Zero-dimensional ideals in polynomial rings 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 an ideal $I := (xy^2+2xz-yz, x^2yz+y^2)$.  Explain why $I$ is not zero-dimensional.

 A: Since the algebraic set  $V(I)$ contains the one-dimensional (infinite!)  line $y=z=0$, it is  not zero-dimensional.
A: Let $I$ be an ideal in $k[x_1,\ldots,x_n]$ ($k$ a field).
Standard theorems in commutative algebra show that TFAE:


*

*$k[x_1,\ldots,x_n]/I$ is finite dim'l over $k$.

*$k[x_1,\ldots,x_n]/I$ has Krull dimension zero.

*$I$ is contained in only finitely many prime ideals.

*$I$ is contained in only finitely many maximal ideals.
If these conditions hold, and if $\mathfrak m_1, \ldots, \mathfrak m_r$ 
are the finitely many maximal ideals containing $I$, then $k[x_1,\ldots,x_n]/I$ 
is the product of  its localizations at the various $\mathfrak m_i$,
and so its dimension is the sum of the dimensions of these localizations.
In the case at hand, we have $J = (y^2 - xy  - 2zx, y^3 + z^2 + 1, (x^2 -1)yz ).$ So if $\mathfrak m $ is a maximal ideal containing $J$, with residue field $k$,
then in $k$ we have the following equations:


*

*$(x^2-1)yz = 0.$

*$y^2 - xy - 2zx = 0.$

*$y^3 + z^2 + 1 = 0.$
It is pretty easy to check that the only solutions to these are


*

*$x = 1$, and $y$ and $z$ satisfy $y^2 - y - 2z = y^3 + z^2 + 1 = 0$.

*$x = -1$, and $y$ and $z$ satisfy $y^2 + y + 2z = y^3 + z^2 + 1 = 0$. 

*$x = y = 0$, $z^2 + 1 = 0$.

*$x  = y$, $y^3 + 1 = 0$, $z = 0$.
(Note in particular that from $y^3 + z^2 + 1 = 0$, at least one of $y$ or $z$ has non-zero image in $k$.)
There are only finitely many solutions in $y$ and $z$ to $y^2 - y - 2z = y^3 + z^2 + 1 = 0$ (indeed, these reduce to an equation of degree $4$ in $y$), and similarly with $y^2 + y + 2 z = y^3 + z^2 + 1 = 0$, and so we see that there 
are only finitely many maximal ideals containing $J$.
Thus $J$ is zero dimensional.

In the case when $I$ is generated by only two elements, the Hauptidealsatz shows that $\mathbb Q[x,y,z]/I$ has Krull dimension $\geq 1$, and so $I$ is not zero-dimensional.
A: Use the Finiteness theorem as explained in Cox et al.: Using Algebraic Geometry.
Zero-dimensional means that the quotient algebra $K[x_1,\dots,x_n]/I$ is finite dimensional. Compute a Groebner basis $G$ of $I$ w.r.t. any global ordering.
Then the quotient is finite dimensional if and only if for each $i$, there is an integer $m_i\ge 0$ such that $x_i^{m_i}$ is the leading monomial of some $g$ in $G$. 
In particular, the size of $G$ is at least $n$, the number of variables; above $n=3$.
This is not the case above.
