How to show an ideal is zero-dimensional [closed]

Question

I have the following past exam paper question, a similar sort of question seems to come up every year.. And I'm completely lost with it...

Let $J$ denote the ideal in $\mathbb{Q}[x,y,z]$ generated by $(y^2-xy-2zx, y^3+z^2+1, x^2yz-zy)$.

Show that $J$ is zero-dimensional.

What is the dimension, as a vector space over $\mathbb{Q}$, of $\mathbb{Q}[x,y,z]/J$?

Define the ideal $I := (xy^2+2xz-yz, x^2yz+y^2)$. Explain why $I$ is not zero-dimensional.

I know the definition for an ideal to be zero-dimensional, but I cant find an example of how to use to it to show an ideal is zero-dimensional...

Defintion. An ideal $I\subset K[X_1,...,X_n]$ is zero-dimensional if the $K$-vector space $K[X_1,...,X_n]/I$ is finite dimensional.

Answer 1

Your question is better suited for math.stackexchange.com and will probably be closed here quickly (don't get discouraged by this -- MathOverflow is exclusively for research level mathematics, while math.stackexchange.com is a great resource for questions of any level).

But, in the mean time: Do you know how to form a Groebner basis? This is an easy way to prove that a finitely presented ideal is zero-dimensional (and is easy to do by hand in this example with a lexicographic monomial ordering $\prec$ s.t. $z \prec y \prec x$).

Think about what it means for $J$ to be zero-dimensional: $J$ is zero-dimensional iff $V_{\mathbb{C}}(J)$ contains only finitely many points. For this to happen, a (monic) Groebner basis for $J$ must contain polynomials $p_1, p_2, p_3$ s.t. $LM(p_1)$ = $x^{k_1}$, $LM(p_2)$ = $y^{k_2}$, and $LM(p_3)$ = $z^{k_3}$, where $LM(p_i)$ is the leading monomial of $p_i$, and $k_i \in \mathbb{N}^{+}$.

In general, to decide whether or not an ideal $I \subset \mathbb{Q}[x_1, \ldots, x_n]$ is zero-dimensional, it suffices to compute a (monic) Groebner basis for $I$ and verify that it contains, for each indeterminate $x_i$, a polynomial $p_i$ s.t. $LM(p_i) = x_i^{k_i}$ for some $k_i \in \mathbb{N}^{+}$.