Ring theory question: $I=\langle x,2 \rangle$ prime/maximal ideal in $\mathbb Z[x]$?

Question

In $\mathbb{Z}[x]$ , let $I = \lbrace f(x) \in \mathbb{Z}[x] : f (0) \text{ is an even integer} \rbrace.$

1. Is $I=\langle x,2 \rangle$ a prime ideal of $\mathbb{Z}[x]$?
2. Is $I=\langle x,2 \rangle$ a maximal ideal?

Answer 1

You can really solve this without using any higher principles. Here is an elementary solution:

Well, what IS $I$? It is an ideal that consists of integer polynomials that have even constant term. $f = a_nx^n + \cdots + a_1x + a_0\in \mathbb{Z}[x]$, then $f\in I$ exactly means that $a_0$ is even.

Okay, $I$ is prime: we need show $f,g\notin I \implies fg\notin I$. Well, if $f$ and $g$ both have a constant term that is not even (i.e. odd), can their product have a constant term that is even? No.

$I$ is maximal: we show that if we pick any element not in $I$, say $f\in \mathbb{Z}[x] - I$, then $I + (f) = \mathbb{Z}[x]$. Well, $f\in \mathbb{Z}[x] - I$ means that $f = a_nx^n + \cdots + a_1x + a_0\in \mathbb{Z}[x]$ with $a_0$ odd. This means that $1\in I+(f)$ (why?), and hence we can't make $I$ any bigger without making it become the whole ring. Hence, $I$ is maximal.

Answer 2

You can go the abstract way directly using the first isomorphism theorems:

$$\Bbb Z[x]/\langle x,2\rangle\cong\left(\Bbb Z[x]/\langle 2\rangle\right)/\left(\langle x,2\rangle/\langle2\rangle\right)\cong\Bbb F_2[x]/\langle x\rangle\cong\Bbb F_2$$

With $\;\Bbb F_2\cong\Bbb Z/2\Bbb Z=\;$ the field with two elements.

Now just use the already mentioned theorem: a proper ideal in a commutative unitary ring is maximal iff its quotient with the ring is a field.

Answer 3

By the description in the first line of the question, $I$ is the kernel of a ring homomorphism $\def\Z{\Bbb Z}\Z[X]\to\Z/2\Z$ that maps $P\mapsto P(0)\bmod 2$. Since $\Z/2\Z$ is a field and the homomorphism surjective, $I$ is a maximal and hence prime ideal. (For maximal: any proper ideal containing $I$ is the inverse image of a proper ideal of $\Z/2\Z$, hence of $\{0\}$; for being a prime ideal, being the kernel of any morphism to an integral domain would have sufficed.)