$\mathbb Z[x]/(x^2-2,5) \cong \mathbb Z/5\mathbb Z[x]/(x^2-2)$

Question

$\mathbb Z[x]/(x^2-2,5) \cong \mathbb Z/5\mathbb Z[x]/(x^2-2) \cong \mathbb Z/5\mathbb Z[x]/(x^2+3)$

I saw this in my textbook (Lovett). I'm guessing that the second isomorphism works becuase we are now in $Z/5\mathbb Z$, and all we did was replace $-2$ with $3$.

But I do not understand how the $5$ "moved up" from the ideal to the ring, making the polynomial ring $\mathbb Z[x]$ into $\mathbb Z/5\mathbb Z[x]$. I do not see it intuitively, nor do I see an obvious proof for this.

Answer 1

We have $$ (5)\subseteq (5,x^2-2) $$ By the third isomorphism theorem: $$ \Bbb Z[x]/(5,x^2-2)\cong (\Bbb Z[x]/(5))/((5,x^2-2)/(5)) $$ Now note that $\Bbb Z[x]/(5)$ is isomorphic to $\Bbb Z/5\Bbb Z[x]$. Under that isomorphism (there are several, but one stands out as canonical) the ideal $(5,x^2-2)/(5)$ becomes $(x^2-2)$, and we are done.

Answer 2

Hint Consider the ring homomorphisms $$\mathbb Z[x] \to \mathbb Z/5\mathbb Z[x] \to \mathbb Z/5\mathbb Z[x]/(x^2-2) $$

What is the kernel of the composition?