The Noether-Deuring Theorem

I have to solve the following exercise taken from the book "Introduction to Representation Theory" by P. Etingof, O. Golberg, S. Hensel, T. Liu, A. Schwendner, D. Vaintrob, E. Yudovina and S. Gerovitch:

Problem 3.8.4

1. Let $V,W$ be finite dimensional representations of an algebra $A$ over a not necessarily closed field $k$. Let $k\hookrightarrow l$ be a field extension and assume that $V\otimes_{k}l\cong W\otimes_{k}l$ as modules over the $l$-algebra $A\otimes_{k}l$. Then $V\cong W$ as $A$-modules.
2. (The Noether-Deuring theorem) In the setting of (1), suppose that $W\otimes_{k}l\cong Y\oplus(V\otimes_{k}l)$ for some $A\otimes_{k}l$-module $Y$. Then $V$ is a direct summand in $W$, i.e. $W\cong V\oplus Z$ (I assume: as $A$-modules, what else? - this is not made more explicit).

The exercise is not part of the online version, unless I have overlooked it.

What have I achieved so far? The book gave the following hint on (1):

Reduce to the case of finitely generated, then finite extension, of some degree $n$. Then regard $V\otimes_{k}l$ and $W\otimes_{k}l$ as $A$-modules, and show that they are isomorphic to $V^{n}$ and $W^{n}$ respectively. Deduce that $V^{n}\cong W^{n}$, and use the Krull-Schmidt theorem (valid over any field by Problem 3.8.3) to deduce that $V\cong W$.

I have managed to prove it - I think - for the case $[l:k]=n$ and that if the statement holds for all finitely generated extensions then it holds for all extensions. The first part was proven by giving an explicit isomorphism of $A$-modules between $V^{n}$ and $W^{n}$. This was deduced from: $$M\otimes_{k}\bigg(\bigoplus_{i\in I}N_{i}\bigg)\cong\bigoplus_{i\in I}\big(M\otimes_{k}N_{i}\big)\quad\text{as $k$ vector spaces}$$ whenever $M,\{N_{i};i\in I\}$ are vector spaces over $k$. For the second part I had to rely on a (in retrospective obvious) hint of my professor: let $\Phi:V\otimes_{k}l\to W\otimes_{k}l$ be the $l$ vector spave isomorphism induced by the module isomorphism assumed, we can pick bases $\{v_{i}\}$ and $\{w_{j}\}$ of $V,W$ respectively to obtain bases of $V\otimes_{k}l$ and $W\otimes_{k}l$ by looking at the simple tensors $\{v_{i}\otimes 1\}$ and $\{w_{j}\otimes 1\}$. Let $M$ be the matrix representation of the isomorphism with respect to these bases and let $m$ be the field extension of $k$ generated by the entries of $M$. Then the restriction of $\Phi$ is an isomorphism $V\otimes_{k}m\to W\otimes_{k}m$ and thus if the statement holds for finitely generated extensions, then it holds for arbitrary extensions. So I end up with the following:

Question 1: Given that $k\hookrightarrow l$ is finitely generated and for all extensions of finite degree the statement (1) holds, how can I deduce that it holds for $k\hookrightarrow l$.

For part (2): here I can tell you only what I have tried. First of all I used the Krull-Schmidt theorem and the fact that tensor products distribute over direct products to write $W\otimes_{k}l$ in two different ways: $$W\otimes_{k}l\cong\bigoplus_{i=1}^{s}n_{i}\hat{W_{i}}\text{ and }W\otimes_{k}l\cong\bigoplus_{j=1}^{t}m_{j}(W_{j}\otimes_{k}l)$$ where the $\hat{W_{i}}$ were irreducible summands of the $A\otimes_{k}l$-module $W\otimes_{k}l$ and the $W_{j}$ were irreducible representations of the $A$-module $W$. Then I wanted to show that in fact the following holds:

$W$ is a finite dimensional, indecomposable representation of the $l$-algebra $A\otimes_{k}l$ if and only if there exists a finite dimensional, indecomposable representation $V$ of the $k$-algebra $A$ such that $W\cong V\otimes_{k}l$.

From this it would have followed that the summands agree up to permutation. Unfortunately this is not true if $k$ is not algebraically closed. I also considered decomposing $W\otimes_{k}l$ as a finite dimensional representation of $A$ but this, I assume, is not the case as $k\hookrightarrow l$ is not necessarily of finite degree. The next thing to look at was to take the quotient with $Y$ in order to actually end up with an isomorphism as (1). So there is the next question:

Question 2: Let $W,V$ be finite-dimensional $A$-modules, where $A$ is a $k$-algebra and let $k\hookrightarrow l$ be a field extension. Let $\Phi:W\otimes_{k}l\twoheadrightarrow V\otimes_{k}l$ be a module homomorphism. Can we deduce that $\operatorname{ker}\Phi\cong U\otimes_{k}l$ where $U$ is a submodule of $W$?

And finally:

Question 3: Am I on the wrong track? Have I overseen something very simple? Am I a complete moron?

Of course hints are very appreciated.

Solution 1:

My answer here will only answer question (1) above. The following is sometimes known as Zariski's Lemma:

Corollary 5.24 (Atiyah - Macdonald): Let $k$ be a field and $B$ a finitely generated $k$ algebra. If $B$ is a field then it is a finite (and hence algebraic) extension of $k$.

