathbb Z/p\mathbb Z)[X]/(X^2 + 3X +1)$.)

For what $p \in \mathbb{Z}$ prime is the ideal $(p,X^2 + 3X +1)$ prime? For $p\equiv 2,3\bmod 5$.