= Solution
Let $d_K$ denote the <field discriminant>, and put $n=[K:\mathbb Q]\geq2$. Let $r_1$ be the number of real <field embeddings> and $r_2$ the number of conjugate pairs of nonreal <field embeddings>, so $n=r_1+2r_2$. The <Minkowski bound for ideal classes> provides a nonzero <integral ideal> $A$ satisfying
$$
1\leq N(A)\leq\left(\frac4\pi\right)^{r_2}\frac{n!}{n^n}\sqrt{|d_K|}.
$$
Rearranging gives the <Minkowski lower bound for a number-field discriminant>
$$
|d_K|\geq\left(\frac\pi4\right)^{2r_2}\left(\frac{n^n}{n!}\right)^2.
$$
Because $2r_2\leq n$ and $\pi/4<1$, this is at least
$$
\left(\frac\pi4\right)^n\left(\frac{n^n}{n!}\right)^2.
$$
Apply the <arithmetic-geometric mean inequality> to $1,2,\ldots,n$:
$$
n!\leq\left(\frac{n+1}{2}\right)^n.
$$
It follows that
$$
|d_K|\geq\left(\frac{\pi n^2}{(n+1)^2}\right)^n
\geq\left(\frac{4\pi}{9}\right)^n>1,
$$
where $n/(n+1)\geq2/3$ for $n\geq2$ was used. Thus
$$
\boxed{|d_K|>1}.
$$
The source writes $|\Delta(K)^2|>1$. If $\Delta(K)$ denotes the embedding <determinant>, its square is $d_K$; if it denotes the <field discriminant>, the displayed result also implies the stated inequality. The conclusion therefore does not depend on that notational convention.
Back to article page
