The Hermite–Minkowski Theorem

We give the standard proof of the Hermite–Minkowski theorem.

Our proof follows Neukrich’s Algebraic Number Theory, where most of the proof is given in Theorem III.2.13.

Theorem 1 (Hermite–Minkowski). Fix positive integers \(d\). Then there are finitely many number fields \(K\) (up to isomorphism) with \(\left|\operatorname{disc}K\right|\le d\).

It is not too hard to describe the general approach: we will bound the collection of possible polynomials which could generate \(K\). This will be done by some clever application of Minkowski theory.

Before doing anything, we note that we are allowed to fix the degree in the proof of Theorem 1.

Theorem 2 (Hermite). Fix positive integers \(d\) and \(n\). Then there are finitely many fields \(K\) (up to isomorphism) of degree \(n\) with \(\left|\operatorname{disc}K\right|\le d\).

Proof that Theorem 2 implies Theorem 1. By looping over all possible degrees, it is enough to show that there are only finitely many degrees \(n\) possible for which there is a number field \(K\) of degree \(n\) and \(\left|\operatorname{disc}K\right|\le d\). Well, in this blog post, we showed that \[\sqrt{\left|\operatorname{disc}K\right|}\ge\frac{n^n}{n!}\left(\frac\pi4\right)^{n/2}.\] The right-hand side goes to \(\infty\) as \(n\to\infty\) because \(n^n\) dominates (e.g., one can see this by taking \(\log\)s), so the claim follows. \(\blacksquare\)

We now begin executing our plan to restrict polynomials \(f\) generating \(K\). Here is our goal.

Lemma 3. Fix positive integers \(d\) and \(n\). Then there is a constant \(C(n,d)\) such that any number field \(K\) with degree \(n\) and discriminant at most \(d\) can be written as \[K\cong\frac{\mathbb Q[x]}{(f(x))},\] where \(f\) has integral coefficients all bounded by \(C(n,d)\).

Proof that Lemma 3 implies Theorem 2. Any number field \(K\) of interest is isomorphic to one of the form \(\mathbb Q[x]/(f(x))\), where \(f\) has integral and bounded coefficients and is of degree \(n\). Theorem 1 follows because there are only finitely many such polynomials. \(\blacksquare\)

Instead of producing a small polynomial, it will turn out to be easier to produce small primitive elements.

Lemma 4. Fix positive integers \(d\) and \(n\). Then there is a constant \(C(n,d)\) such that any number field \(K\) with degree \(n\) and discriminant at most \(d\) admits some \(\alpha\in\mathcal O_K\) for which \(K=\mathbb Q(\alpha)\) and \(\left|\tau(\alpha)\right|\le C(n,d)\) for all embeddings \(\tau\).

Proof that Lemma 4 implies Lemma 3. Given our number field \(K\) as Lemma 3, we are granted some primitive element \(\alpha\) by Lemma 4. Then let \(f(x)\) be the minimal polynomial of \(\alpha\). Here are our checks on \(f\).

• Note \(K\cong\mathbb Q[x]/(f(x))\) because \(\alpha\) is a primitive element.
• Note \(f\) has integral coefficients because \(\alpha\in\mathcal O_K\).
• The coefficients of \(f\) are all bounded by \(\max\{1,C(n,d)\}^n\), which we see by expanding \[f(x)=\prod_{\tau}(x-\tau(\alpha)).\] Here, the product varies over complex embeddings of \(K\). \(\blacksquare\)

The condition that \(K=\mathbb Q(\alpha)\) is a bit awkward. Thankfully, it turns out that we can turn it into an inequality condition.

Lemma 5. Fix a number field \(K\), and choose an embedding \(\tau_0\colon K\hookrightarrow\mathbb C\). Choose a nonzero element \(\alpha\in\mathcal O_K\) for which \[\begin{cases}\left|\tau(\alpha)\right|<1 & \text{if }\tau\notin\{\tau_0,\overline\tau_0\}, \\\left|\operatorname{Re}\tau_0(\alpha)\right|<1 & \text{if }\tau_0\text{ is imaginary}.\end{cases}\] Then \(K=\mathbb Q(\alpha)\).

Proof. We claim that that \(\tau_0(\alpha)\ne\tau(\alpha)\) for each \(\tau\ne\tau_0\). Indeed, this implies that the embedding \(\tau_0|_{\mathbb Q(\alpha)}\) has a unique extension to \(K\), which implies that the separability degree of \(\mathbb Q(\alpha)\subseteq K\) is one, so \(K=\mathbb Q(\alpha)\).

