Quotient ring is field or not? Is $\mathbb{Z}[x]/(x-3)$ a field? I see $x-3$ is not reducible over the integers. So, I guess it should be a field. But the answers says it's not. 
 A: The implication “if $f(x)$ is an irreducible polynomial then the ideal $(f(x))$ is maximal” is valid when the ring of coefficients is a field.
Since $\mathbb{Z}$ is not a field, you can't apply it.
Indeed $x-3$ is an irreducible polynomial over the integers, but the ideal $(x-3)$ is not maximal in $\mathbb{Z}[x]$.
You should be able to prove that the map
$$
e\colon \mathbb{Z}[x]\to\mathbb{Z}
\qquad
e(f)=f(3)
$$
is a surjective ring homomorphism with $\ker e=(x-3)$, thereby proving that the quotient you're dealing with is not a field.
A: The ideal $ (x-3) $ is not maximal in $ \mathbb{Z}[x] $. Indeed, the ideal $ (x-3, x-1) $ properly contains it. To show that this ideal is not $ \mathbb{Z}[x] $ itself, note that
$$ a(x-3) + b(x-1) = 1 $$
implies that $ 2b = 1 $. There is no such $ b \in \mathbb{Z}[x] $, so $ 1 \notin (x-3, x-1) $.
A: As mentioned elsewhere, the ideal generated by $x-3$ is not maximal in ${\bf Z}[x]$. To see this, consider the ideal generated by $x$ and $x-3$. It's not all of ${\bf Z}[x]$, since every element of it evaluates to a multiple of 3 at $x=0$ (so it doesn't contain 1, for example). It's not the same as the ideal generated by $x-3$, since it contains elements (like $x$) that don't vanish at $x=3$. So it's a proper ideal properly containing the ideal generated by $x-3$. 
"Irreducible polynomial implies maximal ideal" works over fields, but not over the integers. 
