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Given an element $m$ in a coalgebra $C$, there always exists a finite-dimensional subcoalgebra $D \subset C$ containing $m$; this is the fundamental theorem for coalgebras. This obviously isn't the case for algebras. However, to make a proper analogue of this theorem for algebras, one should formulate this theorem using arrows and then dualize. An element of $C$ determines a linear map $F:\mathbb{k} \to C$ and vice versa. The fundamental theorem then becomes:

For any linear map $F: \mathbb{k} \to C$ there exists a map $G:\mathbb{k} \to D$, such that $i \circ G=F$, for $D$ some finite dimensional coalgebra, and $i$ an injective map of coalgebras.

Dualizing to algebras, one easily sees that the statement becomes equivalent to the question: is any linear map $f:A \to \mathbb{k}$ zero on an ideal of finite codimension, where $A$ is any algebra. Spelling this out for $k[x]$, one is led to the question: do there exist sequences $(f_{n})_{n}$ that do not satisfy any recursion equation. Does anyone know about a sequence like this, or an existence proof?

A perhaps less sensical question: does the reformulation of the fundamental theorem in terms of algebras have some nice geometrical interpretation? Say in the case of affine algebras?

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Very interesting. I think you have to be a bit more modest about "any recursion equation" though; obviously we can only use algebraic operations. – Zhen Lin Jan 9 '13 at 14:55
Yes, indeed everything is algebraic – Michael Parsons Jan 9 '13 at 19:51

Very interesting. Unfortunately the dual statement you suggest is false. Since $k$ is a field, $k [x]$ is a principal ideal domain. In particular, any non-trivial quotient of $k [x]$ is automatically finite-dimensional. Now, any linear map $f : k [x] \to k$ is freely and uniquely determined by the sequence $f(1), f(x), f(x^2), \ldots$, and $f$ factors through the quotient $k [x] / (p(x))$ only if $$a_0 f(x^n) + a_1 f(x^{n+1}) + \cdots + a_d f(x^{n+d}) = 0$$ for all $n$, where $$p(x) = a_0 + a_1 x + \cdots + a_d x^d$$ so to find an $f$ that does not factor through any non-trivial quotient of $k [x]$, it is enough to find a sequence that does not satisfy any linear recurrence equation (with constant coefficients). But a sequence that satisfies a linear recurrence equation grows at worst exponentially, so we can just take a super-exponential sequence, say $n!$.

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Yes, that's exactly what I was trying to ask. I already had the feeling that such sequences existed but I did not know how to formulate that they exist. Your growth statement seems to do it. – Michael Parsons Jan 9 '13 at 19:52
I even seem to recall that there exist sequences that do not satisfy any algebraic recurrence relation, linear or otherwise, but this may be a fact from logic. Any idea on what the statement would say geometrically if it were right. – Michael Parsons Jan 9 '13 at 19:53

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