Let $A$ be a finite dimensional associative algebra (with unity) over a finite field $F$. Let $L$ be a field extension of $F$. Suppose that after extending scalars to $L$, two elements $a,b$ of $A$ are conjugate, i.e. there is an invertible element $u \in A\otimes\_F L$ such that $a=ubu^{-1}$ . Does it follow that there is an invertible element $s \in A$ such that $a=sbs^{-1}$? 



Conjugation in associative algebras over finite fields
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The finiteness of $F$ is not needed. I have learnt the idea of the following proof from [Torsten Ekedahl](https://mathoverflow.net/questions/28469/hilbert-90-for-algebras).


In the following, all algebras are associative and with unity. The word "finite-dimensional" shall always mean "finite-dimensional as an $F$-vector space".


**Theorem 1.** Let $B$ be a finite-dimensional $F$-algebra. Let $M$ and $N$ be two finite-dimensional $B$-modules. Assume that $M\otimes\_F L \cong N\otimes\_F L$ as $B\otimes\_F L$-modules. Then, $M \cong N$ as $B$-modules.


Theorem 1 is the Noether-Deuring theorem (19.25 in T. Y. Lam, *A First Course in Noncommutative Rings*, 2nd edition 2001).


**Corollary 2.** Let $B$ be an $F$-algebra. Let $M$ and $N$ be two finite-dimensional $B$-modules. Assume that $M\otimes\_F L \cong N\otimes\_F L$ as $B\otimes\_F L$-modules. Then, $M \cong N$ as $B$-modules.


*Proof of Corollary 2.* Corollary 2 differs from Theorem 1 only in the lack of a finite-dimensionality requirement on $B$. We shall prove Corollary 2 by reducing it to Theorem 1.


The action of $B$ on the $B$-module $M \oplus N$ gives rise to an $F$-algebra homomorphism $B \to \End\_F \left(M\oplus N\right)$. The kernel of this homomorphism is an ideal $I$ of $B$, and the $F$-algebra $B/I$ is finite-dimensional (since $B/I$ is isomorphic to the image of this homomorphism, and said image is clearly finite-dimensional). Every element of $I$ acts as $0$ on $M \oplus N$, and thus also on $M$ and $N$; therefore, both $B$-modules $M$ and $N$ can be regarded as $B/I$-modules. Moreover, the $B\otimes\_F L$-module isomorphism $M\otimes\_F L \cong N\otimes\_F L$ (which exists by assumption) can be regarded as a $\left(B/I\right)\otimes\_F L$-module isomorphism, whereas the $B$-module isomorphism $M \cong N$ that we are looking for boils down to a $B/I$-module isomorphism. Thus, Corollary 2 follows from Theorem 1 (applied to $B/I$ instead of $B$).


**Corollary 3.** Let $B$ and $C$ be two $F$-algebras. Let $M$ and $N$ be two finite-dimensional $\left(B,C\right)$-bimodules. Assume that $M\otimes\_F L \cong N\otimes\_F L$ as $\left(B\otimes\_F L,C\otimes\_F L\right)$-bimodules. Then, $M \cong N$ as $\left(B,C\right)$-bimodules.


*Proof of Corollary 3.* Let $C^\op$ be the opposite algebra of $C$. (This is the $F$-algebra which is defined as a "copy of $C$ with the multiplication turned around"; more formally, it is the $F$-vector space $C$ endowed with the multiplication $\*$ defined by $a\*b = ba$.) It is known that $\left(B,C\right)$-bimodules are "the same as" left $B\otimes\_F C^\op$-modules. (More precisely, any $\left(B,C\right)$-bimodule structure on an $F$-vector space can be transformed into a left $B\otimes\_F C^\op$-module structure on the same vector space by first turning the right $C$-module structure into a left $C^\op$-module structure, and then combining the left $B$-module structure with the left $C^\op$-module structure into a left $B\otimes\_F C^\op$-module structure because they commute.) Similarly, $\left(B\otimes\_F L,C\otimes\_F L\right)$-bimodules are "the same as" left $\left(B\otimes\_F L\right) \otimes\_L \left(C\otimes\_F L\right)^\op$-modules, which, conveniently, are the same as left $\left(B\otimes\_F C^\op\right)\otimes\_F L$-modules. Thus, Corollary 3 follows from Corollary 2 (applied to $B\otimes\_F C^\op$ instead of $B$).


