2
$\begingroup$

I'm working with polynomials in $\mathbb{R}[x,y]$; they are $f_1=x^2+y^2$, $f_2=xy^4+y^7$, $f_3=y^8+y^4, f=x^2+tx, t\in\mathbb{R}$. I want to find values of $t$ such that $f\in\sqrt{I}$, where $I=(f_1,f_2,f_3)$.

First of all, I've checked whether $F=\{f_1,f_2,f_3\}$ is a Gröbner basis by reducing all $S$-polynomials, and from my calculations they reduced to $0$. I'm using lexicographical order with $x>y$.

$S(f_1,f_2)=y^4f_1-xf_2=y^6-xy^7\xrightarrow[\text{}]{\text{$f_2$}}y^6+y^{10}\xrightarrow[\text{}]{\text{$f_3$}}0;\\ S(f_2,f_3)=y^4f_2+xf_3=y^{11}-xy^4\xrightarrow[\text{}]{\text{$f_2$}}y^{11}+y^7\xrightarrow[\text{}]{\text{$f_3$}}0;\\ S(f_1,f_3)=y^8f_1-x^2f_3=y^{10}-x^2y^4\xrightarrow[\text{}]{\text{$f_2$}}y^{10}+xy^7\xrightarrow[\text{}]{\text{$f_2$}}0;$

So I have a Gröbner basis, but I'm not sure how to approach the problem from now on. Is the Gröbner basis a good tool to go on, or should I change my approach?

$\endgroup$
  • $\begingroup$ M2 gives $y^5+y, x+y^3$ as a Lex Gröbner basis for $\sqrt{I}$. From help radical : "The method used is the Eisenbud-Huneke-Vasconcelos algorithm. See their paper in Inventiones Mathematicae, 1993, for more details on the algorithm." $\endgroup$ – Jan-Magnus Økland Mar 25 '17 at 12:03
  • $\begingroup$ To continue the above comment: this shows that $f\notin\sqrt I$ for every $t\in\mathbb R$. $\endgroup$ – user26857 Mar 25 '17 at 17:19

Your Answer

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

Browse other questions tagged or ask your own question.