Zero-dimensional ideals in polynomial rings

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.

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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.

Since the algebraic set $V(I)$ contains the one-dimensional (infinite!) line $y=z=0$, it is not zero-dimensional.