Proof: Our proof is different to that given in Atiyah - Macdonald. By Noether Normalization there exists a non-negative integer $n$ and algebraically independent elements $y_1,\ldots,y_n \in B$ such that $B$ is an integral extension of the polynomial algebra $k[y_1,\ldots,y_n]$. By proposition 5.7 of Atiyah - Macdonald we deduce that $k[y_1,\ldots,y_n]$ must be a field. However this can happen iff $n = 0$ which is to say $B$ is a finitely generated $k$ - algebra that is also integral (or algebraic) over $k$. Then

$$B = k(\alpha_1,\ldots,\alpha_n)$$

with each $\alpha_i$ of degree $k_i$ say. Consequently $\dim_k B \leq \prod k_i$; in particular $B$ is a finite extension of $k$.

Solution 2:

to Question 3. Let me sketch a solution to the exercise which circumvents your troubles with infinite field extensions. It is not going to be a nice argument, but then again Noether-Deuring is a technical lemma to begin with, and IMHO should be avoided whenever possible.

Part 1 of the exercise easily follows from Part 2, because Part 2 shows that each of $V$ and $W$ is a direct summand of the other, and therefore $V\cong W$ (because no miracles happen in finite dimensions). So I'll only answer Part 2.

Case 1: The field extension $l/k$ is finite. Let $n=\left[l:k\right]$. Then, $V\otimes_k l\cong V^n$ and $W\otimes_k l\cong W^n$ as $A$-modules, as you proved. Thus, as $A$-modules, $W^n \cong W\otimes_k l \cong Y \oplus \left(V\otimes_k l\right) \cong Y\oplus V^n$.

Now, let us write each of $V$, $W$ and $Y$ as a direct sum of indecomposable $A$-modules. For each indecomposable $A$-module $I$, let $r_I\left(V\right)$ be the number of times $I$ appears in the decomposition of $V$, let $r_I\left(W\right)$ be the number of times $I$ appears in the decomposition of $W$, and let $r_I\left(Y\right)$ be the number of times $I$ appears in the decomposition of $Y$. These three numbers are well-defined because they don't depend on the choice of decompositions (since the Krull-Schmidt theorem says that in any decomposition of a finite-dimensional module into indecomposables, the multiset of the isomorphism classes of the indecomposables involved doesn't depend on the choice of the decomposition). If we can prove that $r_I\left(V\right) \leq r_I\left(W\right)$ for every indecomposable $I$, then we conclude that $V$ is a direct addend of $W$ as $A$-module (because the indecomposables in the decomposition of $V$ all appear in the decomposition of $W$ at least as often), so we are done (in Case 1, at least).

To prove that $r_I\left(V\right) \leq r_I\left(W\right)$ for every $I$, we notice that every $I$ satisfies

$n\cdot r_I\left(W\right) = r_I\left(W^n\right) = r_I \left(Y \oplus V^n\right)$ (since $W^n\cong Y\oplus V^n$) $= r_I\left(Y\right) + r_I\left(V^n\right) \geq r_I\left(V^n\right) = nr_I\left(V\right)$

and thus $ r_I\left(W\right) \geq r_I\left(V\right)$. This solves Part 2 of the exercise in Case 2.

Case 2: The general case.

We are looking for two $A$-module homomorphisms $i:V\to W$ and $p:W\to V$ satisfying $p\circ i=\mathrm{id}$. (In fact, if we can find such $i$ and $p$, then it becomes clear that $V$ is a direct addend of $W$ as $A$-module.) Since $V$ and $W$ are finite-dimensional, we can identify the $k$-vector spaces $V$ and $W$ with $k^n$ and $k^m$ for some nonnegative integers $n$ and $m$, and then $k$-linear maps $V\to W$ can be identified with $m\times n$-matrices over $k$, while $k$-linear maps $W\to V$ can be identified with $n\times m$-matrices over $k$. The conditions that the two maps $i$ and $p$ be $A$-module homomorphisms translate into a system of linear constraints on the entries of these two matrices; and of course, we can WLOG assume that there are only finitely many constraints (because there are only finitely many entries, so if there are infinitely many constraints we can throw away almost all of them for redundancy). The equation $p\circ i=\mathrm{id}$ translates into a system of polynomial equations for the entries of these two matrices. So we are looking for $mn+nm=2nm$ elements of $k$ which satisfy certain linear and certain polynomial equations. Since linear equations are also polynomial, what we are looking for is thus simply a solution for a fixed system of finitely many polynomial equations over $k$ in finitely many indeterminates.

However, because we know that $V\otimes_k l$ is a direct addend of the $A\otimes_k l$-module $W\otimes_k l$, we know that such a solution can be found over $l$ (because tensoring with $l$ means extending scalars to $l$).

Now, a general fact from algebraic geometry says that

(*) if $l/k$ is a field extension, and some given system of finitely many polynomial equations over $k$ in finitely many indeterminates has a solution over $l$, then there exists a finite field extension $m$ of $k$ such that this system has a solution over $m$.

Applying this to our system, we get a finite field extension $m$ of $k$ such that our system has a solution over $m$. This solution translates into a witness that $V\otimes_k m$ is a direct summand of $W\otimes_k m$ as an $A\otimes_k m$-module. But since $m/k$ is finite, we can now apply Case 1 to $m$ in lieu of $l$, and conclude that $V$ is a direct summand of $W$ as an $A$-module. This solves Part 2 in the general case.

to Question 2. I am very interested in this myself; it is similar to MathOverflow question #48909 which is still unanswered.