Proving that a subset of polynomials is an ideal

Question

I want to prove that the subset

I = {a_nxⁿ + · · · + a₁x + a₀ | n ≥ 0, a_n, . . . , a₀ ∈ Z, a₀ = 0}

is an ideal in Z[x].

I know that to start off with, I need to show that the set is non-empty, which is true because a₀ = 0, so 0 ∈ I.

Then, I want to show that for y,z ∈ I, y - z ∈ I. This is that part that I'm stuck on.

I have y = a_mx^m + · · · + a₁x + a₀ and

z = a_nxⁿ + · · · + a₁x + a₀,

for m,n ≥ 0, a_n, . . . , a₀ ∈ Z, a_m, . . . , a₀ ∈ Z, a₀ = 0.

I'm not sure how to reduce these polynomials to show that y - z ∈ I.

Thanks for any help.

Answer 1

If you set $y=a_m x^m+\cdots+a_1x+a0$, ($a_0=0$), then you have to set the polynomial $z$ to have possibly different coefficients (not just different degree). That is, it's coefficients should be called something like $b_k$ or $\hat a_k$, etc.

Let's say $z=b_n x^n+\cdots+b_1 x+b_0$, ($b_0=0$). Suppose $m>n$ (the cases $m<n$ and $m=n$ are analogous), then $$y-z=a_mx^m+\cdots+a_{n+1}x^{n+1}+(a_n-b_n)x^n+\cdots+(a_1-b_1)x+(a_0-b_0).$$

In other words, $$y-z=c_m x^m+\cdots +c_1 x+c_0,$$ and $c_0=a_0-b_0=0-0=0,$ so $y-z\in I$.

Then, you should prove that $y\in I$ and $z\in \mathbb Z[x]$ implies $yz=zy\in I$.

Another way to aboard the proof is to verify that $I=\{x\cdot y\, \colon\, y\in \mathbb Z[x]\}$, which implies that $I$ is the ideal generated by $x$ (a principal ideal), this is $$I=\langle x \rangle.$$

Answer 2

Hint: $I=\{f\in\mathbf{Z}[x]\mid f(0)=0\}$. After this, verifying the ideal properties becomes very easy.

About your attempt, you should be more careful: if you set $y=a_mx^m+\dots+a_1x+a_0$, you have to choose different letters for $z$, say $z=b_nx^n+\dots+b_1x+b_0$.