Maximal ideal in the ring of polynomials over $\mathbb Z$ 
Let $\mathbb Z[x]$ the ring of polynomials with integers coefficients in one variable and $I =\langle 5,x^2 + 2\rangle$, how can I prove that $I$ is maximal ideal.

I tried first see that $5$ and $x^2+2$ are both polynomial in that ring but how can i get that is maximal ? some help please.
 A: Consider the natural homomorphism from the polynomial ring to the five-element field $\mathbb{Z}/5\mathbb{Z}$, extended by $\sqrt 3$:
$$ f: \mathbb Z[x] \longrightarrow (\mathbb{Z}/5\mathbb{Z})\; (\sqrt 3)$$
generated by $f(1) = 1, f(x) = \sqrt{3}$ (you might show this is a homomorphism, if it's not clear). Notice that the kernel of this homomorphism is exactly $I$ (or show it, if it's not clear). The condition that $I$ is the kernel of a ring homomorphism onto a field is equivalent to $I$ being maximal (similarly, if $I$ is the kernel of a ring homomorphism onto an integral domain, then $I$ is prime).
Therefore $I$ is maximal.

Let's speak of the intuition here. It is often easier to show that an ideal is maximal by evidencing a homomorphism to a field with that ideal as the kernel. 
Quotienting out by $5\mathbb Z$ is saying exactly that multiples of $5$ are the same as $0$. As $I$ has one generator that is $5$, this makes a lot of sense.
For $x^2 + 2$ to be in the kernel, we are saying that $f(x^2 + 2) = 0$, or rather that $f(x)^2 = -2 = 3$ (where the last equality comes from us being is $\mathbb{Z}/5\mathbb{Z}$, so we can work mod $5$ instead of with imaginary bits). Well, if $f(x)^2 = 3$, we should make $f(x) = \sqrt 3$.
And that's how we know how to make the $f$ map.
A: If you quotient by the ideal $I$, then 
$$5 \equiv 0 \pmod{I} \text{ and } x^2+2 \equiv 0 \pmod{I}.$$
This also suggests that $x^2 \equiv -2 \equiv 3 \pmod{I}$
This helps you get the idea that perhaps $x \mapsto \sqrt{3}$ and $5 \mapsto 0$
A: Say you add some $f\notin\langle 5, x^2+2\rangle$, and consider the ideal $I'=I+(f)$. We want to show this is the entire ring $\mathbb{Z}[x]$.
Well, dividing by remainder, we can write $f=(x^2+2)g+h$ where $h$ is linear. Reduce $h$ mod $5$ and we write $h=ax+b$ where $0\le a, b\le 4$. By assumption that $f\notin I$, we get $h\neq 0$.
I believe there is some nice algebra to reduce the casework here, to show basically that using $h, 5$ and $x^2+2$, you can get $1$ and hence $I'=\mathbb Z[x]$, but I'm not sure off the top of my head how to prove it in a unified, clean way.
Also, if a reference would help you, in the first few pages of chapter 1, Reid Undergraduate Commutative Algebra, he proves that for any PID $A$, maximal ideals of $A[x]$ are exactly those of the form $(f,p)$ where $p\in A$ is prime and $f\in A[x]$ is irreducible mod $p$. Maybe running through that proof with your special case would be helpful?
A: By the Correspondence Theorem, $(5,X^2+2)$ is maximal in ${\mathbb Z}[X] \Leftrightarrow (5,X^2+2)/(5)$ is maximal in ${\mathbb Z}[X]/(5)$.  Using an isomorphism, this is the same thing as saying that $(X^2 + 2)$ is maximal in $(\mathbb{Z}/(5))[X]$.  This ring is a PID, so we just need to show that $X^2 + 2$ is irreducible mod 5.  Since it has degree 2, this just means checking that it has no zeros mod 5.
A: First look at $\mathbb Z[x]/\langle x^2+2 \rangle$ and some polynomial $p(x)\in \mathbb Z[x]$.
We can find polynomials $q(x), r(x)$ so that $p(x)=q(x)(x^2+2)+r(x)$ and $r(x)=ax+b$. Since $x^2+2$ is irreducible, it has no common factor with $r(x)$ unless $r(x)=0$ and we can find polynomials $s,t$ with $$sp+t(x^2+2)=c$$ where $c$ is an integer. This provides a multiplicative inverse for $p$ only if $c=1$, so we don't yet have a field, although we can nearly invert polynomials - it is just that $\mathbb Z$ doesn't have multiplicative inverses.
Now let's reduce the coefficients modulo $5$. $x^2+2$ remains irreducible*. We can carry through the division, but non-zero $c$ are now invertible modulo $5$, so for non-zero $p$ we can use Euclid's algorithm to find $$sp+t(x^2+2)\equiv 1 $$
So that $s(x)$ is a multiplicative inverse and every non-zero element of the ring is invertible. Hence the ring is a field and the ideal is maximal.
*Note this would not have been the case if we'd been working with $x^2+1$, or if we'd been working with $x^2+2$ modulo $11$ so we have to be careful.
Working modulo $I=\langle x^2+2,5\rangle$ we have $x^2\equiv -2\equiv 3$, so we can compute the inverse of $r=ax+b$ explicitly as $$\frac 1{ax+b}=\frac 1{ax+b}\cdot \frac{ax-b}{ax-b}\equiv\frac a{3a^2-b^2}x-\frac b{3a^2-b^2}$$ where $3a^2-b^2$ is invertible modulo $5$ unless $a,b\equiv 0$.

Working the other way about we can factor $\langle 5\rangle$ first to get the ring of polynomials with coefficients reduced modulo $5$ - so the coefficients are elements of the five element field $F_5$ and $x^2+2$ is irreducible, and we have a standard type of extension field.
