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Let $X$ be the union of the $3$ coordinate axes in $\mathbb{A}^{3}$. I want to compute the tangent space of $X$ at every $p \in \mathbb{A}^{3}$.

One can check that the only singular point of $X$ is the origin so if $p$ is the origin then the tangent space of $X$ at the origin is equal to $\mathbb{A}^{3}$.

Question: how do we express the answer if $p \in \mathbb{A}^{3} \setminus \{(0,0,0)\}$? Would it be correct to say:

Note that $I(X)=(xy,xz,yz)$ so if $p=(a,b,c) \in \mathbb{A}^{3}$ then:

$T_{p}(X):=\{(x_{1},x_{2},x_{3}) \in \mathbb{A}^{3}: \sum_{i=1}^{3} \frac{\partial f}{\partial x_{i}}(a,b,c)(x_{i})=0\ \text{for every f in I(X)}\}$

In case $p$ is not the origin, is there a "nice" way to express $T_{p}(X)$ other than the above?

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2 Answers 2

up vote 1 down vote accepted

Sure there is an easy way to compute $T_p(X)$ for $(0,0,0)\neq p\in X$ !

Just use that $T_p(X)=T_p(Y)$for any subscheme $Y\subset \mathbb A^3$ that coincides with $X$ in an open neighbourhood of $p$: the tangent space $T_p(X)$ only depends on the germ of $X$ at $p$.

So here take for $Y$ the unique coordinate axis axis on which $p$ lies.
For example if $p=(0,b,0)$ with $b\neq 0$, then $T_p(X)\subset T_p(\mathbb A^3)$ is the line $k\cdot (0,1,0)\subset k^3$

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The modern definition of the tangent space at the point $p$ is the dual of the $k$-vectorspace $\mathfrak m /\mathfrak m^2 $. Here $ \mathfrak m$ is the maximal ideal corresponding to the point $p$ and $k=k(p)$ is the residue field at $p$. I personally like the exposition in Qing Liu's book which explains the relation.

For all closed points $p$ the residue field is just the basefield $k$ and the maximal ideal at the point $(a,b,c)$ is $(x-a,y-b,z-c)$ (To be precise the ideal must contain $I$, so at least two out of the three values $a,b,c$ must be zero). So at the origin $(0,0,0)$ the maximal ideal is just $(x,y,z)$ and $\mathfrak m /\mathfrak m^2 $ is three dimensional. Assume wlog $p=(x-a,y,z)$ for any other closed point of $X$. Here we have the relations $(x-a)y=xy-ay=0$ (this is in $\mathfrak m^2$) and also $xy=0$ (we still live in $X$ which is cut out by the ideal $I$). Consequently we have $ay=0$ with a unit $a$, thus $y=0$, similarly $z=0$. Thus the dual of the tangent space is one-dimensional.

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