Finding a group isomorphic to ring of polynomials

Question

let $J$ be an ideal of $F[x]$ such that $J$ contains every polynomial of $\mathbb{Z}[x]$ with constant term a multiple of 4. Im looking for a ring isomorphic to $\mathbb{Z}[x]/J$, with proof.

I was thinking that maybe $\mathbb{Z}[x]/J \cong \mathbb{Z}_4[x]$,

I started by defining a homomorphism $\phi : \mathbb{Z}[x] \to \mathbb{Z}_4[x]$, where $\phi(a(x)) = \overline{a(x)}$ for any $a(x) \in \mathbb{Z}[x]$

Answer 1

Observe that the set of all polynomial with $a_0$ even is exactly the ideal $(2,x)\subset \Bbb Z[x]$. Now you can see that $\Bbb Z[x] /(2,x)\cong \Bbb Z_2[x]/(x)\cong \Bbb Z_2$ that is a field, so $(2,x)$ is maximal and it must be $J=(2,x)$

Answer 2

Note that for polynomials $f,g\in\mathbb{Z}[x]$ we have $f+J=g+J$ if and only if the free coefficient of $f-g$ is even, which happens if and only if the free coefficients of $f,g$ are equal mod $2$. So $\mathbb{Z}[x]/J$ contains just two elements (the class of polynomials with even free coefficient and the class of polynomials with odd free coefficient) and this should make it clear that the ring you are looking for is $\mathbb{Z_2}$.

Formally, define a map $\varphi: \mathbb{Z}[x]\to\mathbb{Z_2}$ by $\varphi(\sum_{i=0}^n a_ix^i)=a_0$ (mod $2$). It is easy to see that $\varphi$ is a surjective homomorphism of rings and its kernel is exactly $J$. Now apply the first isomorphism theorem.

Answer 3

You are, in fact, quite close. Consider the function $\phi : \mathbb{Z}[x] \rightarrow \mathbb{Z}_2$, defined by $\phi(p(x)) \mapsto \overline{a_0} \operatorname{(mod} 2)$.

I) $\phi$ is a homomorphism:

If $p, q \in \mathbb{Z}[x]$, $p(x) = a_0 + ... + a_nx^n$ and $q(x) = b_0 + ... + b_m x^m$, which means the independent term of $(p + q)(x)$ is $(a_0 + b_0)$, and the independent term of $(pq)(x)$ is $a_0b_0$. Therefore, $\phi(p + q) = \overline{(a_0 + b_0)} = \overline{a_0} + \overline{b_0} = \phi(p) + \phi(q)$ and $\phi(pq) = \overline{a_0b_0} = \overline{a_0}\overline{b_0} = \phi(p)\phi(q)$.

II) $\phi$ is clearly surjective

III) $\phi$ has the kernel:

$\operatorname{ker}\phi = \{p \in \mathbb{Z}[x] : \phi(p) = \overline{0}\} = \{p \in \mathbb{Z}[x] : a_0 = \overline{0} \operatorname{(mod}2)\} = J$

Using the First Homomorphism Theorem, $\mathbb{Z}[x]/J \simeq \mathbb{Z}_2$

Does that help?