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.

  • $\begingroup$ 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? $\endgroup$ – Alex Youcis Sep 30 '13 at 8:19
  • $\begingroup$ @Alex, Thank you very much. $\endgroup$ – LJR Sep 30 '13 at 9:54
  • $\begingroup$ So, does that answer your question? $\endgroup$ – Alex Youcis Sep 30 '13 at 10:26
  • $\begingroup$ @AlexYoucis, yes, that answers my question. Thank you very much. $\endgroup$ – LJR Oct 1 '13 at 8:47

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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