# Krull dimension of $\mathbb{C}[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>$

Krull dimension of a ring $R$ is the supremum of the number of strict inclusions in a chain of prime ideals.

Question 1. Considering $R = \mathbb{C}[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>$, how does one calculate the Krull dimension of $R$? This variety is well-known as the twisted cubic in $\mathbb{P}^3$.

Question 2. In general for any ring $R$, how are the Krull dimension of $R$ and the dimension of Spec$(R)$ related?

Thank you.

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Nitpick: the variety is not the twisted cubic but the affine cone over that cubic, which explains why the Krull dimension of $R$ is $2$. – Georges Elencwajg Jul 6 '12 at 6:52
Thank you Georges! Then what are the equations for a twisted cubic? I thought these were the ones (I thought I saw this from several books)... – math-visitor Jul 6 '12 at 6:54
Dear math-visitor: the confusing point is that the equations are exactly the same! However in the case of the twisted cubic they are to be interpreted as defining a curve in $\mathbb P^3$, rather than a cone un $\mathbb A^4$. So to keep things straight, we write $Proj(R)$ for cubic the and $Spec (R)$ for the cone. Let me make an analogy: the single equation $x=0$ defines a line in $\mathbb A^2$; however the very same equation defines a plane in $\mathbb A^3$. – Georges Elencwajg Jul 6 '12 at 7:09
By the way, you can find an elementary, completely explicit calculation that the ideal defining the cubic is prime here. – Georges Elencwajg Jul 6 '12 at 7:15
@YACP, your last edit was amazingly unnecessary. – Mariano Suárez-Alvarez Jan 14 at 17:12
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Question 1: It is a theorem that $\mathrm{dim}\ A[x]=\mathrm{dim}\ A+1$ for any Noetherian ring $A$, where $\mathrm{dim}$ denotes Krull dimension. Thus $\mathrm{dim}\ \mathbb C[x_1,x_2,x_3,x_4]=4$, as $\mathrm{dim}\ \mathbb C=0$ trivially. The easiest way to compute the dimension of $R$ is to verify that $$P=\langle x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\rangle$$ is a prime ideal, i.e. by the methods I employ here, and computing its height. As a caution, I have not proved that $P$ is prime myself but I believe that it is. To determine its height, notice that $\mathbb C[x_1,x_2,x_3,x_4]$ is a UFD so all prime ideals of height $1$ are principal, while $P$ is easily shown not to be principal. By the Generalized Principal Ideal Theorem, $P$ has height at most it's number of generators, i.e. $3$. Hence we need only determine whether $P$ has height $2$ or $3$. In fact, it has height $3$, which can be shown by verifying that $$0\subset \langle x_1x_3-x_2^2\rangle\subset \langle x_1x_3-x_2^2,x_2 x_4-x_3^2\rangle\subset P$$ is a strict chain of prime ideals, e.g. by the methods in the post I linked to.

Question 2: Yes, the dimensions are equal. See for example Wikipedia.

Edit: Not doing calculations can get you in trouble. As Mariano's answer shows, $P$ has height $2$ rather than $3$ (so apparently the second ideal in the chain I wrote is not prime), hence the dimension of $R$ is $4-2=2$.

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Thank you Alex. I will read what you wrote above carefully. And thanks for the additional link reference! – math-visitor Jul 6 '12 at 3:32
@math-visitor If you have Mathematica I recommend using it, but Sage is also a good option. Also, the calculations are tedious so I haven't done them myself, so there's a chance that my intuition is failing me and one of the ideal I claim is prime, isn't. – Alex Becker Jul 6 '12 at 3:34
Thanks Alex! I thought about doing the above calculations by hand, but I will definitely consider using a computer program. =) – math-visitor Jul 6 '12 at 3:36
@math-visitor Computing Groebner bases by hand? That would be horrid. – Alex Becker Jul 6 '12 at 3:40
@YACP This is the wrong tone to use on this site. – Georges Elencwajg Jan 14 at 20:10

Your ideal is generated by binomials, so one can be smart about it.

There is an algebra map $\mathbb C[x_1,x_2,x_3,x_4]\to \mathbb C[s, t]$ such that $x_1\mapsto s^3$, $x_2\mapsto s^2t$, $x_3\mapsto st^2$ and $x_4\mapsto t^3$, and this map maps your ideal $\mathfrak p$ to zero, so it induces $\phi:\mathbb C[x_1,x_2,x_3,x_4]/\mathfrak p\to \mathbb C[s, t]$. This is, in fact, an injection —try to prove this by hand— so that $\mathfrak p$ is prime because the codomain of $\phi$ is a domain.

