Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 101 / 3

Zariski lemma says that a field which is a finitely generated algebra over a field $k$ is a finite algebraic extension of $k$. The Strong Hilbert Nullstellensatz says that, for an ideal $I\subseteq k[T_1,\dots ,T_n]$ over an algebraically closed field,

Let the unique point of $V(\mathfrak{a})$ be $x=(x_1,\dots ,x_n)$ and letThe Nullstellensatz gives $\sqrt{\mathfrak{a}}=\mathfrak{m}$, so $\mathfrak{a}\subseteq \mathfrak{m}$. Each of the finitely many generators $u_i=T_i-x_i$ of $\mathfrak{m}$ has some power $u_i^{e_i}\in \mathfrak{a}$. Ifthen every monomial of total degree $r$ in the $u_i$ is divisible by one of the $u_i^{e_i}$. HenceThus the assertion is true; algebraically, the quotient defines a punctual scheme supported at $x$.

Choose finite generating sets $I=(g_1,\dots ,g_s)$ and $J=(f_1,\dots ,f_t)$, using the Hilbert basis theorem. The assumed inclusion says that each $f_j$ vanishes on $V_{\mathbb{C}}(I)$. By the Strong Hilbert Nullstellensatz, for every $j$ there is an exponent $N_j$ such that

The coefficients in an expression $f_j^{N_j}={\sum }_i{q}_{ij}{g}_i$ solve a finite system of linear equations with rational coefficients. Since it has a complex solution, Gaussian elimination gives a rational solution. Clearing the finitely many denominators produces a nonzero integer $D$ such thatfor every $j$.

For any prime $p\nmid D$, reduce these identities modulo $p$. At a common zero of ${\pi }_p(I)$ in the algebraic closure ${\overline{\mathbb{F}}}_p$, they give ${\pi }_p(f_j{)}^{N_j}=0$, hence ${\pi }_p(f_j)=0$, for all $j$. Thereforefor every prime except the finitely many divisors of $D$. This is the spreading out of an affine zero-set inclusion.

An $A$-module $M$ is a flat module when the tensor functor $-{\otimes }_{A}M$ preserves injections, equivalently when it is exact.

Suppose first that $M$ is flat. For any nonzero $a\in A$, tensor the injection $A\stackrel{a}{\to }A$ with $M$. The resulting map $M\stackrel{a}{\to }M$ is injective, so $am=0$ implies $m=0$. Thus $M$ is a torsion-free module.

Conversely, suppose $M$ is torsion-free over the principal ideal domain $A$. Every finitely generated submodule of $M$ is a finitely generated torsion-free module over a PID, hence a finite free module and therefore flat. The module $M$ is the filtered colimit of these submodules. Tensor products commute with filtered colimits, and filtered colimits of modules preserve exact sequences, so $M$ is flat. This proves that a torsion-free module over a principal ideal domain is flat.