Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Thanks.

share|improve this question
    
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

3 Answers 3

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.

share|improve this answer

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.

share|improve this answer

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.