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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.

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 $fin(B)=0$.

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.

Source Link
Mare
  • 26.5k
  • 6
  • 25
  • 104

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 $fin(B)=0$.