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Suppose $k$ is a field and $A$ is $k$-algebra and $\mathfrak{m}$ a maximal ideal. Consider the $k$-vector space $A/ \mathfrak{m}^N$ for some $N \geq 2$. Can we show that this is finite-dimensional?

For $A = k[X_1,\ldots,X_n]$ this is true. But in this more general case, I am not sure yet. Is there anyone that can shine a light on this matter, albeit a counter-example?

JSchoone
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$\mathbb C$ is a $\mathbb Q$ algebra in which $\{0\}$ is a maximal ideal, and no quotient by any power of this maximal ideal is going to be finite dimensional.

I keep rereading the question to see if you intended to say that $A$ is a finite dimensional algebra as well, but I keep doubting it. Of course, if you did require that, it would be trivially always true since quotients of finite dimensional $k$ modules are again finite dimensional.

It also has nothing to do with $k$ being algebraically closed or not. You could, for example, let $k=\mathbb C$ and then let $A=k(x)$, the field of rational polynomials, and have the same argument as above.

rschwieb
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  • Thank you for your quick reply and nice examples. They also show that there is no class (other than $A$ being finite-dimensional) of algebras that does satisfy this property. – JSchoone May 09 '17 at 13:32
  • @JSchoone No, they don't show that there is no other class, they just show that the entire class of $k$ algebras is too big of a class. As you mentioned, there are infinite dimensional $k$ algebras for which it does hold. For example, in $k[x]$, every quotient by a maximal ideal is finite dimensional over $k$. I'm not immediately sure how to capture this "maximal ideals have finite codimension" property, but there is probably a good way. – rschwieb May 09 '17 at 14:30
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    To name a positive result: It holds true if $A$ is a finitely generated $k$ algebra. – MooS May 09 '17 at 14:34
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    In fact, while searching the internet, I do see several references alluding to showing maximal ideals have finite codimension in special cases. You might pursue this. – rschwieb May 09 '17 at 14:34
  • Also, if I am not mistaken, in the noetherian (which is wider than the finite type $/k$ case) case it suffices to consider the case $N=1$, i.e. $A/\mathfrak m^N$ is finite dimensional for all $N \geq 1$ iff $A/\mathfrak m$ is. – MooS May 09 '17 at 14:39