What do we mean by a variety being arithmetically CohenMacaulay? Is every such variety also Gorenstein?

1$\begingroup$ Determinental varieties (whose defining ideal is genenerated by minors of a matrix) are ACM but usually not Gorenstein. $\endgroup$ – J.C. Ottem Nov 28 '10 at 18:05
Arithmetically CohenMacaulay means, depending on the source/context, either:
 The homogeneous coordinate ring (with respect to a given embedding into $\mathbb{P}^n$) is CohenMacaulay. This seems to be more common.
 The section ring (with respect to a given ample line bundle) of the variety is CohenMacaulay.
Of course if you are projectively normal in $\mathbb{P}^n$ and the ample line bundle is a very ample line bundle of that embedding, these two definitions coincide.
It doesn't imply anything about Gorensteinness. In fact, any CohenMacaulay projective variety with $H^i(X, \mathcal{O}_X) = 0$ for $0 < i < \dim X$ is arithmetically CohenMacaulay with respect to some embedding into projective space.
To see this, take a sufficiently ample line bundle $L$ such that $H^i(X, \omega_X \otimes L^n) = 0$ and $H^i(X, L^n) = 0$ for all $n \geq 1$ and all $0 < i < \dim X$. In the previous version of this answer, I forgot the CohenMacaulay hypothesis on $X$, in which case the first vanishing can't be forced to hold.
If I recall correctly, these notions appear prominently in the study of Linkage (see Eisenbud's book for an introduction).
A related notion is that of arithmetic Macaulayfication of a ring. This means that there exists an ideal $I$ such that the Rees algebra of $I$ (the ring you blowup to get the blowup of $I$) is CohenMacaulay. These were shown to exist in the last decade by Kawasaki. If I recall correctly, a corollary of this result is that every ring with a dualizing complex is a quotient of a Gorenstein ring (this was previously a conjecture of Sharp). Someone correct me if I'm wrong on this.
EDIT: Added the CM hypothesis on the variety and added an explanation (thanks to Long). EDIT2: Added the two possible definitions (section ring vs coordinate ring). Thanks to J. C. Ottem.

$\begingroup$ A stupid question: When $X$ is projectively normal under the embedding given by $L$, the section ring $A=\bigoplus_{n\ge 0}H^0(X,L^n)$ is isomorphic to the coordinate ring $S(X)$ of X under this embedding. In general $A$ can be regarded as a $S(X)$module. When $X$ is not projectively normal, is there any essential difference between ACM and $S(X)$ being CM? $\endgroup$ – J.C. Ottem Nov 28 '10 at 18:41

3$\begingroup$ Karl, don't you need the middle cohomology of all the twists to vanish? $\endgroup$ – Hailong Dao Nov 28 '10 at 20:25

3$\begingroup$ @Ottem. They are different. Take $\mathbb P^1$ embedded in $\mathbb P^4$ by $x^4$, $x^3y$, $xy^3$, $y^4$. This is a closed immersion, but not projectively normal. The section ring is CohenMacaulay, but the homogeneous coordinate ring is not, because it is regular in codimension 1, but not normal and therefore not S2. @Pascal. It is the section ring which is losing information. In the example above, the section ring doesn't distinguish between the given embedding and the embedding by all sections of $\mathcal O(4)$. $\endgroup$ – Dustin Cartwright Nov 28 '10 at 22:36

1$\begingroup$ @Long, all the middle twists vanish by Serre vanishing and Serre duality. Just take a very high veronese subring. $\endgroup$ – Karl Schwede Nov 28 '10 at 22:48

2$\begingroup$ @J. C. Ottem, I think you are right, the coordinate ring definition is much more common. Before posting the original answer, I found some places where it was defined the other way, and went with that. But I probably should not have $\endgroup$ – Karl Schwede Nov 29 '10 at 5:40