Decide belonging of an element to an ideal

Question

Let $f_1=x^2-y, f_2=xy-z,f_4=xz-y^2$ be polynomials with coefficients in some field $k$. I want to prove that $f_2\notin (f_1,f_4)$.

My attempt: by contraddiction, let $f_2\in (f_1,f_4)$. Then there exist $\alpha,\beta\in k[x,y,z]$ such that $f_2=\alpha f_1+\beta f_4$. Computations give $\alpha=\frac{y}{x}$ and $\beta=-\frac{1}{x}$, wich are not polynomials.

Do you think this is correct? However, what is the straight way to proceed in these cases?

How did "computations give" that? If you're allowing rational functions, you could just have easily gotten $\alpha = 0$ and $\beta = \frac{f_2}{f_4}$ or any other crazy thing. — , Feb 10 '14 at 14:56

score 0 · Answer 1 · answered Feb 10 '14 at 14:42

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The ring is k[x,y,z].

I don't think this is correct. You have only found some rationally functions that satisfy the representation.

One approach is to argue based on the degree of the polynomials on both sides.

answered Feb 10 '14 at 14:42

What I want to prove is that $f_2$ does NOT belong to the ideal – Danae Kissinger Feb 10 '14 at 14:50
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Yes. And you do that by arguing it's impossible by comparing terms (degrees) on both sides. See Patrick's solution. – Feb 10 '14 at 15:24

score 0 · Accepted Answer · answered Feb 10 '14 at 14:56

You cannot have such an expression $f_2 = \alpha f_1 + \beta f_4$. This is because if $$ xy - z = \alpha(x,y,z) (x^2 - y) + \beta(x,y,z) (xz-y^2), $$ then you can see that $$ z = xy - \alpha(x,y,z)(x^2 - y) - \beta(x,y,z) (xz-y^2) $$ The only (possible) monomials terms on the right-hand side is $\alpha(0,0,0)y$, which cannot be simply $z$. Therefore such an expression cannot exist.

Hope that helps,

Decide belonging of an element to an ideal

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