4
$\begingroup$

Given an ideal $I$ of $\mathbb{R}[X_1,X_2,X_3,X_4,X_5]$ generated by two unknown polynomials. I know two homogenous polynomials $p_1 \in I$ and $p_2 \in I$ such that

  1. $p_1$ is of degree 2 and up to a multiplicative constant the polynomial of smallest degree
  2. $p_2$ is of degree 3 and up to a linear combination with $p_1$ the only polynomial of degree 3 in $I$.

Can I conclude that $p_1$ and $p_2$ generate $I$?

$\endgroup$
4
  • 2
    $\begingroup$ Please clarify what you mean by the ring $\mathbb{R}^5$. To me, the notation suggests 5-tuples of real numbers, but then there would not be any polynomials to consider. $\endgroup$ Feb 24, 2015 at 19:50
  • $\begingroup$ Sorry for the typo. That was utterly stupid of me. I have edited the question accordingly. $\endgroup$
    – warsaga
    Feb 24, 2015 at 21:36
  • $\begingroup$ can't you take $p_1, p_2$ to be as described in $\mathbb{R}[x_1,\ldots, x_4]$ and take $p_3=x_5^4$ ? $\endgroup$ Feb 24, 2015 at 22:13
  • $\begingroup$ @DavidLehavi Such an example will not (or, at least, not usually) satisfy the requirement in the first sentence that the ideal is 2-generated. $\endgroup$ Feb 24, 2015 at 22:20

1 Answer 1

5
+50
$\begingroup$

As stated, I believe the answer is no. Set $q_1=x_1^2$, $q_2=x_2^4$, and consider $I=(q_1,q_2)$. Let $p_1=q_1$, and let $p_2=x_1p_1$. Then $p_1$ is (up to a constant) the only degree 2 polynomial in $I$, and $p_2$ is (up to a linear combination of multiples of $p_1$) the only polynomial of degree 3 in $I$. But they do not generate $I$.


If we add the requirement that $p_2$ is not a multiple of $p_1$, as Dave Witte Morris suggests, we can use the following example. Let $q_1=x_1x_2$ and $q_2=x_2^4+x_1^2$. Let $p_1=x_1x_2$ and $p_2=x_1^3=x_1q_2-x_2^3q_1$. The ideal $I=(q_1,q_2)$ is not generated by $p_1$ and $p_2$ since $q_2$ not divisible by $x_1$. Using normal forms, we can check directly that $x_1x_2=0$, $x_1^3=0$, $x_2^4=-x_1^2$ is a reduction system for the ring $\mathbb{R}[x_1,x_2,x_3,x_4,x_5]/I$, and so there are no more polynomials in degrees 2 or 3 in $I$ than those given by linear combinations of multiples of $p_1$ and $p_2$.

$\endgroup$
6
  • 1
    $\begingroup$ Adding the requirement that $p_2$ is not a multiple of $p_1$ would make the problem more interesting. $\endgroup$ Feb 24, 2015 at 22:38
  • 2
    $\begingroup$ It's interesting that the polynomials of smallest degree can have a nontrivial gcd, even though $\mathrm{gcd}(I) = 1$. Next (and final?) question: what if $p_1$ is required to be relatively prime to $p_2$? $\endgroup$ Feb 25, 2015 at 18:59
  • $\begingroup$ @DaveWitteMorris: I've tried to answer your new question, without any success. I can prove that any potential counter-example cannot have one of the generators equal to $p_1$ or $p_2$, but haven't made much further progress, sorry. $\endgroup$ Feb 26, 2015 at 22:44
  • $\begingroup$ @Pace: Thanks a lot for your answers, they were very helpful. Do you have any advice how to tackle the case gcd(I)=1 $\endgroup$
    – warsaga
    Mar 2, 2015 at 12:12
  • $\begingroup$ @warsaga: I tried quite a few different techniques, but made no progress in the $gcd(I)=1$ case. There are some good Groebner basis packages out there, which can quickly tell you whether or not you have a counter-example. Besides just a brute-force search, I don't know what else to suggest. $\endgroup$ Mar 2, 2015 at 16:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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