I don’t understand the application of ring isomorphism theorems in this answer, which states $\mathbb{Z}[x]/(x^2 - 2, 17) \cong \mathbb{Z}_{17}[x]/(x^2-2)$.
Shouldn’t it rather be
$$ \begin{align} \mathbb{Z}[x]/(x^2 - 2, 17) &= \mathbb{Z}[x]/((x^2 - 2) + (17)) \\ &\cong \mathbb{Z}_{17}[x]/(((x^2 - 2) + (17)) / (17)) \\ &\cong \mathbb{Z}_{17}[x]/((x^2 - 2) / ((x^2 - 2) \cap (17)))? \end{align} $$
The second line because of the third isomorphism theorem, the third line because of the second isomorphism theorem. Why is $(x^2 - 2) / ((x^2 - 2) \cap (17)) = (x^2 - 2)$?
The Third Isomorphism Theorem states that if $I,J$ are ideals of a ring $R$ and $J \subseteq I$, then $$ \frac{R}{I} \cong \frac{R/J}{I/J} \, . $$ In your example, we have $R = \mathbb{Z}[x]$, $I = (x^2 - 2, 17)$, and $J = (17)$. Then \begin{align*} \frac{\mathbb{Z}[x]}{(x^2 - 2, 17)} \cong \frac{\mathbb{Z}[x]/(17)}{(x^2 - 2, 17)/(17)} \cong \frac{(\mathbb{Z}/17\mathbb{Z})[x]}{(x^2 - 2)} \, . \end{align*} (Note that $(x^2 - 2, 17)/(17) = (x^2-2)$ in $(\mathbb{Z}/17\mathbb{Z})[x]$ since $17 = 0$.)