Isomorphism in Projective modules over the algebraic closure of a finite field Fp Let $F \otimes Q \cong F\otimes R$. Does it imply $Q\cong R$?
where $R$ and $Q$ are modules over a field containing $\mathbb{F}_p$ and F is the algebraic closure of $\mathbb{F}_p$.
 A: Let $K\subseteq L$ be an extension of fields, and let $M$ and $N$ be two $K$-modules such that there is an isomophism of $L$-modules $L\otimes_KM\cong L\otimes_KN$. Since $L$-modules are vector spaces, this implies that the dimensions of $L\otimes_KM$ and $L\otimes_KN$ as $L$-vector spaces are equal. But $\dim_L(L\otimes_KM)=\dim_KM$ and $\dim_L(L\otimes_KN)=\dim_KN$, so the dimensions of $M$ and $N$ as $K$-vector spaces are equal: it follows that $M$ and $N$ are isomorphism $K$-modules.
A: As a counterexample, let $Q,R$ denote the modules over $\Bbb F_2$ given by $Q = \Bbb F_2$ and $R = \Bbb F_2[t] \cong \Bbb F_4$, where $t^2 + t + 1 = 0$.
$Q$ and $R$ are non-isomorphic free (and therefore projective) modules over $\Bbb F_p$ of dimensions $1$ and $2$ respectively. On the other hand, $F \otimes Q \cong F\otimes R \cong F$ (as $\Bbb F_2$-modules).

As another example, take $Q = \Bbb F_4$ and $R = \Bbb F_4[u] \cong \Bbb F_{16}$, where $u^4 + u + 1 = 0$. As before, $Q$ and $R$ are non-isomorphic modules, but $F \otimes Q \cong F \otimes R \cong F$.
