If $\mathbb K$ is a field I would like to find $$\dim \mathbb K[x,y,z]/(xy,xz,yz).$$

I'm starting to study the concept of dimension of rings and I don't know the basic tools and techniques to discover the dimensions of non-trivial rings like this one. I would be very grateful if someone could help me to solve this question which it will open my mind regarding this new subject.

Thanks in advance


Since I didn't understand Martin's solution, I'm using another strategy, in order to solve this question, I'm trying to find an isomorphism between $\mathbb K[x,y,z]/(xy,xz,yz)$ and a ring which we know its dimension. I didn't find any possible candidate to this isomorphism, I would like to know if it's possible to solve this question using this strategy.

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2 Answers2


The algebraic variety $V(xy,xz,yz) \subseteq k^3$ is the union of the three coordinate axes $k \times 0 \times 0$, $0 \times k \times 0$ and $0 \times 0 \times k$. These are the irreducible components, and the dimension is the maximum dimension of an irreducible component. Hence, the dimension is $1$. This proof works if $k$ is algebraically closed and assumes some basic knowledge in algebraic geometry. Therefore, let me give a more direct proof which also works when $k$ is an arbitrary field.

A prime ideal in $k[x,y,z]/(xy,xz,yz)$ corresponds to a prime ideal $\mathfrak{p}$ of $k[x,y,z]$ which contains $xy$, $xz$ and $yz$. Since $xy \in \mathfrak{p}$, we have $x \in \mathfrak{p}$ or $y \in \mathfrak{p}$. Assume w.l.o.g. that $x \in \mathfrak{p}$. Then $\mathfrak{p}$ corresponds to a prime ideal of $k[y,z]/(yz)$. Thus, a prime ideal chain in $k[x,y,z]/(xy,xz,yz)$ starting with a prime ideal containing $x$ is really the same as a prime ideal chain in $k[y,z]/(yz)$. But this ring has dimension $1$: A prime ideal chain starts with a prime ideal containing, say $y$ (the case of $z$ works the same), hence corresponds to a prime ideal chain of $k[z]$, which has dimension $1$.

The same proof shows: If $k$ is an arbitrary commutative ring, then $\dim \, k[x_1,\dotsc,x_n]/(x_i x_j : 1 \leq i<j \leq n) = 1.$

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Martin Brandenburg
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  • I didn't understand this part: "Then $\mathfrak{p}$ corresponds to a prime ideal of $k[y,z]/(yz)$." why? For me it would be $k[x,y,z]/(yz)$, instead of $k[y,z]/(yz)$ – user75086 Jun 10 '14 at 09:58
  • We have $(k[x,y,z]/(xy,xz,yz))/(x) = k[y,z]/(yz)$. – Martin Brandenburg Jun 10 '14 at 10:05
  • I'm sorry disturbing you Martin I have so many questions, I didn't understand this equality and why you take $(k[x,y,z]/(xy,xz,yz))/(x)$. – user75086 Jun 10 '14 at 15:21
  • The following Lemma is used all the time: If $I \subseteq A$ is an ideal, then there is an order-preserving correspondence between prime ideals of $A$ containing $I$ and prime ideals of $A/I$. – Martin Brandenburg Jun 10 '14 at 17:12

An usual dimension formula says that $$\dim K[X_1\dots,X_n]/I=n-\operatorname{ht} I,$$ where $\operatorname{ht} I$ is the minimum of $\operatorname{ht} P$, with $P$ a minimal prime ideal containing $I$.

In your case there are exactly three minimal primes containing $I$, namely $(x,y)$, $(x,z)$ and $(y,z)$, all having height two. In conclusion the height of $I$ is two and therefore the dimension of your ring is one.

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  • Why the fact that there are exactly three minimal primes containing $I$ all having height two implies that the height of $I$ is 2? – user75086 Jun 10 '14 at 12:49