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 $I$?



Generators vs minimal degree polynomials of ideals
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 $I$.




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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 $p\_2$.


