**Noether-Deuring theorem** (not in the strongest form, but in the one I usually need)**:**

Let $L\diagup K$ be a field extension. Let $A$ be a $K$-algebra which is finite-dimensional as a vector space over $K$. Let $U$ and $V$ be two left $A$-modules which are finite-dimensional as vector spaces over $K$. If $U\otimes_K L \cong V\otimes_K L$ as $A\otimes_K L$-modules, then $U\cong V$ as $A$-modules.

This was nicely proven by Torsten Ekedahl in Hilbert 90 for algebras . The same proof shows something more general: If $U\otimes_K L$ is isomorphic to a direct summand of $V\otimes_K L$ as $A\otimes_K L$-modules, then $U$ is isomorphic to a direct summand of $V$ as $A$-modules. (Actually I have seen this also being called the Noether-Deuring theorem. Yes, this is stronger than the original Noether-Deuring theorem.)

But what about this: If there exists an $A\otimes_K L$-linear map $U\otimes_K L\to V\otimes_K L$ of some $L$-rank $r$, must it follow that there exists an $A$-linear map $U\to V$ of $K$-rank at least $r$ ? Again, the only hard case is when $K$ and $L$ are both finite fields, but Krull-Remak-Schmidt doesn't come to our help now...

Of course, the interesting cases are when $r=\dim_K U$ and when $r=\dim_K V$, so we are talking about the existence of injective resp. surjective maps.

**Remarks:**

**(1)** The condition that $A$ be finite-dimensional as a vector space over $K$ can be removed: we can always replace $A$ by the image of $A$ in $\mathrm{End}_K U \times \mathrm{End}_K V$, and that image is finite-dimensional if $U$ and $V$ are so.

**(2)** When $A$ is the polynomial $K$-algebra $K\left[X\right]$, then this question generalizes the conjugacy rank question.

at least$r$?" Otherwise, you could take $U=V$ an irreducible, but not absolutely irreducible, $A$-module. Then`$\operatorname{End}_A(U)$`

is a division ring and all nonzero maps have the same rank,`$\dim_K U$`

, while`$\operatorname{End}_{A\otimes_K L}(U\otimes_K L)$`

contains maps with smaller rank. – Frieder Ladisch Feb 19 '11 at 16:30