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Given the ring, $\mathbb{Z}_6[x,y]/\langle x \rangle$. What is the Krull dimension of the ring? Isn't the following a chain of prime ideals in the ring, $\langle \overline{2}\rangle \subsetneq \langle \overline{2},\overline{3}\rangle \subsetneq \langle \overline{2}, \overline{3}, \overline{y}\rangle$? Therefore the dimension is 2?

Zoey
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Your chain is not a chain of prime ideals, since $1 \in \langle 2,3 \rangle$.

Your ring is clearly isomorphic to $\mathbb Z_6[y]$. One easily shows, that this ring is one-dimensional:

On the one hand, we have chain of length $1$: $\langle 2 \rangle \subsetneq \langle 2,y\rangle$.

Now let $\mathfrak p$ be a prime ideal of height $1$, $\mathfrak p \cap \mathbb Z_6$ is a prime in $\mathbb Z_6$, hence $\mathfrak p \cap \mathbb Z_6 =\langle a \rangle$ with $a \in \{2,3\}.$ Furthermore, we have $\mathfrak p \neq \langle a \rangle$, since the latter is of height $0$. We can of course assume $a=2$, it works the same with $a=3$:

$$\mathbb Z_6[y]/\mathfrak p \cong \mathbb Z_6[y]/\langle 2, \mathfrak p \rangle = \mathbb Z_2[y]/(\mathfrak p/\langle 2 \rangle).$$

Since $\mathbb Z_2[y]$ is a PID and $\mathfrak p/\langle 2 \rangle$ is a non-zero prime-ideal, the latter is a field. Hence $\mathfrak p$ was maximal and we are done.

MooS
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  • Thank you. I was wrong in identifying prime ideals. So dimension of $\mathbb{Z}_6$ is $0$. – Zoey Apr 26 '16 at 09:22
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    Yes, any finite commutative ring is zero-dimensional, since if we take the quotient by any prime ideal, we end up with a finite integral domain, which is well known to be a field. Thus any prime ideal is in fact maximal. – MooS Apr 26 '16 at 13:55