Let H be a radical ideal that defines an algebraic set. A chain of irreducible regions within H′ pulls back to a chain of prime ideals containing H. This corresponds to a chain of prime ideals in the quotient ring K[x1,x2,…xn]/H, which is the coordinate ring. The dimension of an algebraic set is equal to the dimension of its coordinate ring.
Let's have a look at this coordinate ring. It is a finitely generated K algebra, which can be normalized into a transcendental extension followed by an integral extension. An integral extension always has the same dimension as its base ring. This is because chains of prime ideals correspond. Therefore the dimension of the coordinate ring equals the dimension of its transcendental subring.
Since K is a field it is a ufd. Adjoin l indeterminants and the resulting dimension is l. Put this all together and the dimension of the algebraic set H′ is the transcendent degree of the coordinate ring. It doesn't matter which chain of prime ideals we select; the answer is l. The dimension of an algebraic set is well defined.
The entire space Cn is the region associated with the 0 ideal. Mod out by 0 and the coordinate ring is the ring of polynomials. This has transcendent degree n, hence the dimension of Cn is n. This is reassuring; the dimension of n space should be n.
Suppose a proper algebraic set with nonzero ideal H has dimension n. This implies a chain of length n+1 in the polynomial ring, starting with H. Bring in the 0 ideal and the chain has length n+2, which is impossible. Every proper closed set has dimension less than n.
At the other end of the scale, let H′ be a 0 dimensional variety. Assume K = C, that is, K is algebraically closed. A 0 dimensional region corresponds to a maximal ideal, and these have been characterized. The 0 dimensional regions are the points of Cn, just as you would expect.
At this point the rationale for the dimension of a ring becomes clear. It corresponds to the traditional definition of dimension in n space. It (roughly) represents the number of free variables, the transcendentals that are not nailed down by the ideal. For dimensions between 0 and n, the algebraic set looks like l dimensional space, locally, though it may curve about in higher dimensions. For example, a paraboloid looks 2 dimensional at the local level, though it curves in 3 space. The corresponding equation z = x2+y2 is an irreducible polynomial, generating a prime ideal, and the quotient ring, replacing z with x2+y2, is 2 dimensional.