Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 15 / 3 / Solution

The usual field norm $N_{L/K}$ is defined for a finite field extension. The printed hypothesis “algebraic” alone allows an infinite extension, for which this determinant norm is not defined. We therefore assume $[L:K]<\infty$. For an arbitrary algebraic extension the meaningful elementwise statement is instead $N_{K(b)/K}(b)\in A$, and we prove this as well.

First a unique factorization domain is an integrally closed domain. Indeed, write an integral fraction as $u/v$ with relatively prime numerator and denominator. A monic integral equation, after multiplication by $v^r$, shows that $v$ divides $u^r$. Unique factorization forces $v$ to be a unit.

Let $m_b(T)=T^r+a_{r-1}{T}^{r-1}+\cdots +a_0\in K[T]$ be the monic minimal polynomial of $b$. Because $b$ is an integral element over $A$, it satisfies a monic polynomial $f\in A[T]$, and $m_b$ divides $f$ in $K[T]$. Every root of $m_b$ in an algebraic closure is therefore integral over $A$. This assertion includes repeated roots when the extension is inseparable. The coefficients of $m_b$ are elementary symmetric expressions in these roots. The integral elements form a ring: finitely many integral elements generate a finite $A$-module algebra, and the determinant trick shows every element of that algebra is integral. Thus all $a_i$ are integral over $A$. Since $a_i\in K$ and $A$ is integrally closed, $a_i\in A$.

Multiplication by $b$ on $K(b)$, in the basis $1,b,\dots ,b^{r-1}$, has the companion matrix of $m_b$, whose determinant is $(-1{)}^r{a}_0$. If $e=[L:K(b)]$, take a $K(b)$-basis of $L$. Multiplication by $b$ preserves each summand in the resulting direct sum of $e$ copies of $K(b)$, so its determinant over $K$ isIn particular $N_{K(b)/K}(b)=(-1{)}^r{a}_0\in A$. The argument requires no separability; in fact it holds for every integrally closed domain, not just a unique factorization domain.

For the connection with dimension, let $B$ be the coordinate ring of an irreducible affine variety. By Noether normalization there is a finite integral inclusion $A=k[t_1,\dots ,t_d]\subset B$, with $d=\dim B$. Their fraction fields form a finite extension. For any $0\ne b\in B$, its minimal-polynomial constant term $a_0$ is nonzero and lies in $A$. Its equation givesThus a nonzero function on the normalized variety produces a nonzero function on the base that vanishes wherever $b$ vanishes. Up to sign, the field norm is a power of this $a_0$, so it too belongs to $bB\cap A$.

Under the finite morphism to ${\mathbb{A}}^d$, the image of $V_B(b)$ is therefore contained in the proper Zariski-closed set $V_A(a_0)$ of dimension at most $d-1$. Finite morphisms preserve the dimension of a closed subvariety and its image. Applying this to a nonzero element of the ideal of any proper closed subvariety proves proper closed subvarieties have strictly smaller dimension. This is the useful role of the norm argument: nonzero equations upstairs give nonzero equations downstairs under a finite normalization.