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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.

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Mathsstudent147
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    Some possibly relevant facts: $\mathbb{Q}[x,y,z]$ has dimension 3. A field has dimension 0. An integral domain that's not a field has dimension >0. – vadim123 May 03 '13 at 22:29
  • @MattE the definition I know is.... An ideal $I \subset k[x_1,....,x_k]$ is zero dimensional if and only if the k- vector space dimension $dim_k R/I$ is finite. – Mathsstudent147 May 05 '13 at 00:03
  • Dear Mathstudent, Okay, so this is pretty clearly equivalent to the Krull dimension of the quotient being zero. I'll write an answer in a minute. Cheers, – Matt E May 05 '13 at 02:21
  • https://mathoverflow.net/questions/129634/how-to-show-an-ideal-is-zero-dimensional – user26857 Feb 07 '19 at 22:25

3 Answers3

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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.

Matt E
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  • Thank you.. I know there is another way of doing this by finding the Groebner basis.. Are you aware of this method? I can compute The Groebner basis but I'm not sure what to do with that to then show if is zero dimensional or not.. – Mathsstudent147 May 05 '13 at 10:04
  • @MattE If I understand well you want to show that there are only finitely many maximal ideals containing $I$. How this follows from proving that the defining equations of $I$ have only finitely many solutions modulo any maximal ideal? –  May 05 '13 at 14:23
  • @YACP: Dear YACP, Showing that there are finitely many maximal ideals containing $J$ is the same as showing that there only finitely many homomorphisms $\mathbb Q[x,y,z] \to \overline{\mathbb Q}$. This what the above computation shows. Regards, – Matt E May 06 '13 at 01:45
  • @YACP I tried to compute the Grobner Bases by hand, but I struggled.. I have since tried using Maple just to find the Grobner Basis.. and have found that with lex x>y>z it has lots of terms with powers $z^6$ etc...

    But with lex z>y>z it is... $G:= (x^3+x-x^4+x^6, -x^3+x-y+x^2y, 4y^3 -4x^3 +x^4 +3x^2+4x^5+y^2-2y^3+y^4, -y^2+xy+2zx, x^4 -x^2 +y^2-y^3x+2zy,y^3+z^2+1)$

    I'm a little confused as I wouldn't be able to compute this in an exam..

     – Mathsstudent147 May 06 '13 at 12:52
  • Theorem 3.5.1 (7) in "A Singular Introduction to Commutative Algebra" states that an ideal is zero-dimensional iff a (global) Grobner basis contains, for every variable $x_i$, an element whose leading term is $x_i^{a_i}$, for some $a_i$ – Brent Baccala May 27 '13 at 04:28
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Since the algebraic set $V(I)$ contains the one-dimensional (infinite!) line $y=z=0$, it is not zero-dimensional.

Georges Elencwajg
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  • Dear Georges, This is certainly more elegant and direct than citing the Hauptidealsatz! Best wishes, – Matt E May 06 '13 at 01:45
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    Dear Matt, I can't begin to say how I appreciate the fair-play of your comment, especially coming from a mathematician of your calibre. – Georges Elencwajg May 06 '13 at 07:31
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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.

user26857
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