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I encountered the term "Artinian local $k$-algebra," where $k$ is a field. I think the author meant an Artinian local ring which is a $k$-algebra, but is it by any chance equivalent to a local ring which is an Artinian $k$-algebra? Since Artian, Noetherian, finite-(vector space)-dimensional are all equivalent for vector spaces, I guess the question is whether a local ring $A$ which is an Artinian $k$-algebra is a finite-dimensional $k$-vector space.

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The answer is no. Pick a field $k$ and a trascendental field extension $L$ of $k$ (for example $L = k(x)$); obviously $L$ is an artinian local ring (it is actually a field!), but the dimension of $L$ as $k$-vector space is not finite.

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Dear ashpool: You write:

I encountered the term "Artinian local $k$-algebra", where $k$ is a field. I think the author meant an Artinian local ring which is a $k$-algebra,...

The answer here is yes. The point is this. Let $K$ and $B$ be commutative rings. A structure of $K$-algebra on $B$ is just a ring morphism $\phi:K\to B$. Assuming such a morphism has been given, consider the following conditions on a subset $\mathfrak a$ of $B$:

$(1)\ \mathfrak a$ is an ideal of $B$ viewed as a ring,

$(2)\ \mathfrak a$ is an ideal of $B$ viewed as a $K$-algebra.

These conditions are equivalent, because if $\mathfrak a$ is preserved by $B$, it is a fortiori preserved by $\phi(K)$.

You write:

... but is it by any chance equivalent to a local ring which is an Artinian $k$-algebra?

The answer is again yes, for the same reason.

You write:

Since Artinian, Noetherian, finite-(vector space)-dimensional are all equivalent for vector spaces, I guess the question is whether a local ring $A$ which is an Artinian $k$-algebra is a finite-dimensional $k$-vector space.

If $B$ is a $K$-algebra as above, you can consider the sub-$K$-modules of $B$. Ideals are such sub-modules, but in general there are many sub-$K$-modules of $B$ which are not ideals. They are additive subgroups of $B$ which are preserved by $\phi(K)$, and they have no reason of being preserved by the whole of $B$. Andrea gave a nice answer to this part of your question.

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