Conjugation in associative algebras over finite fields 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 $a=sbs^{-1}$? 
 A: This is true, and it follows from Serge Lang's proof of Serre's "Conjecture I" for finite fields.  The analogous result is true for any perfect field $F$ of cohomological dimension $1$ by Steinberg's solution of the general conjecture.
Denote by $E_A$ the $F$-group scheme that is an affine space of dimension $\text{dim}_F(A)$ and such that $E_A(R)$ is identified with $A\otimes_F R$ for every commutative, unital $F$-algebra $R$.  The subfunctor of the Yoneda functor of $E_A$ parameterizing invertible elements $u\in A\otimes_F R$ is represented by a Zariski open subset $\textbf{GL}_A$ of $E_A$, namely the inverse image of the Zariski open subset $\text{Aut}_F(E_A)\subset \text{End}_F(E_A)$ under the $F$-linear morphisms of affine $F$-spaces, $$\alpha: E_A \to \text{End}_F(E_A), \ u \mapsto (v\mapsto u\cdot v).$$
For a fixed element $b$ of $A$, consider the locally closed subscheme $C_b$ of $\textbf{GL}_A$ such that for every commutative, unital $F$-algebra $R$, $C_b(R)\subset \textbf{GL}_A(R)$ is the set of invertible elements $u\in A\otimes_F R$ satisfying $ub = bu$.  The point is, there is a linear subspace $L_b \subset E_A$ parameterizing $u$ (invertible or not) such that $ub=bu$.  Thus the intersection $C_b = \textbf{GL}_A\cap L_b$ is a nonempty Zariski open subset of an $F$-affine space, and thus $C_b$ is geometrically integral.  By construction, $C_b$ is a linear algebraic $F$-group.  Thus, $C_b$ is a smooth connected affine $F$-group.
For an element $a$ of $A$, there is a closed subscheme $T_{b,a}\subset \textbf{GL}_A$ parameterizing $v$ such that $bv = va$.  As above, since this is a linear condition, it is a Zariski closed subscheme of $\textbf{GL}_A$.  Assuming this is not the empty scheme, $T_{a,b}$ is a torsor for the smooth connected affine $F$-group $C_b$ via the action, $$ \mu: C_b \times_F T_{b,a} \to T_{b,a}, \ \ (u,v) \mapsto u\cdot v.$$  For $F$ a finite field, Serge Lang proved that every torsor over a finite field $F$ for every smooth connected affine $F$-group admits an $F$-point.
MR0086367 (19,174a) 
Lang, Serge
Algebraic groups over finite fields. 
Amer. J. Math. 78 (1956), 555–563. 
http://www.jstor.org/stable/2372673?origin=crossref&seq=1#page_scan_tab_contents
For $F$ an arbitrary perfect field of cohomological dimension $1$, this was proved by Robert Steinberg.
MR0180554 (31 #4788) 
Steinberg, Robert 
Regular elements of semisimple algebraic groups.  
Inst. Hautes Études Sci. Publ. Math. No. 25 1965 49–80. 
http://www.numdam.org/item?id=PMIHES_1965__25__49_0
This is also Appendix 1 of Serre's book on Galois cohomology.
Thus the $C_b$-torsor $T_{b,a}$ has an $F$-point $v \in T_{b,a}(F) \subset A$ such that $bv$ equals $va$.
Edit. Darij Grinberg is correct that this result holds over all fields.  For every infinite field $F$, the affine $F$-space $\overline{T}_{b,a} \subset E_A$ has a Zariski dense set of $F$-points.  Thus, the dense, Zariski open subset $T_{b,a}$ has an $F$-point.
A: $\newcommand{\End}{\operatorname{End}}$
$\newcommand{\op}{{\operatorname{op}}}$
The finiteness of $F$ is not needed. I have learnt the idea of the following proof from Torsten Ekedahl.
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.
