Is it true that $\mathbb{Z}[\sqrt{79}]/(7, 1+\sqrt{79}) \cong \mathbb{Z}_7[X]/(\hat{1}+x)$? I am trying to see whether $(7, 1+\sqrt{79})$ is a prime ideal in $\mathbb{Z}[\sqrt{79}]$. I tried doing this by looking at the factor ring $\mathbb{Z}[\sqrt{79}]/(7, 1+\sqrt{79})$. We have $\mathbb{Z}[\sqrt{79}]\cong \mathbb{Z}[X]/(X^2-79)$, so $$\mathbb{Z}[\sqrt{79}]/(7, 1+\sqrt{79}) \cong \frac{\mathbb{Z}[X]/(X^2-79)}{(\hat{7}, \hat{1}+X)}=\frac{\mathbb{Z}[X]/(X^2-79)}{(7, 1+X)/(X^2-79)}\cong \mathbb{Z}[X]/(7, 1+X)\cong \frac{\mathbb{Z}[X]/(7)}{(7, 1+X)/(7)}\cong\frac{\mathbb{Z}_7[X]}{(\hat{1}+X)}.$$
Is this correct? Since $\hat{1}+X$ is irreducible in $\mathbb{Z}_7[X]$, it would follow that my ideal $(7, 1+\sqrt{79})$ is prime in $\mathbb{Z}[\sqrt{79}]$.

Are you sure you have the problem right? Call $I=(7,\sqrt{79}+1)$. Since it’s an ideal, it also contains $(\sqrt{79}+1)(\sqrt{79}-1)=78$. Since $78$ and $7$ are relatively prime, $1\in I$, so you’re dealing with the unit ideal.
