# Questions about tangent spaces of $A^n$.

Let $A^n=k\times \cdots \times k$ be the affine $n$-space, $k$ is a field. Define $d_xf = \sum_{i=1}^n \frac{\partial f}{\partial T_i}(x)(T_i-x_i)$, where $x=(x_1, \ldots, x_n) \in A^n$.

Suppose that $X \subset A^n$ is defined by polynomials $f(T)=f(T_1, \ldots, T_n)$. Define $$\operatorname{Tan}(X)_x = \{x \in A^n \mid d_xf(x) = 0, \forall f(T) \in I(X) \}.$$

Suppose that $I(X)$ is generated by $f_1(T), \ldots, f_m(T)$. Then each element in $I(X)$ is of the form $a_1f_1+\cdots +a_nf_n$ for some $a_i \in I(X)$. How to show that $d_xf_1, \ldots, d_xf_m$ generated by ideal of $\operatorname{Tan}(X)_x$? Thank you very much.

I think that the ideal of $\operatorname{Tan}(X)_x$ is $$\{f \mid f(x) = 0, \forall x \in \operatorname{Tan}(X)_x \}.$$ If $x \in \operatorname{Tan}(X)_x$, then $d_xf(x)=0$ for all $f\in I(X)$. That is, $d_xf(x)=0$ for all $f$ which are of the form $a_1f_1+\cdots a_mf_m$. Therefore the ideal of $\operatorname{Tan}(X)_x$ is contained in the ideal generated by $d_xf_1, \ldots, d_xf_m$. If $f$ is generated by $d_xf_1, \ldots, d_xf_m$, we also have $f(x)=0$ for all $x \in \operatorname{Tan}(X)_x$. Hence $x$ is in the ideal of $\operatorname{Tan}(X)_x$. Is this correct? Thank you very much.

• If I am understanding your question correctly, which I might not be, then evidently $\text{Tan}_x(X)=V(d_x f_1,\ldots,d_x f_m)$, and you're looking for $I(\text{Tan}_x(X))$? But, the above shows that this is just $\sqrt{(d_x f_1,\ldots,d_x f_m)}$. Since ideals generated by linear polynomials are radical, it follows that $I(\text{Tan}_x(X))=(d_xf_1,\ldots,d_x f_m)$ as desired. Yeah? – Alex Youcis Sep 30 '13 at 8:19
• @Alex, Thank you very much. – LJR Sep 30 '13 at 9:54
• So, does that answer your question? – Alex Youcis Sep 30 '13 at 10:26
• @AlexYoucis, yes, that answers my question. Thank you very much. – LJR Oct 1 '13 at 8:47

## 1 Answer

So, you're defining $\text{Tan}_x(X)$ to be $V(\{d_x f: f\in I(X)\})$. You're asking then to prove that $I(\text{Tan}_x(X))=(d_x f_1,\ldots,d_x f_n)$ if $I(x)=(f_1,\ldots,f_n)$. But, evidently $\text{Tan}_x(X)=V(d_x f_1,\ldots,d_x f_n)$ and so

$$I(\text{Tan}_x(X))=I(V(d_x f_1,\ldots, d_x f_n))=\sqrt{(d_x f_1,\ldots,d_x f_n)}=(d_x f_1,\ldots, d_x f_n)$$

where the last equality follows from the fact that ideals generated by linear polynomials in $k[x_1,\ldots,x_m]$ are radical (this should be easy to see).

I believe this is what you wanted to prove.