Krull dimension of $\mathbb{C}[x,y,z,w]/(xw-yz)$

Question

I have the following exercise:

Consider the rings $A:=\mathbb{C}[x,y,w,z]/(xw-yz)$ and $B:=A/(\bar{x}, \bar{y})$.

(i) Calculate the Krull dimensions of $A$ and $B$.

(ii) Consider the prime ideal $P=(\bar{w}, \bar{z})\subset A$ and let $Q$ be its image in $B$. Calculate the height of $P$ and the height of $Q$.

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(i)

• Dimension of A

I know that $\mathbb{C}[x,y,w,z]$ has Krull dimension$=4$, because $\mathbb{C}$ is a field. So I expect dim$A<4$ (upper bound).

The first problem is that $A$ is a integral domain...so can I be sure that dim$A<4$ and not dim$A\leq 4$? I think no: if $A$ had been a integral domain, a chain in $\mathbb{C}[x,y,w,z]$ could always have extended to a chain in $A$ by addition of $(0)$.

However, if really dim$A<4$, I would like to find a chain of prime ideals of $A$ with lenght 3. In this case a prime ideal of A has height$=3$ and by definition of Krull dimension, dim$A\geq3$ (lower bound) so I can conclude dim$A=3$. (Though this is just a supposition, I'm not really sure that dim$A=3$). But I can't find the ideals that form the chain!

• Dimension of B

Because of $(0)\subset(\bar{x})\subset(\bar{x,}\bar{y})$, can I immediately conclude dim$B=2$? I'm not sure...Can dim$B$ not depend on dim$A$?

(ii)

• Height of P

$A$ is a integral domain, so we have $(0)\subset(\bar{w})\subset(\bar{w},\bar{z})=P$ and height of $P$ is 2, isn't it?

• Height of Q

I don't know how to calculate its height. Perhaps because $B$ is obtained by quoting $A$ with $(\bar{x},\bar{y})$, instead $P$ is generated by the two remaining $\bar{z},\bar{w}$,its image in $B$ has the same height.

If you can help me, thank you so much.

Answer 1

There's a theorem to the effect that if $f(X_1,\ldots,X_n)$ is a non-zero polynomial over a field $K$ then $K[X_1,\ldots,X_n]/f(X_1,\ldots,X_n)$ has Krull dimension $n-1$. It's a special case of the theorem that if $R=K[X_1,\ldots,X_n]/I$ for some ideal $I$, then the Krull dimension of $R$ is the transcendence degree of the field of fractions of $R$ over $K$.

To find a chain of prime ideals in $R=K[X_1,\ldots,X_n]/f(X_1,\ldots,X_n)$ think geometrically and find a solution $(a_1,\ldots,a_n)$ of $f(X_1,\ldots,X_n)=0$ and consider the chain $(0)\subseteq(X_1-a_1)\subseteq(X_1-a_1,X_1-a_2)\subseteq\cdots \subseteq(X_1-a_1,X_1-a_2,\ldots,X_{n-1}-a_{n-1})$. If you choose $(a_1,\ldots,a_n)$ carefully this will be a suitable chain of prime ideals.

Note that $$B\cong\frac{\Bbb C[x,y,w,z]}{(x,y,xw-yz)}.$$ There is a convenient alternative way to write the ideal $(x,y,xw-yz)$ which makes the structure of $B$ more evident.

Answer 2

Generally, for any domain $R$ that is a finitely generated algebra over a field and any ideal $I$ of $R,$ we have that $$\operatorname{height} I + \dim(R / I) = \dim R.$$ Every polynomial ring in finitely many indeterminates over a field $k$ is a domain and a finitely generated $k$-algebra, hence this equation holds. By Krull's Height Theorem, we have that $\operatorname{height} I = \operatorname{height}(wx - yz) \leq 1.$ But $I$ is a nonzero prime ideal of $\mathbb C[w, x, y, z],$ hence we have that $\operatorname{height} I = 1.$ Consequently, we have that $$\dim(\mathbb C[w, x, y, z] / (wx - yz)) = \dim \mathbb C[w, x, y, z] - \operatorname{height} I = 4 - \operatorname{height} I = 3.$$ Observe that a maximal chain of prime ideals in $\mathbb C[w, x, y, z] / (wx - yz)$ is given by $$(\bar 0) = (\overline{wx - yz}) = (\bar w \bar x - \bar y \bar z) \subsetneq (\bar x, \bar y) \subsetneq (\bar x, \bar y, \bar z) \subsetneq (\bar w, \bar x, \bar y, \bar z).$$

By the Third Isomorphism Theorem, we have that $B \cong \mathbb C[w, x, y, z] / (x, y).$ Once again, we have that $\operatorname{height} (x, y) \leq 2,$ and we have a maximal chain of prime ideals $(0) \subsetneq (x) \subsetneq (x, y),$ from which it follows that $\dim B = \dim \mathbb C[w, x, y, z] - 2 = 4 - 2 = 2$ with a maximal chain of prime ideals $(\bar 0) \subsetneq (\bar x) \subsetneq (\bar x, \bar y).$

For the ideal $P = (\bar w, \bar z)$ of $A,$ you are correct that the height is $2,$ as evidenced by your maximal chain of prime ideals. For the ideal $Q,$ observe that by our above isomorphism, we have that $Q = (\bar w, \bar z)$ in $\mathbb C[w, x, y, z] / (x, y) \cong \mathbb C[w, z],$ hence $Q$ is maximal, and its height is $\dim \mathbb C[w, z] = 2.$