I know that generally the answer is no, for example the weyl algebra。 But is this true for commutative algebra？ or we may restrict to affine commutative algebras。
Maybe ，it is a classical result. So, please let me know the reference. Thanks a lot.
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Suppose $A$ is non-negatvely graded and connected, with $A_0=k$ your base field. Then the same is true of the enveloping algebra $A^e$.
As with all algebras, we have $\def\gldim{\operatorname{gldim}}\def\pdim{\operatorname{pdim}}\gldim A\leq\pdim_{A^e}A$.
Suppose $\pdim_{A^e}A=n<\infty$. By taking a minimal resolution of $A$ as an $A$-bimodule of length $n$ to compute, we see that $\def\Ext{\operatorname{Ext}}\Ext^n_{A^e}(A,k)\neq0$. Now $k\cong\hom_k(k,k)$ as a bimodule, and $\Ext^\bullet_{A^e}(A,\hom_k(k,k))\cong\Ext^\bullet_A(k,k)$: this tells us that $\gldim A\geq n$, so your equality holds in this case. If the dimension is infinite, the same works.
Something similar can be done if $A$ is local noetherian with the base field as residue field.
On the other hand, the equality does not hold always, even in the affine commutative case. For example, suppose $L/k$ is a finite field extension of your ground field $k$ which is not separable. Since separability of $L/k$ is the same thing as $L$ being a separable $k$-algebra, we have that $L$ is not a projective $L$-bimodule and therefore $$\pdim_{L^e}L\geq1>0=\gldim L.$$
answered Dec 30, 2012 at 10:59
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$\begingroup$ Of course, this counterexample will not work if $k$ is perfect. Then just take $L=k(t)$ with $t$ trascendental over $k$; then $L$ is not a separable $k$-algebra (because separable algebras over fields are finite dimensional) and the same reasoning applies. $\endgroup$ Commented Dec 30, 2012 at 11:21
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$\begingroup$ A sillier counterexample is: pick any algebra $A$ whose left and right global dimensions are different (such this exist); since the projective dimension of $A$ as an $A$-bimodule is side-agnostic, it must be different to at least one of the two global dimensions! $\endgroup$ Commented Dec 30, 2012 at 11:26
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$\begingroup$ Well，thank you for the explainations. So, if the gldim is finite，then how about the bimodule dimension of an affine commutative k-algebra？ Is that still finite since your examples implies that these two numbers may not be the same. PS: Is the bimodule dimension of L in your example finite? Thanks again. Happy new year to you！^_^ $\endgroup$– iffCommented Dec 31, 2012 at 4:08
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$\begingroup$ I know that for commutative affine algbera over a field of characteristic zero, then global dimension is finite iff the bimodule projective dimension is finite which is also equivalent to say that the algebra is regular. $\endgroup$– iffCommented Jan 5, 2013 at 4:52
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$\begingroup$ Can you explain why $Ext_{A^e}^n (A,k) \neq 0$ and why $Ext_{A^e}(A,hom_k(k,k))=Ext_A(k,k)$? $\endgroup$ Commented Sep 7, 2017 at 13:43