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I have read in multiple places that a field $K$ has a Krull dimension of $0$. How is this true? Isn't $(0)\subset K$ a prime ideal in $K$? Obviously $K$ is an integral domain.

Thanks in advance!

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Sure, but $K$ is not a prime ideal, so the inclusion $0\subset K$ does not count towards the dimension. – anon Jul 23 '13 at 6:51
up vote 6 down vote accepted

The Krull dimension is defined to be the length of the longest chain of prime ideals. In other words, a ring $R$ is said to have a Krull dimension $n$, if the longest possible chain of prime ideals $$ \mathfrak{p}_{0}\subsetneq\mathfrak{p}_1\subsetneq\cdots\subsetneq\mathfrak{p}_n$$ has length $n$.

Since $0$ is the only prime ideal in a field (in fact the only proper ideal in a field), it follows that Krull dimension of a field is 0 (since there is no way to have $0\subsetneq\mathfrak{p}$ for another prime ideal $\mathfrak{p}$).

Added. In case it wasn't made clear, $K$ itself is not a prime ideal of $K$, because by definition prime ideals must be proper ideals.

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Oh. Thanks. I had thought that if there is a single prime ideal in the chain, then the length of the chain is one. – Ayush Khaitan Jul 23 '13 at 7:01
@AyushKhaitan: You are welcome. Yeah I think I had that confusion at some point too. Thing to take from this: The length of the chain is always one less than the number of terms in the chain. – Prism Jul 23 '13 at 7:04

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