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!

  • 1
    $\begingroup$ if $g$ doesn't divide $f$ and $f$ doesn't divide $g$, $f(\alpha) = g(\alpha) =0$ and $deg(f) = deg(g)$ then there exists two integers $b,c$ such that $b f(\alpha) - c g(\alpha) = 0$, $bf-cg$ is not the zero polynomial, and $deg(b f - cg) < deg(f)$, contradicting that $f,g$ are of minimal degree $\endgroup$
    – reuns
    Commented Mar 29, 2016 at 0:01
  • $\begingroup$ So basically, $bf-cg$ would be the minimal degree polynomial... But how is that showing that $bf-cg$ is unique? what if there is some other polynomial with root alpha and same degree as $bf-cg$? $\endgroup$
    – yung_Pabs
    Commented Mar 29, 2016 at 0:06
  • $\begingroup$ among the (nonzero) polynomials having $\alpha$ as one of their roots, there is a minimal degree. the question is if there is only one polynomial of that degree, or two or more. $\endgroup$
    – reuns
    Commented Mar 29, 2016 at 0:08
  • $\begingroup$ OP what do you mean by unique minimal polynomial? $X^2 - 2$ and $2X^2 - 4$ are polynomials in $\mathbb{Z}[X]$ of minimal degree of which $\sqrt{2}$ is a root, and they generate different principal ideals. $\endgroup$
    – D_S
    Commented Mar 29, 2016 at 1:17

2 Answers 2


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.

  • $\begingroup$ Ohh I see. Should I also claim that the two polynomials are monic? Say $\deg f(x) = \deg g(x) = n$, and it they are monic, $\varphi(f(x)-g(x))=\varphi(f(x))-\varphi(g(x))=f(\alpha)-g(\alpha)=0-0=0$. Then $\deg f(x)-g(x) < n$ since the leading term is eliminated. So then there is some irreducible polynomial of degree less than $n$ with root alpha? This causes an issue though because I can continue claiming that there is some polynomial with degree less than $n$ whose root is $\alpha$... $\endgroup$
    – yung_Pabs
    Commented Mar 29, 2016 at 0:03
  • $\begingroup$ But the problem was not that the minimal polynomial was of degree $n$, just that there were two distinct minimal polynomials. $\endgroup$ Commented Mar 29, 2016 at 0:10
  • $\begingroup$ I guess I am getting confused because did I not contradict that there were 2 polynomials with my original proof? I don't see how else I could contradict minimality using the substitution map. $\endgroup$
    – yung_Pabs
    Commented Mar 29, 2016 at 0:14
  • $\begingroup$ In an integral domain, the generator of a principal ideal is unique, up to a unit. $\endgroup$
    – Bernard
    Commented Mar 29, 2016 at 0:19
  • $\begingroup$ @bernard : that $\mathbb{Z}[x]$ is an integral domain is needed to state that $f$ doesn't divide $g$ and $g$ doesn't divide $f$ ? $\endgroup$
    – reuns
    Commented Mar 29, 2016 at 0:20

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 $f(x)$.


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