The dimension of a field is 0 and the dimension of a dedekind domain is 1.
An ideal H can also have a dimension, refering to the longest chain inside H, including H itself if H is prime. the dimension of R is at least the dimension of R/H plus the dimension of H. This conservative bound assumes H is prime, and has been counted twice.
If S is an integral extension of R, use chain correspondence to show they have the same dimension. If one ring has no chains longer than n, then the other ring cannot either, and the (finite) dimensions coincide. If chains are arbitrarily long in one ring, they are arbitrarily long in the other, hence R has infinite dimension iff S has infinite dimension.
Suppose u and v are polynomials in R[x], with uwv in Q for every polynomial w, even though u and v are not in Q. Furthermore, select u and v so that the sum of their degrees is minimal. Since uwv lies in Q for all polynomials w, restrict w to R, whence urv lies in Q. The lead coefficient of the product always winds up in P. Since P is prime, one of our factors, say u(x), has a lead coefficient c in P. Subtract this lead term, giving another polynomial u′ that is not in Q. However, u′wv = uwv - cxiwv is still in Q for every w. This contradicts our selection of u and v. Therefore Q = P[x] is a prime ideal in R[x].
By adjoining x, a chain of prime ideals in R becomes a chain of prime ideals in R[x]. (I will still talk about primes lying over primes, even though this is not an integral extension.) In addition, we can place one more ideal at the top. When P is a maximal ideal in R, i.e. at the top of its chain, the ideal generated by P and x properly contains P[x], and is maximal in R[x], with quotient R/P. Thus the dimension of R[x] is greater than the dimension of R.
Of course we need not adjoin x; any irreducible polynomial over R/P will do. The quotient becomes a field, a finite extension of the field R/P.
If R is noncommutative, and P is maximal (as above), and R/P happens to be a division ring (which need not be the case), and if g is irreducible, then any h(x) with degree less than g can be combined with g, using euclid's gcd algorithm, to get 1. Backtrack, and write 1 as a linear combination of g and h, which makes h invertible. Thus the extension gives another division ring, and adjoining g gives a maximal ideal.
The same results hold for P in the power series ring R[[x]]. Review the proof that P[[x]] is a prime ideal. This time u and v are chosen so that the sum of the degrees of their least terms is minimal.
Push the chain of prime ideals in R out to a chain of prime ideals in R[[x]]. At the top, when R/P is a field, or a division ring, the power series in R/P becomes a complete dvr, and the only thing you can adjoin to create the maximal ideal is x.
Don't worry, I'm done with division rings. For the rest of this page, rings are assumed to be commutative.
For convenience, scale g(x) so that its coefficients all lie in R. If they have a common factor, divide it out. If R is a factorization domain, we can't keep dividing factors forever, thus g becomes a primitive irreducible polynomial in R[x].
The prime ideal implied by g is the product of g and all other polynomials, divided by all elements of R, provided the result lies in R[x]. Thus the ideal is fractionally generated by g.
If R happens to be a ufd, and gh/d lies in R, then gh has content d or greater, and since g is primitive, h has content d or greater. (See gauss' lemma for more details.) In other words, h/d is a polynomial in R. Thus the prime ideal is principal, generated by g. Conversely, an irreducible polynomial g becomes prime in the ufd R[x], and generates a prime ideal.
Remember that the coefficients of g have been scaled, so that they lie in S, rather than the fraction field of S. If S is a factorization domain, e.g. when R is a ufd (see below), or noetherian, divide the coefficients of g by any common factors, until g is primitive. Finally, you have the option of representing the coefficients of g with elements of R, since P is already part of the ideal.
If Q is a minimal nonzero prime ideal in S, let z lie in Q and factor z into primes. If f is a prime factor of z it generates a prime ideal inside Q, which has to be Q. Thus Q is principal, generated by f.
Let R have dimension n-1, so that the longest chain in R has length n. We know R[x] has a chain of length n+1. Suppose R[x] admits a chain that is at least n+2. Contract this back to an ascending sequence of prime ideals in R. Some of the primes may repeat, and some may be missing.
As we step through the chain of primes in R, each prime P in the chain acts as a base for one or more primes in the chain in R[x], with P[x] being the smallest possible prime lying over P. Mod out by P, and (R/P)[x] becomes an integral domain that supports at most two prime ideals, as shown above. Each prime P in R implies 0 1 or 2 prime ideals in R[x], P[x] being the base, and perhaps another prime ideal beyond P[x] that does not bring in any new constants. The chain in R[x] is finite, bounded by 2n.
March up the chain of primes in R, and note the number of primes in R[x], being 0 1 or 2. 0 is losing ground, 1 is keeping even, and 2 is gaining ground. In order to exceed n+1, we have to find 2, 1, 1, 1, … 2. Without this sequence, we can't produce a chain of length n+2.
Let P be the first prime in the aforementioned sequence. The two primes lying over P are P[x] and P[x] adjoin g(x), where g is irreducible and primitive, with coefficients in R/P. Let J be the prime ideal containing P and g.
Step up to the next prime Q containing P. Within R, Q is minimal over P. Since R is a ufd, Q becomes principal in R/P, generated by some f. If the coefficients of g all lie in Q they are all divisible by f, and that contradicts the fact that g is primitive. Thus g has coefficients that do not belong to Q. This means Q[x] cannot contain J, and cannot be part of our chain. The presence of g pushes Q[x] up to a higher prime ideal. This prime looks like Q[x] adjoin h(x), where h is irreducible and primitive over R/Q. This continues up the chain, until you run into the second prime ideal in R with two primes lying over it; which is impossible.
In summary, the dimension of R[x] is one more than the dimension of R when R is a ufd. By induction, the dimension of R adjoin l indeterminants is l more than the dimension of R.
If the higher prime ideal is not given, and you want to construct it, let h be any irreducible factor of g in the new ring/field. This creates a new prime that contains the original, and it implements the chain lifting property that is sometimes called going up.