**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](https://mathoverflow.net/questions/28469/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](https://mathoverflow.net/questions/9162/conjugacy-rank-of-two-matrices-over-field-extension).



Noether-Deuring for injections and surjections?
Hendrik W. Lenstra has just informed me that the answer to my question is "no", both in the case of $r = \dim U$ (so we are looking at injective $A$-linear maps) and in the case of $r = \dim V$ (so we are looking at surjective $A$-linear maps). Counterexamples can be found in Exercise 21 (c) and Exercise 22 of his notes [*The unit theorem for finite-dimensional algebras* (arXiv:1703.07273v1)](https://arxiv.org/abs/1703.07273v1).


Thanks, Hendrik, for resolving this nagging question!


