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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 $(x^2 - 2) / ((x^2 - 2) \cap (17)) = (x^2 - 2)$?

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    The title does not reflect the principle underlying the quotient ring example that motivates your Question. Note that a quotient of ideals in ring $R$ is apt to correspond to an ideal in a quotient ring, not an ideal in $R$ itself. – 2017-01-20
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    In $\mathbb Z_{17}[x]$ we have $17$ is equal to zero. So what does it mean to take $(x^2-2)\cap (17)$? The real key is that you are dealing with essentially different $x$s - that is, you should replace the $x$s on one side with a $y$ to get an isomorphism which is clearer. – 2017-01-20
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    Thank you everybody for clearing out the mess in my head and sorry if I sometimes have to ask stupid questions. – 2017-01-20

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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 $17 = 0$.)

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    Thank you, I was missing your last line... – 2017-01-20
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    Still, I wonder if your approach would work. Since $17 = 0$ in $(\mathbb{Z}/17\mathbb{Z})[x]$, then $(x^2 - 2) \cap (17) = (x^2 - 2) \cap (0) = (0)$, so you get the right answer. But you should be more precise: the Second Isomorphism Theorem says $\frac{I+J}{J} \cong \frac{I}{I \cap J}$, not that they are equal. Anyway, just because $I_1 \cong I_2$ does not mean that $R/I_1 \cong R/I_2$ (consider $R = I_1 = \mathbb{Z}$ and $I_2 = 2 \mathbb{Z}$), but it seems to work out in your case... – 2017-01-20
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    That’s a very good point. I did not mean to say that the OP was correct, it was only the result of my toying around with the isomorphism theorems, overlooking your last remark... – 2017-01-20