# Prove $I$ $\in$ $\mathbb{Z}[x]$ s.t. $I=\{f(x)\mid f(0) \text{ is even}\}$ is a maximal ideal, and find the number of elements in $\mathbb{Z}[x]/I$ [duplicate]

I have already shown (in a previous question) that $I=\langle x,2\rangle$, and that I is a prime ideal, and I believe that allows me to show that $\mathbb{Z}[x]/I$ is an integral domain, which if it is finite would allow me to conclude that $\mathbb{Z}[x]/I$ is a field, which would imply that I is a maximal ideal. But I don't know how to determine the number of elements in $\mathbb{Z}[x]/I$, and thus do not know how to determine whether or not it is finite.

Can someone tell me if this thinking is correct, and if so how to determine the number of elements in $\mathbb{Z}[x]/I$?

Note: we have not covered irreducibility in this course.

• How did you show that $I$ was prime? If you did it by finding a homomorphism from $\Bbb Z[x]$ to an integral domain with kernel $I$, then $|\Bbb Z[x]|/I$ will the size of the image of that homomorphism. Jul 30, 2017 at 18:52

Define the homomorphism surjective $\phi:\mathbb{Z}[x]\to \mathbb{Z}$ by $\phi(f(x))=f(0)$; Indeed, for any $m\in\mathbb{Z}$, consider $f(x)=x+m$ and we have $\phi(f)=m$. Consider the projection $\pi:\mathbb{Z}\to \mathbb{Z}_2$. Note that $I=\ker(\phi\circ\pi)$, thus $\mathbb{Z}[x]/I=\mathbb{Z}[x]/\ker\left(\phi\circ\pi\right)\simeq\mathbb{Z}_2$. Then $I$ is maximal and $\mathbb{Z}[x]/I$ has 2 elements.
f(x)= xg(x) +2h(x), we have f(0) =2h(0), so that $f(x)\in I$. If $f(x)\in I$ , then $f(x)=a_nx^n+...+a_1x+2k=x(a_nx^{n-1}+...+a_1)+2k \in<x, 2>$ . I is prime and maximal. $\frac {Z[x]}{I}$ has two elements. . If h(x) & g(x) are two polynomials when u multiply them and get some f(x) which belongs to I then it's obvious that any one of h(x) & g(x) will belong to I. That's why I is prime. And $\frac {Z[x]}{I}$ will be isomorphic to $Z_2$. Moreover $\frac {Z[x]}{<x,p>}$ is always isomorphic to $Z_p$ for prime p. And as you said that you read that book so you should be aware of the fact that For some positive integer n You will have
$<x> \subset<x,2^n>\subset<x,2^{n-1}>\subset......<2,x> \in Z[x]$ so it is maximal ideal.