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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 $I$?

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    $\begingroup$ 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. $\endgroup$ – Dave Witte Morris Feb 24 '15 at 19:50
  • $\begingroup$ Sorry for the typo. That was utterly stupid of me. I have edited the question accordingly. $\endgroup$ – warsaga Feb 24 '15 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 $p_3=x_5^4$ ? $\endgroup$ – David Lehavi Feb 24 '15 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. $\endgroup$ – Dave Witte Morris Feb 24 '15 at 22:20
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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 $p_2$.

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    $\begingroup$ Adding the requirement that $p_2$ is not a multiple of $p_1$ would make the problem more interesting. $\endgroup$ – Dave Witte Morris Feb 24 '15 at 22:38
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    $\begingroup$ 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 $p_2$? $\endgroup$ – Dave Witte Morris Feb 25 '15 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. $\endgroup$ – Pace Nielsen Feb 26 '15 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 '15 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. $\endgroup$ – Pace Nielsen Mar 2 '15 at 16:05

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