The polynomial of minimal degree with root $\alpha$ is unique.

Question

So I am working on the following proof:

Problem Statement: Let $\alpha$ be a complex number. Prove that the kernel of the substitution map $\mathbb{Z}[x] \rightarrow \mathbb{C}$ that sends $x \mapsto \alpha$ is a principal ideal, and describe its generator.

To assist me with the proof, I need to show that the polynomial $f(x)$ of minimal degree with root $\alpha$ is unique, so that I can then show that $f(x)$ generates the kernel of the substitution map.

I am trying to do the following:

Let $\varphi: \mathbb{Z}[x] \rightarrow \mathbb{C}$ defined by $x \mapsto \alpha$. We will first show that there is only one unique polynomial of minimal degree with root $\alpha$ in $\mathbb{Z}[x]$.

Let's claim that $\exists$ $2$ irreducible polynomials of minimal degree with root $\alpha$ in $\mathbb{Z}[x]$, namely $f(x)$ and $g(x)$. Then consider $h(x)=f(x)g(x)$. Then we have $$\varphi(h(x))=\varphi(f(x)g(x))=\varphi(f(x))\varphi(g(x))=f(\alpha)g(\alpha)=0\cdot0=0$$ Since $\mathrm{im}\ \varphi$ is an integral domain, then $fg=0$ implies that $f(x)\equiv 0$ or $g(x)\equiv 0$. Thus, we cannot cannot have more than one polynomial of minimal degree with root $\alpha$ in $\mathbb{Z}[x]$.

I think I am getting a little confused here. My professor told me "The image of a substitution map is an integral domain, but if the ideal is generated by $fg$ them $fg=0$ in the quotient, so the quotient is not an integral domain. Thus, you definitely want the kernel to be generated by an irreducible polynomial."

I think I am just bothered because $f(\alpha)=g(\alpha)=0$, but is my claim that since $\mathrm{im}\ \varphi$ is a domain, that one of $f$ or $g$ are identically $0$ invalid? Because in the image, would they not both be identically $0$?

If someone could verify this or help lead me in the right direction, I would appreciate it!

Answer 1

You only have that one of them is zero at $\alpha$, not identically. But remember you have two things you are assuming about $f,g$, that they are zero at $\alpha$, and of minimal degree. So you should try and contradict the minimality in order to show uniqueness.

Answer 2

First, there may be no "minimal polynomial" that vanishes at $\alpha$ - namely, if $\alpha$ is transcendental. In that case, the kernel is $(0)$, which is indeed a principal ideal, so you are done.

Then suppose $\alpha$ is algebraic. For every polynomial $f$ that vanishes at $ \alpha$, $-f$ ALSO vanishes there, so it doesn't make sense to speak of "the" minimal polynomial (or perhaps it does, but it would not be unique).

Rather, for $\alpha$ algebraic, let $f(x)$ be the non-zero polynomial of minimal degree that vanishes at $\alpha$ AND having the least leading coefficient among ALL polynomials of the same degree and with positive leading coefficient, vanishing at $\alpha$. You may show that this characterization defines a UNIQUE polynomial $f(x)$, but that is not part of the proof. Rather, if $a$ is the leading coefficient of $f$, and $g(x)$ is another polynomial of the same degree, with positive leading coefficient $b$ and that also vanishes at $\alpha$, FIRST show that $a|b$ by the minimal choice of $a$, and then prove that this implies $f|g$ due to the minimal degree. You have to do it in these two separate steps. Then show that $f$ divides all polynomials of higher degree that vanish at $\alpha$. This proves that the kernel is the principal ideal generated by $f(x)$.