Let $I[x]$ be subset of $\mathbb{Z}[x]$ where $I[x]$ is the set of polynomials with constant term is even.
Show $I[x]$ is an Ideal of $\mathbb{Z}[x]$
Def of $I[x]$ Ideal of $\mathbb{Z}[x]$ that is $I \subset \mathbb{Z}[x]$
where
$$ \begin{align*}
&i)\forall a,b \in I[x]\Rightarrow a-b \in I[x] \\
&ii)\forall r \in Z[x], \forall i \in I[x]:ri \in I[x] \wedge ir \in I[x] \
\end{align*} $$
[Showing i)]
Assume $a\in I[x], b \in I[x]$ that is
$$ a=a_nx^n+\dots+a_0 \text{ and } b=b_nx^n+\dots+b_0$$
Now, $a-b=\dots+(a_0-b_0)$ is even since $\exists k_1,k_2 \in Z:a_0=k_1*2,b_0=k_2*2$.
That is $$\begin{aligned} a-b=&\dots+(a_0-b_0)
\\=&\dots+2k_1-2k_2\\
=&\dots+2(k_1-k_2)
\\=&\dots+2(k_3) && \text{ where } k_1-k_2=k_3 \end{aligned}$$
So, a-b has an even constant.
[Showing ii)]
Assuming $a\in Z[x],b\in \mathbb{Z}[x]$ where
$$ a*b=\dots +a_0b_0$$
$b_0$ is even and $a_0$ is even or odd. Either way even times even is even and even times odd is even. So $ab \in I$
appreciate critique on the work, or some weird way of answering the question
Best Answer
You way works, but it is better to think in terms of morphisms. Specifically, to show that something is an ideal, show that it is the kernel of a morphism.
In this case, let's consider the map $\phi : \mathbb Z[x] \to\mathbb Z : f \mapsto f(0)$ composed with canonical projection $\pi : \mathbb Z \to \mathbb Z/2\mathbb Z$. It maps a polynomial $f(x) = a_nx^n+\dots+a_0$ to the constant term $a_0 \bmod 2$.
Since $\phi$ and $\pi$ are morphisms, so is their composition $\pi\circ\phi$. Now we use that the kernel of a morphism is an ideal and we get for free that $\ker(\pi\circ\phi) = I[x]$ is an ideal.
To show that $I[x]$ cannot be principal, argue by contradiction: assume that $I[x]=(c)$ for some $c\in\mathbb Z[x]$ and observe that $(2,x)\subseteq I[x]$. This implies that $c\mid x$ and $c\mid 2$. From this, conclude that $c$ is a unit. (*) But then $I[x]=(c)=\mathbb Z[x]$, a contradiction.
[(*) Hint: If $fg=h$ in $\mathbb Z[x]$, then $\deg f+ \deg g = \deg h$. Therefore, if $f\mid h$, then $\deg f \leq \deg h$.]