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Is $I=(x,y)$ seen as an $\mathbb C[x,y]-$module is free ?

I would say that $\{x,y\}$ is a basis but since $x$ and $y$ are in $\mathbb C[x,y]$, may be it's not true.

user386627
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  • ${x,y}$ is not a basis, because they satisfy a linear relationship: e.g. $y \cdot x + (-x) \cdot y = 0$. –  May 17 '17 at 09:16
  • @Hurkyl: So how can I conclude ? – user386627 May 17 '17 at 09:23
  • I don't think it's free, otherwise it woul be isomorphic to some $R^n$ (where $R= \Bbb{C}[x,y]$) – Crostul May 17 '17 at 09:51
  • @Crostul: Thank you but how can it be proved ? – user386627 May 17 '17 at 11:08
  • Even though the question is not exactly the same, I think you can find a proof for your question by looking at this: https://math.stackexchange.com/questions/388176/prove-2-x-is-not-a-free-r-module – Hans May 17 '17 at 12:29
  • A non principal ideal $I$ of any commutative non-zero ring $R$ cannot be free over $R$, since a basis would have at least two elements ($I$ is not principal), say $a,b \in I$, but then $ab-ba=0$ provides a non trivial $R$-linear dependence relation between $a$ and $b$ ; contradiction. – Watson May 23 '17 at 18:14

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A non principal ideal $I$ of any commutative non-zero ring $R$ cannot be free over $R$, since a basis over $R$ would have at least two elements ($I$ is not principal), say $a,b \in I$, but then $ab-ba=0$ provides a non trivial $R$-linear dependence relation between $a$ and $b$ ; contradiction.

Notice that in your case, $(x,y)$ is not even a flat $R$-module, where $R = k[x,y]$ and $k$ is a field, because the morpshim $M := (x,y) \otimes_R (x,y) \longrightarrow (x,y) \otimes_R R \cong (x,y)$ is not injective (you just have to prove that $x \otimes y - y \otimes x$ is non zero in $M$, because then this element is a non-trivial element of the kernel of this morphism).

Watson
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