Q:

Global dimension of Artin algebras over a perfect field

Let $A$ be an Artin algebra over a perfect field $k$. Suppose that the global dimension of $A$ is finite. How one can prove that
  $$
\operatorname{gl}(A)=\max\{i \geq 0\mid \operatorname{Ext}^i(DA, A) \neq 0\},
$$
  where $DA=\operatorname{Hom}_k(A,k)$.

Property that $k$ is perfect can be used to go to the Hochschild cohomology because then global dimension and Hochschild dimension coincide and this claim is trivial, but I want some elementary proof for this elementary observation.

A:

It doesn't need "perfect".
All the modules in this answer will be finitely generated.
Suppose $\operatorname{gl}(A)=n<\infty$. Then there are modules $X$ and $Y$ with $\operatorname{Ext}^n(X,Y)\neq0$.
Take a short exact sequence $0\to\Omega Y\to P\to Y\to0$ with $P$ projective. The long exact sequence of $\operatorname{Ext}(X,-)$ contains
$$\operatorname{Ext}^n(X,P)\to\operatorname{Ext}^n(X,Y)\to\operatorname{Ext}^{n+1}(X,\Omega Y)=0,$$
so $\operatorname{Ext}^n(X,P)\neq0$.
Now take a short exact sequence $0\to X\to I\to\Omega^{-1}X\to0$ with $I$ injective. The long exact sequence of $\operatorname{Ext}(-,P)$ contains
$$\operatorname{Ext}^n(I,P)\to\operatorname{Ext}^n(X,P)\to\operatorname{Ext}^{n+1}(\Omega^{-1}X,P)=0,$$
so $\operatorname{Ext}^n(I,P)\neq0$.
Since $P$ is a direct summand of a direct sum of copies of $A$, and $I$ is a direct summand of a direct sum of copies of $DA$, it follows that $\operatorname{Ext}^n(DA,A)\neq0$.

