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I am reading the first chapter of 'methods of representation theory, Volume I' section 9A, written by Charles W.Curtis & Irving Reiner. Let $A$ denotes a finite dimensional algebra over an arbitrary field $K$ and let $\beta : A \times A \rightarrow K$ be a bilinear form. The author stated without proof that '$\beta$ is nondegenerate for $A$/$K$ if and only if it is nondegenerate for $E \otimes_K A/E$'. However I cannot fulfil the whole story behind this claim. Here is what I have tried:

\begin{align} \{\text{bilinear form } \beta: A \times A \rightarrow K\} &\leftrightarrow Hom_K(A, A^*)\\ \beta &\rightarrow (a \mapsto (b \mapsto \beta(a,b)))\\ f(a)(b) &\leftarrow f \end{align}

So nondegenerate bilinear forms corresponds to injective maps in $Hom_K(A, A^*)$ (hence isomorphism as assumed finite dimension). It might be helpful to consider the map

\begin{align} Hom_K(A, A^*) \xrightarrow{\cdot \otimes_K E} Hom_K(A^E, (A^*)^E). \end{align}

Oe-thau
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Your notation is a little confusing; I assume you mean "over $K$" and "over $E$" respectively and that $K \hookrightarrow E$ is a field extension, but as written those could be interpreted as quotients of some kind.

Anyway, the multiplication on $A$ is irrelevant here so this is really a fact about bilinear forms on vector spaces. In the finite-dimensional case there is a nice trick: if $a_1, \dots a_n$ is a basis of $A$ then it can be used to write down the Gram determinant $\det(\beta(a_i, a_j)) \in K$. This determinant depends on a choice of basis, but you can show that applying a change of basis only multiplies the determinant by an invertible square, so its class in the quotient by $(K^{\times})^2$ is well-defined. This class is the discriminant of $\beta$.

Now show that $\beta$ is nondegenerate iff it has nonzero discriminant. This condition is clearly preserved and reflected under field extensions.

Qiaochu Yuan
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  • However, I believe this is still true in infinite dimensions and this technique is unavailable. Already the analogous statement for linear maps is not entirely obvious without the use of the determinant: that a linear map $T : V \to W$ is invertible (resp. injective, surjective) iff its extension of scalars $T_E : V_E \to W_E$ is invertible (resp. injective, surjective). And in this case one also has to deal with the interaction with dualization which is messy if $E$ is an infinite extension. – Qiaochu Yuan Jun 24 '24 at 17:12
  • Your assumption about my notation is correct. I should be more formal rather than using my own simplified version. It suffices to prove the finite case here though the author did not emphasize it. Thanks for the answer! – Oe-thau Jul 04 '24 at 10:28