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

  • $\begingroup$ What is your question? $\endgroup$
    – Anurag A
    Nov 26 '15 at 19:36
  • $\begingroup$ @anaragA Next question is to show why it is not a principle ideal $\endgroup$ Nov 26 '15 at 19:39
  • $\begingroup$ An ideal has no principles ;o) $\endgroup$
    – Bernard
    Nov 26 '15 at 19:42
  • $\begingroup$ Literally, toke out like 3 words. This question is meant to show that it is an ideal from the beginning and show alternate ways to determine that conclusion. Showing it is not principle is the next logical step but that is just a small part. $\endgroup$ Nov 26 '15 at 21:29
  • $\begingroup$ Reopening now that the question is limited to the question whether such polynomials form an ideal. I agree with user26857 that $I(x)$ is a bit unusual way of denoting this set. Normally that would be reserved for the set of polynomials with ALL their coefficients belonging to that ideal. Whatever :-) Anyway, be cautious about edits that make a major change to the question. Here's the link in case you still need it. $\endgroup$ Nov 26 '15 at 21:31

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 $\deg f \leq \deg h$.]


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