Decide if certain polynomial is in an Ideal Let $I$ be the ideal $ I = (x^3y-x^2y^2,x^3z+z^2yx,x^2-xz) \subset \mathbb{Q}[x,y,z]$.
I have to decide if $x$ is part of $I$ or $\sqrt I$. My first take was computing the Groebnerbasis $G$ of $I$ by using Buchberger's algorithm (by hand) but that gave no useable result.
The idea was to look if $x$ is part of $G$. If yes, $x$ would be in $I$. If not, I would look if there is some $x^n \in G$ for $n > 1$. Then $x$ would be in $\sqrt I$.
Is my general idea wrong? Are there other ways to determine if $x \in I$ or $x \in \sqrt I$?
 A: Your idea to use a Groebner basis is good, but it only works if you choose a suitable monomial ordering.
To find a polynomial in $x$ that belongs to $I$, you want to eliminate $y$ and $z$, so you want to choose an elimination ordering.
By choosing the lexicographical ordering with $x<y<z$ we get that every monomial containing $y$ or $z$ is greater than a monomial containing only $x$, so cancelling leading terms (as in the computation of $S$-polynomials and the reduction steps in Buchberger's algorithm) will mean (often) getting rid of $y$'s and $z$'s.
In the new ordering, the given basis is $I = (-y^2x^2+yx^3,z^2yx+zx^3,-zx+x^2)$.
The $S$-polynomial of $z^2yx+zx^3$ and $-zx+x^2$ is $z^2yx+zx^3 + zy(-zx+x^2) = zyx^2 + zx^3$. We reduce this modulo the elements of the basis so far. The leading monomial is $zyx^2$; the only element of the basis whose leading monomial divides this one is $-zx+x^2$, so we add $yx(-zx+x^2)$ to get $zx^3 + yx^3$. The leading monomial is $zx^3$; the only element of the basis whose leading monomial divides this one is $-zx+x^2$, so we add $x^2(-zx+x^2)$ to get $yx^3 + x^4$. No more reduction is possible, so we add this to our basis.
The $S$-polynomial of $-y^2x^2 + yx^3$ and $yx^3 + x^4$ is $x(-y^2x^2+yx^3)-y(yx^3+x^4)=-yx^4-yx^4=-2yx^4$. The leading monomial is divisible by the leading monomial of $yx^3 + x^4$, so we add $2x(yx^3+x^4)$ to get $2x^5$. No more reduction is possible, so (normalizing the coefficient) we add $x^5$ to our basis.
In fact we are done: $I = (-y^2x^2+yx^3,z^2yx+zx^3,-zx+x^2, yx^3+x^4, x^5)$ is a Groebner basis.
Since the ordering is an elimination ordering, $I \cap \mathbb{Q}[x] = (x^5)$.
So $x^5 \in I$, hence $x\in\sqrt{I}$, but no power of $x$ lower than $5$ belongs to $I$, so $x \not\in I$.
