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Can you please describe how to use the intrinsic definition of tangent space to show that the tangent space of the curve $Z \left( y^2-x^3 \right)$ at the point $x = \left(1,1\right)$ is one-dimensional?

That is, if $m_x/m_x^2$ is one dimensional, then so is the tangent space. Show that $m_x/m_x^2$ is one dimensional.


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Is your intrinsic definition $Hom_k(m/m^2, k)$? – Matt Jan 29 '13 at 4:17
Yes, my intrinsic definition is that the tangent space is (isomorphic to) the linear forms on $m/m^2$. – GuyStalks Jan 29 '13 at 5:25
Thank you very much to the answerers, but especially @Sanchez for reading my mind. I did indeed want a direct calculation of the cotangent space. – GuyStalks Jan 30 '13 at 21:37
up vote 5 down vote accepted

I would assume that you work with $\mathbb{C}$ for simplicity. Clearly, $m_x/m_x^2$ is spanned by $x-1$ and $y-1$. Now $$\begin{eqnarray}y^2-x^3 =& (y-1+1)^2 - (x-1+1)^3 \\ =& 2(y-1) - 3(x-1) + (y-1)^2 - 3(x-1)^2 - 3(x-1)^3\end{eqnarray}$$ So in $m_x/m_x^2$, $2(y-1) - 3(x-1) = 0$. Therefore $m_x/m_x^2$ is spanned by $x-1$.

It then suffices to show that $x-1 \notin m_x^2$. Suppose the contrary, then $x-1$ is a $\mathbb{C}$-linear sum of $(x-1)^2$, $(x-1)(y-1)$, $(y-1)^2$ and some multiple of $(y^2-x^3) $. Consider this equality in the polynomial ring $\mathbb{C}[x-1,y-1]$, we see that it is impossible. (The linear term can never be the same)

I don't think anything would change for a general field, but you would have to be slightly careful for $\mathrm{char} k = 2$, where the generator for $m_x/m_x^2$ would be $y-1$ instead.

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The morphism $\mathbb{A}^1 \to Z(y^2-x^3), t \mapsto (t^2,t^3)$ restricts to an isomorphism $\mathbb{A}^1 \setminus \{0\} \cong Z(y^2-x^3) \setminus \{(0,0)\}$. Tangent spaces of $\mathbb{A}^n$ have dimension $n$: By translation, it suffices to observe this at the origin, and $(x_1,\dotsc,x_n)/(x_1,\dotsc,x_n)^2$ has $k$-basis $x_1,\dotsc,x_n$. Of course, the base field $k$ (which could be any base scheme!) doesn't play a role at all.

This implies that the tangent space of $Z(y^2-x^3)$ at any point $\neq (0,0)$ is $1$-dimensional. But the tangent space at the origin turns out to be $2$-dimensional, which geometrically means that the origin is a singularity (a so-called cusp).

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Another way to see this is by using the high-tech notion of cotangent sheaf $\Omega_{X/\mathbb{C}}.$

Let $R=\mathbb{C}[x,y]/(y^2-x^3)$ be the affine coordinate ring of $X=Z(y^2-x^3)$. We have

$$\Omega_{X,p} \otimes \mathcal{O}_{X,p}/\mathfrak{m}_{X,p}\cong \mathfrak{m}_{X,p}/\mathfrak{m}^2_{X,p}$$

where $\mathfrak{m}_{X,p}$ is the unique maximal ideal of the stalk $\mathcal{O}_{X,p}.$

Claim: $\Omega_{X,p} \cong \mathcal{O}_{X,p}$ for $p \ne (0,0).$

Let's compute the cotangent sheaf. By definition $\Omega_X$ is the coherent sheaf associated to $\Omega_{R/\mathbb{C}}$ the module of relative differentials of the $\mathbb{C}$-algebra $R.$

In fact, $\Omega_{R/\mathbb{C}}=(Rdx \oplus R dy)/(2ydy-3x^2dx).$ Let $P \ne (0,0),$ with non-zero $x$-coordinate, then the localization at $\mathfrak{m}_p$ is $(\Omega_{R})_{\mathfrak{m}_p}=(R_{\mathfrak{m}_p}dx \oplus R_{\mathfrak{m}_p} dy)/(2ydy-3x^2dx) \cong R_{\mathfrak{m}_p}$ via $fdx+gdy \mapsto 2yf+3x^2g$ for $f,g \in R_{\mathfrak{m}_p}$ with inverse $h \mapsto (h/3x^2)dy$ for $h \in R_{\mathfrak{m}_p}.$ Therefore, $\Omega_{X,p} \cong \mathcal{O}_{X,p}$ for $p \ne (0,0).$

If $p=(0,0),$ we have $(\Omega_{R})_{\mathfrak{m}_0}=R_{\mathfrak{m}_0}dx \oplus R_{\mathfrak{m}_0} dy$ thus $\Omega_{X,0} \ncong \mathcal{O}_{X,0}.$

Indeed we have the following SES of sheaves

$$0 \to \mathbb{C}_0 \to \Omega_X \to \mathcal{O}_X \to \mathbb{C}_0 \to 0$$

where $\mathbb{C}_0$ is the skyscraper sheaf supported at the origin which is the kernel and cokernel of the $\Omega_X \to \mathcal{O}_X.$

Hence $\text{dim}_{\mathbb{C}}(\mathfrak{m}_{X,p}/\mathfrak{m}^2_{X,p})=\text{dim}_{\mathbb{C}}(\Omega_{X,p} \otimes \mathcal{O}_{X,p}/\mathfrak{m}_{X,p})=1$ for $p \ne (0,0)$ and is $2$ at the origin.

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