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Let $A$ be a commutative ring with identity. Let $p, q$ be two distinct prime ideals. Is it possible that $p \subseteq q$?

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Krull dimension of a ring is the maximum $n$ such that there exists a chain of prime ideals $\mathfrak{p}_0 \subsetneqq \mathfrak{p}_1 \subsetneqq \cdots \subsetneqq \mathfrak{p}_n$. – Andrea May 15 '12 at 17:52
up vote 7 down vote accepted

Sure. In $\mathbb{Z}$ we have $(0) \subseteq (p)$ for $p$ any prime number. In fact, arbitrarily long, or even infinite chains of containments, are possible. For instance, in $k[x_1, x_2, x_3, \ldots]$, the polynomial ring in infinitely many variables, the ideals $(x_1)$, $(x_1, x_2)$, $(x_1, x_2, x_3), \ldots$ are all prime and each is contained in the next.

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