Let us now return to the problem.


For every $u \in A$, we let $L\_u$ denote the map $A \to A, \ x \mapsto ux$ (known as "left multiplication by $u$"). The map $L\_u$ is a homomorphism of right $A$-modules.


Now, we can define an action of the $F$-algebra $F\left[X\right]$ on $A$ by letting $X$ act as $L\_a$. Combined with the canonical right $A$-module structure on $A$ (by right multiplication), this gives rise to an $\left(F\left[X\right],A\right)$-bimodule structure on $A$. Let us denote this $\left(F\left[X\right],A\right)$-bimodule $A$ by $A\_a$.


Similarly, we can define an $\left(F\left[X\right],A\right)$-bimodule $A\_b$, which differs from $A\_a$ in that $X$ acts as $L\_b$ rather than $L\_a$.


It is easy to see that the right $A$-bimodule homomorphisms $A\_b \to A\_a$ are precisely the maps $L\_u$ for $u \in A$. Thus, the $\left(F\left[X\right],A\right)$-bimodule homomorphisms $A\_b \to A\_a$ are precisely the maps $L\_u$ for $u \in A$ which satisfy $ub = au$. Thus, the $\left(F\left[X\right],A\right)$-bimodule isomorphisms $A\_b \to A\_a$ are precisely the maps $L\_u$ for **invertible** $u \in A$ which satisfy $ub = au$. Hence, an invertible $u \in A$ which satisfies $ub = au$ exists if and only if $A\_a \cong A\_b$ as $\left(F\left[X\right],A\right)$-bimodules. In other words,


**(1)** the elements $a$ and $b$ of $A$ are conjugate in $A$ if and only if $A\_a \cong A\_b$ as $\left(F\left[X\right],A\right)$-bimodules.


A similar argument (applied to the elements $a\otimes\_F 1$ and $b\otimes\_F 1$ of $A\otimes\_F L$ over the field $L$ instead of the elements $a$ and $b$ of $A$ over the field $F$) shows that the elements $a\otimes\_F 1$ and $b\otimes\_F 1$ of $A \otimes\_F L$ are conjugate in $A \otimes\_F L$ if and only if $\left(A\otimes\_F L\right)\_{a\otimes\_F 1} \cong \left(A\otimes\_F L\right)\_{b\otimes\_F 1}$ as $\left(L\left[X\right],A\otimes\_F L\right)$-bimodules (where $\left(A\otimes\_F L\right)\_{a\otimes\_F 1}$ and $\left(A\otimes\_F L\right)\_{b\otimes\_F 1}$ are defined similarly to $A\_a$ and $A\_b$). Since we know that the elements $a\otimes\_F 1$ and $b\otimes\_F 1$ of $A \otimes\_F L$ are conjugate in $A \otimes\_F L$, we thus conclude that $\left(A\otimes\_F L\right)\_{a\otimes\_F 1} \cong \left(A\otimes\_F L\right)\_{b\otimes\_F 1}$ as $\left(L\left[X\right],A\otimes\_F L\right)$-bimodules. In other words, $A\_a \otimes\_F L \cong A\_b \otimes\_F L$ as $\left(F\left[X\right]\otimes\_F L, A\otimes\_F L\right)$-bimodules. Thus, Corollary 3 (applied to $B = F\left[X\right]$, $C = A$, $M = A\_a$ and $N = A\_b$) shows that $A\_a \cong A\_b$ as $\left(F\left[X\right], A\right)$-bimodules. By **(1)**, this shows that the elements $a$ and $b$ of $A$ are conjugate in $A$. Qed.


