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First, I'd like to show $\mathbb{C}[x,y] / (y^2-x^3)$ is an integral domain. Then I need to find out whether or not it is a PID.

For the first part, I want to show $y^2-x^3 \: | \: fg \implies y^2-x^3 \: | \: f$ or $y^2-x^3 \: | \: g$ in $\mathbb{C}[x,y]$. I could not get past this.

For the second part, I was thinking of maybe showing $(y^2-x^3)$ is maximal; then the quotient would be a field and hence a PID. Again, I couldn't get further than this.

Any help is appreciated!

user26857
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StrangerLoop
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2 Answers2

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Any PID is a UFD, and any UFD is integrally closed. But $k[x,y]/(y^2-x^3)$ is not integrally closed, since $t:=y/x$ satisfies $t^2=x$ and $t^3=y$.

Martin Brandenburg
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  • How would you argue that $t$ is not contained in $k[x,y]/(y^2-x^3)$? – Dune Jul 29 '14 at 16:20
  • $k[x,y]/(y^2-x^3)$ is a free $k[x]$-module with basis $1,y$, hence a free $k$-module with basis $x^{\mathbb{N}} \cup x^{\mathbb{N}} y$. This basis is invariant under multiplication with $x$. Then it is clear that there is no element $t$ with $tx=y$. – Martin Brandenburg Jul 29 '14 at 17:47
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    Alternatively, consider $k[x,y]/(y^2-x^3) \to k[t]$, $x \mapsto t^2$, $y \mapsto t^3$ and observe that $t$ is clearly not contained in the image (for example by degree considerations). – Martin Brandenburg Jul 29 '14 at 17:49
  • Great, thanks! By the way, the second argument also immediately shows that $(x,y)$ is not principal. – Dune Jul 29 '14 at 18:04
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The zero set of $Y^2-X^3$ is a one-dimensional curve with a cusp at $(0,0)$. From the picture it is clear that the tangent spaces at roots $(a,b) \neq (0,0)$ are one-dimensional while the tangent space at $(0,0)$ is two-dimensional. Translating these geometric properties into algebraic properties, it means that $\dim_\mathbb{C}(\mathfrak{m}/\mathfrak{m}^2) = 1$ for $\mathfrak{m} = (X-a,Y-b)$ with $(a,b) \neq (0,0)$ and $\dim_\mathbb{C}(\mathfrak{m}/\mathfrak{m}^2) = 2$ for $\mathfrak{m} = (X,Y)$, where these ideals $\mathfrak{m}$ are considered to be ideals of the coordinate ring $R = \mathbb{C}[X,Y]/(Y^2-X^3)$. This in turn is equivalent to saying that all maximal ideals $\mathfrak{m}$ of $R$ but $(X,Y)$ are principal in $R_\mathfrak{m}$, and $(X,Y)$ is not. In particular, $(X,Y)$ cannot be principal in $R$.

Of course we cannot argue by looking at the picture here, so it is your turn to give an algebraic proof that $\mathfrak{m} = (X,Y)$ is not principal, or equivalently, that $\dim_\mathbb{C}(\mathfrak{m}/\mathfrak{m}^2) \geq 2$.

Dune
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