Dimension of quotient ring

Question

What is the dimension of the following quotient ring, $\mathbb{Z}[x,y,z]/\langle xy+2, z+4 \rangle$, where $\mathbb{Z}$ is the ring of integers?

I realized this is isomorphic to $\mathbb{Z}[x,-2/x]$. How does $-2/x$ affect the dimension since the ring is $\mathbb{Z}$.

My apologies if this question sounds repetitive but I am asking only because I am trying to understand an important concept and this example will help me a lot. — Zoey, Aug 17 '15 at 04:56
Can someone give me a hint on how to work at the solution before putting it on hold? — Zoey, Aug 17 '15 at 10:06
@Zoey Have you tried to use similar arguments to the ones from this answer to another question posted by you? — user26857, Aug 17 '15 at 15:39
@user26857 here I thought $2,4$ will create problems. We have $\mathbb{Z}[x,-2/x,-4]$. Therefore dimension is 2? — Zoey, Aug 17 '15 at 15:49

score 2 · Accepted Answer · answered Aug 17 '15 at 18:32

2

Let $R=\mathbb Z[X,Y]/(XY+2)$. We have $\dim R\le2$. Furthermore, since $x$ is a non-zero divisor on $R$ we have $\dim R\ge\dim R/(x)+1=2$. (Note that $R/(x)\simeq(\mathbb Z/2\mathbb Z)[Y]$.)

answered Aug 17 '15 at 18:32

user26857

52,094

Why should $\mathrm{dim} R \leq 2$ ? Also how are these rings related tensor product of $A$- algebras? – Zoey Aug 18 '15 at 01:43
1

@Zoey If $a\in A$ is a non-zero divisor, then $\dim A/(a)\le\dim A-1$. (In our case note that $\dim\mathbb Z[X,Y]=3$.) I've never claimed that such rings are related to the tensor product of algebras. – user26857 Aug 19 '15 at 15:54

Dimension of quotient ring

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