Minimality of the Generator of a Kernel

Question

Take the map $f: \mathbb{Z}[x] \to \mathbb{R}$ defined $f(x) = \sqrt{3}+2$. I have found an element of the kernel to be $x^4-2x^2-11$, since it is equal to $0$. I am unsure why/if this generates the kernel, since could we not have a polynomial of degree $3$? Can I somehow use division algorithm?

Answer 1

Note that $\sqrt{3}$ is a root of $x^2-3$, and so $\sqrt{3}+2$ is a root of $(x-2)^2 - 3 = x^2 - 4x + 1$. So this element is certainly in the kernel, and cannot be a multiple of your degree $4$ polynomial.

Moreover, I’m not sure your calculations are correct. Since $$\begin{align*} (\sqrt{3}+2)^2 &= 3 + 4\sqrt{3} + 4 = 7+4\sqrt{3}\\ (\sqrt{3}+2)^4 &= (7+4\sqrt{3})^2\\ &= 49 + 56\sqrt{3} + 48\\ &= 97+56\sqrt{3}, \end{align*}$$ I get that $$\begin{align*} (\sqrt{3}+2)^4 -2 (\sqrt{3}+2)^2 -11 &= 97+56\sqrt{3} -14-8\sqrt{3} - 11\\ &= 72 + 48\sqrt{3} \neq 0 \end{align*}$$ I don’t see why $x^4 - 2x^2 - 11$ is in the kernel.

(You can tell your polynomial cannot be in the kernel just with back of the envelope calculations: $3\lt \sqrt{3}+2\lt 4$. Letting $\alpha=\sqrt{3}+2$, we have that $81\lt \alpha^4 \lt 256$, and $-32\lt -2\alpha^2\lt -18$, so $$38 \lt \alpha^4 - 2\alpha^2-11 \lt 227.$$ So it cannot evaluate to $0$)

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How do you tell if a given polynomial generates the kernel?

You can go to $\mathbb{Q}[x]$, but even then verifying irreducibility of a given polynomial is not necessarily trivial. However, if you do establish irreducibility over $\mathbb{Q}$, then that will definitely generate the kernel over $\mathbb{Q}[x]$. The kernel for $\mathbb{Z}[x]$ will then correspond to $K\cap\mathbb{Z}[x]$, which will be generated by a multiple of the generator over $\mathbb{Q}[x]$ that has trivial content (the gcd of the coefficients is $1$); this follows from Gauss’s Lemma, which implies that if $f(x)\in\mathbb{Z}[x]$ with trivial content is reducible in $\mathbb{Q}[x]$, then it is reducible in $\mathbb{Z}[x]$. (If the content is not trivial, then it is easily seen to be not irreducible in $\mathbb{Z}[x]$).

Irreducibility over $\mathbb{Q}$ is much more difficult than irreducibility over $\mathbb{R}$; for the latter, we know a polynomial is irreducible if and only if it is linear or irreducible quadratic (which can be checked via the discriminant). Over $\mathbb{Q}$, there are irreducible polynomials of arbitrarily high degree, and that also means that you can have a polynomial with no roots which is nonetheless reducible (the same is true over $\mathbb{R}$, of course... the product of two irreducible quadratics, for example).

In the instant case, the fact that you know the polynomial has no roots over $\mathbb{Q}$ and we have found a quadratic that lies in the kernel tells you that this is the generator.

That you can first work over $\mathbb{Q}$ and then “clear denominators” may not be obvious. Here’s the argument: the kernel of the evaluation map $\varepsilon_{\alpha}\colon \mathbb{Z}[x]\to\mathbb{C}$ is the intersection of the kernel $K$ of the evaluation map $\varepsilon_{\alpha}\colon \mathbb{Q}[x]\to\mathbb{C}$ with $\mathbb{Z}[x]$ (here $\alpha\in\mathbb{C}$. Since $\mathbb{Q}[x]$ is a PID, and the evaluation map is not trivial (it maps $1$ to $\alpha)$), we have two cases.

1. If $\varepsilon_{\alpha}$ is one-to-one, that is, if $\alpha$ is transcendental, then $K=(0)$ and the kernel of the restriction to $\mathbb{Z}[x]$ is also $(0)$.

2. If $\alpha$ is algebraic, then $K$ is generated by the minimal polynomial over $\mathbb{Q}$, call it $q(x)$. Let $Q(x)$ be a polynomial with integer coefficients and trivial content that is a constant multiple of $q(x)$ (clear denominators, then factor out the gcd of the coefficients). Note that $K=(Q(x))$, since $Q(x)$ is an associate of $q(x)$ in $\mathbb{Q}[x]$. I claim that $Q(x)$ is the generator of the kernel over $\mathbb{Z}[x]$. Indeed, let $m(x)\in\mathbb{Z}[x]$ be an element of the kernel. Then there exists a polynbomial $r(x)\in\mathbb{Q}[x]$ such that $m(x) = Q(x)r(x)$. There exists a rational $s$ and a polynbomial with integer coefficients $R(x)$ such that $R(x)$ has trivial content and $r(x) = sR(x)$. Thus, $m(x) = sQ(x)R(x)$. By Gauss’s Lemma, $Q(x)R(x)$ has trivial content, so $s$ is actually the content of $m(x)$, and hence is an integer. Thus, $r(x)$ has integer coefficients, so $m(x)$ lies in the ideal generated by $Q(x)$ in $\mathbb{Z}[x]$, as desired.

So it comes down to being able to recognize that a polynomial with integer/rational coefficients is irreducible over $\mathbb{Q}$. There are lots of tests one can perform to check, but that is not a trivial problem.