# Showing a polynomial does not belong to an ideal in $k[x,y,z]$

Consider the ring of polynomials $k[x,y,z]$, $k$ a field.

Why is it clear that $x \not \in (x^{2},xz,z^{2},xy-z^{2})$ ?

I mean, how do we show this rigorously? so assume it belongs, then:

$x = f(x,y,z)x^{2} + g(x,y,z)xz + h(x,y,z)z^{2} + t(x,y,z)(xy-z^{2})$ where $f,g,h,t$ are elements of $k[x,y,z]$, yes?

Now from here I'm not sure how to proceed and derive a contradiction. I tried setting $x=0, y=0$ and $z$ fixed so:

$0=h(0,0,z)z^{2} + t(0,0,z)z^{2}$ and thus $0=z^{2}p(z)$ where $p(z)=h(0,0,z)+t(0,0,z)$.

Now this can only happen when $p(z)=0$ but how we know this cannot happen?

What is an elegant way (or quick one) to see that $x$ does not belong to such ideal?

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Consider the degrees of the monomials in the summands. None of them contains any linear (degree 1) terms at all, so... – Alon Amit Apr 28 '11 at 19:59

Hint: Look at the degrees of the polynomials on both sides of the equation you get after assuming that $x$ is in the ideal.