# Subalgebra of free associative commutative algebra which is not free

I'm trying to provide a counterexample for analogue of Nielsen–Schreier theorem for the variety of associative commutative algebras (not necessary with unity) over a filed $F$.
A counterexample for noncommutative case is well known: ideal generated by commutators in a free algebra on 2 variables $F<x,y>$ is not itself a free algebra.
In commutative case the above construction obviously does not work, but I'm sure that there is a subalgebra of free commutative algebra which is not free.
I think that ideal generated by $x^2$ (or $F[x^2,x^3],$ it's the same subalgebra) in $F[x]$( $F[x]$ here means commutative polynomial algebra over $F$ without unity) is not a free commutative algebra.
Is my hypothesis correct? If yes, how do I prove it?

Firstly, I think that every free proper subalgebra $A$ of $F[x]$ which is also an ideal of $F[x]$ must be generated (in sense of free algebras) by single element:
Let's consider two nonequal generators $f$ and $g$ of $A$. $F[x]$ is an unique factorization domain implies that $f\cdot h_1 = g\cdot h_2,$ where $h_1 \neq g,$ $h_2\neq f$ because $gcd(f,g)\neq 1$ (It's well known that $F[x]$ is also a Principal ideal domain).
Next, multiplying on any $i \in A$ (in our case we can take $i=x^2$), we get $f\cdot (h_1\cdot i) = g\cdot (h_2\cdot i),$ and $h_1\cdot i, h_2\cdot i \in A$ implies that they can be expressed as terms in generators of $A$, so we have a nontrivial relation in $A$ and $A$ is not free (However, I'm not sure that this relation is a nontrivial one).
Is this correct? If it is, it can be applied to the case above, and it should be then easy to prove that $F[x^2,x^3]$ can't be freely generated by one element.
If not, it still seems to be true that every free subalgebra of $F[x]$ which satisfies the above conditions must be 1-generated in free sense.
So, can you help me with this?
Pardon my poor English and LaTeX skills.

EDIT:
It appears that I was correct with my assumption. P. M. Cohn in his article Subalgebras of free associative algebras provides exactly $F[x^2,x^3]$ as an example of subalgebra of $F[x]$ which is not free. A theorem in article states that subalgebra of $F[x]$ is free if and only if it's integrally closed, and $F[x^2,x^3]$ is not integrally closed.
Also, proving necessity, he states that that every free subalgebra $R$ of $F[x]$ is either $F$ (in this article only algebras with unity are considered) or $F[y]$ for some $y$ transcendental over $F$, and in each case $R$ is integrally closed.
It's stated as obvious that the number of free generators of $R$ is 1.
So, is my proof correct?

• Your proof is not correct. $F[x^2, x^3]$ is not an ideal. The ideal generated by $x^2$ doesn't contain the identity. – Qiaochu Yuan May 9 '12 at 7:30
• Well, judjing by your answer it seems we have a definition conflict or something. $F[x^2,x^3]$ is definetly a subalgebra; multiplying any poylynomial on polynomial like $a_2 x^2+\dots +a_n x^n$ (which is a typical member of $F[x^2,x^3]$) we still get a polynomial of that type. Maybe I haven't made clear that I'm considering the variety of all associative commutative algebras, not only unital ones. Of course $F[x^2,x^3]$ is a non-unital algebra, just as $F[x]$. Maybe I need to note this somehow? – Yury Popov May 9 '12 at 7:42
• For me $F[x^2, x^3]$ denotes the unital subalgebra generated by $x^2$ and $x^3$. I think it is not standard for this notation to refer to $(x^2)$. What do you mean by $F[x]$? – Qiaochu Yuan May 9 '12 at 7:43
• Okay, I will emphasize mentioned above in my question, thank you. – Yury Popov May 9 '12 at 7:45
• Yes, it was my fault. By $F[x]$ i mean free associative commutative algebra on one generator, i.e commutative polynomial algebra over F without unity. – Yury Popov May 9 '12 at 7:55

The only nonzero ideal of $F[x]$ which is also a subalgebra is $F[x]$, since any such ideal needs to contain a nonzero idempotent and the only one in $F[x]$ is the identity. You are correct that $F[x^2, x^3]$ is not a polynomial algebra; this should be clear from degree considerations (a generator must have degree exactly $2$ or else it can't generate $x^2$, but then the subalgebra it generates consists of polynomials of even degree so can't contain $x^3$).
Geometrically $\text{Spec } F[x^2, x^3]$ is the singular curve $u^2 = v^3$, so cannot be isomorphic to the affine line which is nonsingular. To convert this into an algebraic proof we consider the dimension of the space of derivations $D : F[x^2, x^3] \to F$ satisfying $$D(ab) = D(a) \phi(b) + \phi(a) D(b)$$
where $\phi : F[x^2, x^3] \to F$ sends $x$ to $0$. This space can be identified with the Zariski tangent space at the origin and is $2$-dimensional, which is not true of the corresponding tangent space at any point of the affine line.
• Yes, it's obvious that $F[x^2,x^3]$ can't be one-generated freely. But why should there be only one generator? For example (maybe not so appropriate, but still) free group on two generators contains a free subgroup of countable rank. According to article mentioned above, it is true, but I don't know if my proof is correct. – Yury Popov May 9 '12 at 7:53
• $F[x^2,x^3]$ is noetherian, so if its free it is free on finitely many generators. But it is not integrally closed, so it is not free. – Mariano Suárez-Álvarez May 9 '12 at 7:59
• @Yury: its fraction field has transcendence degree $1$. – Qiaochu Yuan May 9 '12 at 8:07
• @Mariano: yes, the number of generators must be finite because $F[x^2,x^3]$ is noetherian, it's helpful note. But I'm trying to understand why should it be exactly one? And yet another question (it can be dumb, but I've never worked with integral extensions): is commutative polynomial algebra over $F$ without unity still an integral domain? It seems that Euclidean algorithm still holds in that algebra, am I correct? – Yury Popov May 9 '12 at 8:29