Can someone explain me what’s going on in the Artin-Tate lemma

abstract-algebracommutative-algebra

Let $P\subset Q\subset R$ be rings with $P$ noetherian and $R$ simultaneously finitely generated as a $P$-Algebra and finitely generated as a $Q$ module. Then $Q$ is finitely generated as a $P$ algebra.

That's our version of the Artin-Tate lemma. But somehow I can't think of an explicit example. Could you please help me?

If I remember correctly we say that an $R$-module $M$ is an $R$-algebra if we have a multiplication map $$\cdot~~:M\times M\rightarrow M$$ such that $(x+y)m=xm+ym$, $x(y+m)=xy+xm$, $\lambda(xy)=(\lambda x)y=x(\lambda y)$ for all $x,y,m\in M$ and $\lambda \in R$.

Thanks for your help.

Best Answer

I will reproduce Atiyah and MacDonald's proof of this result, but try to motivate the steps.

  1. Note that we'll be done if we can produce a subring $Q_0$ between $P$ and $Q$ $$ P\subseteq Q_0\subseteq Q $$ such that $Q$ is f.g. (finitely generated) as a $Q_0$-module and $Q_0$ is f.g. as a $P$-algebra. (Why?)

  2. Now, on to constructing $Q_0$:

    • $R$ is given to be f.g. as a $Q$-module. Thus we may write $R = \operatorname{span}_Q\{ r_1, \ldots, r_m \}$. We try to refine the scalars of $Q$ to come from a subring as follows.
    • Since $R$ is also f.g. as $P$-algebra, we can write $R = P[\tilde r_1, \ldots, \tilde r_n]$. Now, we express $\tilde r_j = \sum_{i = 1}^m q^{(j)}_i r_i$. Substituting these in a typical monomial of $P[\tilde r_1, \ldots, \tilde r_n]$ will give (after expanding it out) a sum whose terms will look like so: $$ \text{(an element of $P$) (product of $q^{(j)}_i$'s) (product of $r_i$'s)} $$
    • We further write $r_j r_k = \sum_{i = 1}^m q^{(j, k)}_i r_i$ (remember, $R = \operatorname{span}_Q\{ r_1, \ldots, r_m \}$) so that the last term in the above product can be decomposed, after "repeated substitutions", into the following sum: $$ \sum_{i = 1}^m \text{(product of $q^{(j, k)}_{i'}$'s)} \; r_i $$
    • Thus, we have been successful in showing that $P[\tilde r_1, \ldots, \tilde r_n]\subseteq \operatorname{span}_{Q_0}\{ r_1, \ldots, r_m \}$ where $Q_0$ denotes $P$ adjoined with $q^{(j)}_i$'s and $q^{(j, k)}_i$'s. It thus follows that $R = \operatorname{span}_{Q_0}\{ r_1, \ldots, r_m \}$. Note that $Q_0\subseteq Q$, and $Q_0$ is f.g. as a $P$-algebra (why?).

  3. Now it suffices to show that $Q$ is f.g. as a $Q_0$-module. Stare at this: $$ P\subseteq Q_0\subseteq Q\subseteq R = \operatorname{span}_{Q_0}\{r_1, \ldots, r_m\} $$ while reading the following. Note that $Q_0$ is Noetherian being f.g. as an algebra over the Noetherian $P$ (finally using Noetherian-ness of $P$). Also, $R$ being f.g. as $Q_0$-module, is Noetherian as a $Q_0$-module. Finally, since $Q$ is a $Q_0$-submodule of the Noetherian module $R$, we have that $Q$ is a Noetherian $Q_0$-module and hence f.g. over $Q_0$.

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