It follows, in particular, that $A=\mathbb C[x_1,x_2,x_3,x_4]/\mathfrak p$ is a domain, so its Krull dimension is equal to the trascendence degree of its fraction field $F=\operatorname{Frac}(A)$. In $F$ we have $x_1=x_2^2/x_3$ and $x_4=x_3^2/x_2$, so the inclusion $\mathbb C(x_2,x_3)\to F$ is actually surjective. This makes it obvious that $F$ is of trascendence degree $2$.

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Wow, cool! Thank you Mariano! I wish all quadratic polynomial problems are cool and simple like this. – math-visitor Jul 6 '12 at 4:21
@math-visitor, every variety can be cut by cuadrics (in some sense: mathoverflow.net/questions/11488/varieties-cut-by-quadrics) so you are out of luck :) – Mariano Suárez-Alvarez Jul 6 '12 at 4:38
Thank you for the link-- a lot of info there! Well, I was hopeful for the time-being... – math-visitor Jul 6 '12 at 4:40

I'll prove the following result:

$$K[x_1, x_2, x_3, x_4]/\left< x_1x_3-x_2^2,x_2 x_4-x_3^2,x_1x_4-x_2 x_3\right>\simeq K[s^3,s^2t,st^2,t^3],$$ where $K$ is a field.

Let $\varphi: K[x_1,x_2,x_3,x_4]\to K[s, t]$ be the ring homomorphism that maps $x_1\mapsto s^3$, $x_2\mapsto s^2t$, $x_3\mapsto st^2$ and $x_4\mapsto t^3$. Obviously $\operatorname{Im}\varphi=K[s^3,s^2t,st^2,t^3]$; this is a subring of $K[s,t]$ and the extension $K[s^3,s^2t,st^2,t^3]\subset K[s,t]$ is integral, hence $\dim K[s^3,s^2t,st^2,t^3]= \dim K[s,t]=2.$

It remains to prove that $\ker\varphi=\left< x_1x_3-x_2^2, x_2 x_4-x_3^2, x_1x_4-x_2 x_3\right>$. By definition $\varphi(f(x_1,x_2,x_3,x_4))=f(s^3,s^2t,st^2,t^3)$. In particular, this shows that the polynomials $g_1=x_1x_3-x_2^2$, $g_2=x_2 x_4-x_3^2$ and $g_3=x_1x_4-x_2 x_3$ belong to $\ker\varphi$.

Conversely, let $f\in\ker\varphi$, i.e. $f(s^3,s^2t,st^2,t^3)=0$. We want to show that $$f\in\left< x_1x_3-x_2^2, x_2 x_4-x_3^2, x_1x_4-x_2 x_3\right>.$$ The initial monomials of $g_1$, $g_2$, resp. $g_3$ with respect to the lexicographical order are $x_1x_3$, $x_2x_4$, resp. $x_1x_4$. The remainder on division of $f$ to $G=\{g_1,g_2,g_3\}$, denoted by $r$, is a $K$-linear combination of monomials none of which is divisible by $x_1x_3$, $x_2x_4$, resp. $x_1x_4$. This shows that the monomials of $r$ can have one the following forms: $x_1^ix_2^j$ with $i\ge 1$ and $j\ge 0$, $x_2^kx_3^l$ with $k\ge 1$ and $l\ge 0$, respectively $x_3^ux_4^v$ with $u,v\ge 0$. In order to prove that $f\in\left< x_1x_3-x_2^2, x_2 x_4-x_3^2, x_1x_4-x_2 x_3\right>$ it is enough to show that $r=0$. But we know that $f(s^3,s^2t,st^2,t^3)=0$ and therefore $r(s^3,s^2t,st^2,t^3)=0$. The monomials (in $s$ and $t$) of $r(s^3,s^2t,st^2,t^3)$ are of the following types: $s^{3i+2j}t^j$, $s^{2k+l}t^{k+2l}$, respectively $s^ut^{2u+3v}$. Now check that there is no possible cancelation between these monomials (because they can't be equal), so $r=0$.

Now it follows that $\dim R=2$.

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This proof can be easily generalized in order to show that, in general, the coordinate ring of a rational normal scroll is isomorphic to $K[s^d,s^{d-1}t,\dots,st^{d-1}t^d]$. – YACP Jan 9 at 20:10
Only the first comment seems relevant to the current answer, so I have removed the rest (YACP has saved me some work). – robjohn Jan 14 at 21:58