Spec R, Examples

Fields

The spectrum of a field or division ring, or any simple ring, is a single point. The only proper ideal is maximal, and prime.

Dedekind Domain

Let R be a dedekind domain. The maximal ideals are closed points in spec R. Any finite union of these is closed, and the generating ideal is equal to their product. If the 0 ideal is part of a closed set, the result is all of spec R. Thus 0 is the generic point for spec R, and spec R is a generic T₀ space.

As a special case, a dvr is dedekind with one maximal ideal. Its spectrum has two points, and all sets are closed except {0}.

Polynomials over a Field

The ring F[x] is a pid, which is dedekind, hence its prime ideals and its spectrum are well characterized. In this case the prime ideals correspond to the irreducible polynomials over F. These include, but are not limited to, x-a for every a in F.

If F is algebraically closed, such as the complex numbers, there are no other irreducible polynomials, and spec R comprises a closed point for each element of F, plus the generic zero ideal.

If F is the reals there are additional closed points, corresponding to x²+bx+c where 4c > b².

Polynomials over a PID

If R is a pid, the ring R[x] is noetherian, and a ufd. Any prime ideal P has a finite set of generators g₁ g₂ g₃ etc. If a generator is the product of two polynomials in R[x], we can replace it with one of the two factors, since P is prime. Thus we can assume each generator is a prime element of R, or a primitive irreducible polynomial in R[x].

Suppose one of the generators is q, a prime in R. So far R could have been a ufd, but at this point we need it to be a pid. Thus q generates a maximal ideal, and R/q is a field F. Now P includes all the polynomials whose content is divisible by q. Mod out by this ideal and P remains prime in F[x]. Since F[x] is a pid, the generator g is either 0 or an irreducible polynomial. Therefore P could be generated by q, or q and g(x), where g is irreducible mod q.

If P contains two primes in R, use bezout's identity to span 1; hence P does not contain more than one element from R.

Now assume there are no generators from R. Since P contains nothing in R, take the fractions by R^*, and find a corresponding prime ideal. Once again F[x] is a pid. Let g be its generator. Multiply through by a common denominator to find g primitive in R[x]. This is the generator for P.

In summary, the points in spec R[x] correspond to the primitive irreducible polynomials in R[x], or {q,g(x)} where q is prime in R and g(x) is irreducible mod q, or {q}, or {0}.

If {q,g(x)} is contained in a larger ideal, mod out by q and find something larger in F[x]. Yet g generates a maximal ideal in the quotient ring, hence {q,g(x₀} is a closed point in spec R.

The closure of {q} or {g} brings in {q,g}. Also, the closure of {h(x)} includes {q,g} if h is no longer irreducible mod q, and g is one of its irreducible factors. Finally, 0 is the generic point, as usual.

To illustrate, consider Z[x]. Let h = x²+1. This generates a prime ideal, which implies a point in spec R. Let's take the closure of this point. If p = 3 mod 4 then h remains irreducible. The prime ideal {p,h} is brought into the closure. If p = 1 mod 4 then h splits into x+c and x-c mod p. Two prime ideals are brought in: {p,x+c} and {p,x-c}. (These coincide if p = 2.)