It remains to show the claim. To begin, we note that \(\alpha\) being nonzero implies that \[\prod_\tau\left|\tau(\alpha)\right|=\left|\operatorname N_{K/\mathbb Q}(\alpha)\right|\ge1,\] so with \(\left|\tau(\alpha)\right|<1\) for all \(\tau\notin\{\tau_0,\overline\tau_0\}\), we conclude that \[\left|\tau_0(\alpha)\right|>1>\left|\tau(\alpha)\right|\] for each \(\tau\notin\{\tau_0,\overline\tau_0\}\).

Now, if \(\tau_0\) is a real embedding, then we are done. Otherwise, we see that \(\left|\operatorname{Re}\tau_0(\alpha)\right|<1\) while \(\left|\tau_0(\alpha)\right|>1\) forces \(\operatorname{Im}\tau_0(\alpha)\ne0\), so \(\tau_0(\alpha)\ne\overline\tau_0(\alpha)\) follows. \(\blacksquare\)

Thus, we see that we are on the hunt for a nonzero element \(\alpha\in\mathcal O_K\) such that the sizes of its various archimedean places are controlled. This is exactly what Minkowski theory is made for! We refer to this post for details, but we will recall pieces when we need them. Here is the main theorem of Minkowski theory.

Theorem 6 (Minkowski). Let \(V\) be a real inner product space of dimension \(n\). Let \(\Lambda\subseteq V\) be a lattice of rank \(n\), and let \(\Omega\subseteq V\) be a measurable, convex, and centrally symmetric subset. If \[\operatorname{vol}\Omega>2^n\operatorname{covol}\Lambda,\] then \(\Omega\cap\Lambda\) has a nonzero element.

Proof. Omitted. See this post. \(\blacksquare\)

We are now ready to complete the proof of Theorem 1 by proving Lemma 4.

Proof of Lemma 4. By Lemma 5, we are interested in finding a constant \(C(n,d)\) such that any number field \(K\) of interest admits an embedding \(\tau_0\colon K\hookrightarrow\mathbb C\) and some nonzero \(\alpha\in\mathcal O_K\) for which \[\begin{cases}\left|\tau(\alpha)\right|<1 & \text{if }\tau\notin\{\tau_0,\overline\tau_0\}, \\\left|\operatorname{Re}\tau_0(\alpha)\right|<1 & \text{if }\tau_0\text{ is imaginary}, \\\left|\tau_0(\alpha)\right|\lt C(n,d).\end{cases}\] We go ahead and choose some specific \(K\), and because it will not matter, we choose \(\tau_0\) arbitrarily.

For the proof, we apply Theorem 6.

• For our inner product space, we will use \(K_{\mathbb R}=K\otimes_{\mathbb Q}\mathbb R\), with its inner product given by the restriction of the canonical Hermitian inner product on \[K_{\mathbb C}=K\otimes_{\mathbb Q}\mathbb C=\prod_\tau\mathbb C.\]
• Our lattice will be the image of \(\mathcal O_K\) in \(K_{\mathbb R}\); as shown in this post, the covolume of this lattice is \(\sqrt{\left|\operatorname{disc}\mathcal O_K\right|}\).
• Lastly, we let \(\Omega\subseteq K_{\mathbb R}\) be the subset cut out by the \((x_\tau)\in K_{\mathbb R}\) for which \[\begin{cases}\left|x_\tau\right|<1 & \text{if }\tau\notin\{\tau_0,\overline\tau_0\}, \\\left|\operatorname{Re}x_{\tau_0}\right|<1 & \text{if }\tau_0\text{ is imaginary}, \\\left|x_{\tau_0}\right|\lt C(n,d),\end{cases}\] where \(C(n,d)\) is some constant depending only on \(n\) and \(d\) to be chosen shortly. Quickly, observe that \(\Omega\) is convex and centrally symmetric by the nature of the inequalities cutting it out.

Now, the volume of \(\Omega\) is some (positive) function of the signature of \(K\) multiplied by \(C(n,d)\), so we may select \(C(n,d)\) large enough so that \[\operatorname{vol}\Omega>2^n\sqrt{\left|\operatorname{disc}\mathcal O_K\right|}.\] Thus, Theorem 6 applies, and the produced nonzero \(\alpha\in\mathcal O_K\cap\Omega\) is exactly the required element. \(\blacksquare\)

Remark 7. Most proofs of Theorem 1 reduce to the case where \(K\) is totally imaginary because it removes the “casework on \(\tau_0\)” required in the final argument. In short, this reduction is done by passing from \(K\) to \(K(i)\); here are sketches of the checks on this mapping.

• This mapping has fibers of size bounded by a function depending only on \(d\) and \(n\); e.g., \(2^{(2n)!}\) will work.
• The degree only goes up by at most a factor of \(2\).
• Lastly, the discriminant increases to at most \(4^nd^2\).

Thus, if there are only finitely many totally imaginary number fields of bounded discriminant and degree, the same will be true of all number fields by looping over all fibers of the map \(K\mapsto K(i)\).