Is Krull dimension equal to maximal length of chain of irreducible varieties over arbitrary field?

Given an affine variety $X$, we know the prime ideals in its coordinate ring $\mathbf{K}[X]$ correspond to irreducible subvarieties in $X$ when the underlying field $\mathbf{K}$ is algebraically closed. Thus the length of maximal chain of prime ideals $P_d \subsetneq P_{d-1} \subsetneq ...\subseteq P_0$ equals the length of maximal chain of irreducible subvarieties $X_0 \subsetneq X_1 \subsetneq ... \subsetneq X_d$. Thus these two definitions of dimension coincide.

But when $\mathbf{K}$ is not algebraic closed, the correspondence between prime ideals and irreducible subvarieties no longer holds. Do we still have the equality of two definitions of dimension?

• Yes, if you use schemes (and you should!). – KReiser Nov 12 '16 at 22:30
• Yes, even if you do not use schemes, but your varieties include (by definition) the points that are not $\mathbf{K}$-rational (i.e. zeros of polynomials are cut from $\mathbb{A}^n_{\overline{\mathbf{K}}}\equiv \overline{\mathbf{K}}^n$ and not only $\mathbb{A}^n_{\mathbf{K}}\equiv \mathbf{K}^n$). – Pavel Čoupek Nov 13 '16 at 3:46