$GL(-)$ as a k-group functor

232 Views Asked by At

My question is essentially may lye simply in a notational obstruction.

For a k-algebra M, Jantzen J. defines the k-group functor $GL(M)$ as: $GL(M)(A):=(End_A(M\otimes_{\mathbb{k}} A)^*$. My question is what does the subscripted A mean?

I would understand $GL(M)(A)$ to be defined as $(End_{\mathbb{k}-Mod}(M\otimes_{\mathbb{k}} A)^*$.

1

There are 1 best solutions below

1
On BEST ANSWER

It means $A$-linear endomorphisms. If $M = k^n,$ then $M \otimes_k A = A^n$, and then $End_A(M\otimes_k A)^{\times} = Aut_A(A^n) = GL_n(A).$


Note that taking $End_k(M\otimes_k A)^{\times}$ gives the wrong thing, since if $A$ has dimension $d$ over $k$, then $M\otimes_K A$ has dimension $nd$, which isn't independent of $A$ (and is not even finite if $A$ is not a finite-dimensional $k$-algebra!). (Not to mention that this expression is not functorial in $A$.)