I have a commutative algebra $A$ over the complex numbers. I know that $\mathbb{C}[t]$ is a subalgebra of $A$, and that for any natural number $n$, the quotient $A/(t-n)$ is finite dimensional (and non zero). I do not know anything about the quotients $A/(t-c)$ where $c$ is any other complex number.

Can I deduce that $A$ is finitely generated as an algebra over $\mathbb{C}$? Is there anything which can be said about the sequence of numbers $a_n=dim_{\mathbb{C}}(A/(t-n))$?

Many thanks!

  • 2
    $\begingroup$ Take $\mathbb{C}[t]$ and invert all polynomials whose roots are not integers. This is not finitely generated and has the dimension of all quotients by $t-n$, $n$ an integer equals one. $\endgroup$
    – Mohan
    Oct 21, 2013 at 18:09

1 Answer 1


No, e.g. $A$ could be $\mathbb{C}^{\mathbb{N}}$ where $t$ is the sequence $(1, 2, 3, ...)$. This is not even Noetherian or countable-dimensional over $\mathbb{C}$. A slight modification of this construction shows that there are no restrictions on the sequence $\dim A/(t-n)$ and also that matters need not improve if you know $\dim A/(t-c)$ for $c$ complex.

(This example is universal; $A$ is the terminal object in the category of $\mathbb{C}[t]$-algebras with $\dim A/(t-n) = 1$. You can think of it as a product of skyscraper sheaves on $\mathbb{A}^1$.)


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