Is every nonzero integer the discriminant of some algebraic number field?

I do know that if $m \equiv 1 \pmod 4$ and squarefree, it is probably the discriminant of $\mathbb{Q}[\sqrt{m}]$, and I also know some negative multiples of 27 are discriminants of cubic number fields.

But are there integers that are not the discriminant of any number field? If so, are they listed in Sloane’s?

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The answer to the title question is no and the answer to the body question is yes.

First, Stickelberger’s theorem asserts that the discriminant $\Delta_K$ of a number field $K$ is congruent to $0, 1 \bmod 4$. So that already rules out half of the possibilities.

Second, Minkowski’s bound implies that if $K$ has degree $n$ then

$$\sqrt{|\Delta_K|} \ge \left( \frac{\pi}{4} \right)^{n/2} \frac{n^n}{n!}.$$

The RHS is always strictly bigger than $1$, so the discriminant can also never be equal to $1$; that is, $\mathbb{Q}$ has no unramified extensions. Moreover it follows that number fields of a given discriminant cannot have arbitrarily high degree, and in fact there are only finitely many number fields of a given discriminant, so for particular small discriminants one can rule them out via casework.

To start, let’s record the value of the Minkowski bound for some small values of $n$.

  • For $n = 2$ it is about $1.57$, so a number field of degree at least $2$ has discriminant of absolute value at least $3$.
  • For $n = 3$ it is about $3.13$, so a number field of degree at least $3$ has discriminant of absolute value at least $10$.
  • For $n = 4$ it is about $6.58$, so a number field of degree at least $4$ has discriminant of absolute value at least $44$.

Let’s also recall that for a squarefree integer $d$ the discriminant of $\mathbb{Q}(\sqrt{d})$ is $d$ if $d \equiv 1 \bmod 4$ and $4d$ otherwise. Now let’s go through the smallest few discriminants in order, skipping the ones that are impossible by Stickelberger to see what Minkowski has to say about them.

  • $1$: impossible by Minkowski.
  • $-3$: realized uniquely by $\mathbb{Q}(\sqrt{-3})$.
  • $4$: impossible by Minkowski (can only be realized by a quadratic field and isn’t).
  • $-4$: realized uniquely by $\mathbb{Q}(i)$.
  • $5$: realized uniquely by $\mathbb{Q}(\sqrt{5})$.
  • $-7$: realized uniquely by $\mathbb{Q}(\sqrt{-7})$.
  • $8$: realized uniquely by $\mathbb{Q}(\sqrt{2})$.
  • $-8$: realized uniquely by $\mathbb{Q}(\sqrt{-2})$.
  • $9$: impossible by Minkowski (can only be realized by a quadratic field and isn’t).
  • $-11$: realized by $\mathbb{Q}(\sqrt{-11})$.
  • $12$: realized by $\mathbb{Q}(\sqrt{3})$.

$-12$ is the first discriminant whose status I can’t determine from just Stickelberger and Minkowski. If it occurs as a discriminant it must be the discriminant of a cubic field. I think it’s known that in fact the smallest possible discriminant (in absolute value) of a cubic field is $-23$, realized by $\mathbb{Q}(x)/(x^3 – x – 1)$, but I don’t know how to prove this.

In any case, here’s a discriminant I can rule out using an additional technique: $25$ is congruent to $1 \bmod 4$ and, by the Minkowski bound, could only be the discriminant of a cubic field. However, I claim it isn’t the discriminant of a cubic field, and so can’t be a discriminant at all. The reason is that it’s a square, which means that the corresponding cubic field is Galois with Galois group $A_3 \cong C_3$. By the Kronecker-Weber theorem it must therefore be a subfield of the cyclotomic integers $\mathbb{Q}(\zeta_n)$, and it’s known that we should in fact be able to take $n = 5$. However, $\mathbb{Q}(\zeta_5)$ has no cubic subfields since it has degree $4$.