Finistic dimensions under scalar extensions

Question

Let $A$ be a finite dimensional algebra over some field $K$. Denote the finistic dimension of A by fin($A$), that is, the supremum of the projective dimensions of finite generated modules whose projective dimensions are finite. Let $K\subset L$ be a field extension. If we tensor $A$ with $L$ over $K$. Then we get a new algebra $A'$ over $L$. This is the so-called scalar extension. My question is the following:

What is the relationship between fin($A$) and fin($A'$)?

If having finite finistic dimension is preserved under scalar extensions. Then one could first tensor with an algebraically closed field and then take the basic algebra, thus reduce the question to that of path algebras. But it seems that no one says it suffices to verify the conjecture for path algebras. I am interested in this conjecture recently. So I am just wondering what's the gap here?

Answer 1

We have $fin(A)=fin(A')$ at least for finite extensions by theorem 16 of https://www.cambridge.org/core/journals/nagoya-mathematical-journal/article/on-the-dimension-of-modules-and-algebras-viii-dimension-of-tensor-products/58116B52E52F0F6165E84AE11284CCF6 .

Namely for two algebras $A$ and $B$ over a field $K$, we have $fin(A \otimes_K B)=fin(A)+fin(B)$.

Now if $B$ is a field, then $fin(B)=0$.

Thus it is enough to prove the conjecture for quiver algebras as any finite dimensional algebra has a finite field extension such that the algebra is split and thus Morita equivalent to a quiver algebra.