- Zariski Topology
- Noetherian
- Lownil(R)
- Spec R is Compact
- Irreducible Sets and Dimension
- Noetherian Spectrum is Zariski
- Direct Product
- Examples
- A Contravariant Functor
- The Dimension of R[x]
- Going Up

As promised, I'm going to place a topology on a ring, whereupon a ring homomorphism will induce a continuous function between the two topological spaces. To be precise, the topology is placed on the prime ideals of R. In other words, the prime ideals are the points of the space, and certain prime ideals clump together into open sets. This is the zariski topology, or the spectrum of R, or spec R. Since maximal ideals are prime, spec R is nontrivial, i.e. it has at least one point.

I'll define the closed sets in this topology; take complements for the open sets. For any set of elements E in R, the collection of prime ideals containing E, denoted vE, is a closed set. The complement oE is open. Note that v1 is the empty set, and v0 is every prime ideal of R. The empty set and the entire space are closed and open, as they should be.

Recall that the radical of E, written rad(E), is the intersection of the prime ideals containing E. If H = rad(E), then vH = vE. You can always ratchet E up to its radical ideal without changing vE.

Of course we must prove this is a valid topology. Let E1 E2 E3 etc be sets of elements in the ring R, and let U be their union. If a prime ideal P contains U it contains each Ei, and lives in each of the corresponding closed sets. Conversely, if P lives in each of the closed sets then it contains every Ei, and it contains their union. Therefore the intersection of arbitrarily many closed sets is closed via vU.

Next consider the union of two closed sets vE and vF. Ratchet E and F up to ideals, and let H = E∩F. Let P be a prime ideal in vE∪vF. Now P contains E or F, hence it contains H. The intersection contains the product, hence P contains EF. Conversely, a prime ideal P containing EF contains E or F. Put this all together and vE∪vF = vEF. The finite union of closed sets is closed, and the topology is valid.

In fact this topology has a base, namely ox, the complement of vx, for each element x in R. Consider an arbitrary open set oE with complement vE. vE is the intersection of vx for every x in E. Thus oE is the union over ox. Every open set is the union of base sets.

The intersection of two open sets is open, hence ox∩oy is open, and covered by base open sets. We have a valid base for the topology.

If R is commutative, then ox is empty iff x is nilpotent (x belonging to every prime ideal), and ox is everything iff x is a unit.

If a closed set vE properly contains vF, where E and F are radical ideals, Then there are additional prime ideals containing E that do not contain F. Therefore E is properly contained in F. Chains of closed sets map back to chains of radical ideals, running in the other direction.

If R is noetherian, then spec R is also noetherian. (In this case noetherian refers to the ideals of R, not left or right ideals.) A descending chain of closed sets implies an ascending chain of ideals, which is impossible. Thus there is no infinite chain of descending closed sets in spec R.

Don't be confused about the term noetherian as it is applied to spec R. A topology is noetherian if it has no infinite descending chains of closed sets. In every other context, noetherian refers to ascending chains, but not here. Why the switch? So that R noetherian implies spec R noetherian. That's all.

Similarly, R artinian implies spec R is artinian, with no infinite ascending chains of closed sets.

Don't assume the converse.
Let R be **Q** (the rationals) adjoin infinitely many commuting indeterminants x1 x2 x3 …,
such that each xi raised to the fifth power is 0.
Every indeterminant is nilpotent, and belongs to the single prime ideal P,
which is maximal.
Bring in or take out indeterminants one at a time,
and build an ascending or descending chain of ideals inside P,
hence R is neither noetherian nor artinian.
Yet spec R is a single point,
and is certainly noetherian and artinian.

Let H be the ideal that belongs to every prime ideal of R. In other words, H is the intersection of all the prime ideals of R. This is denoted lownil(R).

A closed set is defined by a set of elements E, which can be raised to an ideal, and then a radical ideal, which is the intersection of all the prime ideals containing E. These prime ideals necessarily contain H. Thus any set E, when raised to a radical ideal, contains H. If no prime ideals contain E then E is the entire ring, which also contains H. In any case, we can assume the closed sets correspond to ideals that contain H.

Mod out by H to get the quotient ring Q. By correspondence, prime ideals in Q correspond to prime ideals in R containing H, but every prime ideal in R contains H, so the map is a bijection. Furthermore, the ideals that generate closed sets also correspond, since they all contain H. The bijection is a homeomorphism. spec R and spec Q are the same topological space. Any information in H is lost.

The previous section asserted spec R is noetherian if R is noetherian. A stronger version states that spec R is noetherian if R/H is noetherian.

Remember that a space is compact iff it is compact relative to its base sets, so assume a collection of base sets covers spec R. The intersection of the complements of the designated base sets is empty. Let E be the set of elements that generate these base sets. E is contained in no proper ideal, and E generates R. Write 1 as a finite linear combination of generators g1 g2 g3 … gn from E, with coefficients from R on either side. The base open sets associated with g1 through gn cover spec R. This is a finite subcover, hence spec R is compact.

If spec R is noetherian, e.g. as when R/lownil(R) is noetherian, then every subspace of spec R is noetherian, and every subspace of spec R is compact.

If R is commutative, the base open set ox can be characterized. If x is nilpotent then it belongs to every prime ideal. This makes ox empty, which isn't very interesting, so assume x is not nilpotent. Let S be the fraction ring of R with powers of x in the denominator. By correspondence, the prime ideals of S are the prime ideals of R that miss x, which are the points of ox. Extention into the fraction ring S is a bijection between ox and spec S.

Let E be the radical ideal that is the intersection of certain prime ideals missing x. vE gathers these prime ideals together into a closed set. Fractions and intersection commute, thus the image of E in S is the intersection of the corresponding prime ideals in S. The image of a closed set is closed. Conversely, start with a radical ideal F in S, the intersection of certain prime ideals in S, and pull this system back to R. Let E be the preimage of F. E is contained in the preimages of all the prime ideals that contain F. Raise E up to the intersection of these prime ideals in R. Push this forward to S and the resulting ideal contains F, and is still contained in all the prime ideals. Thus E is still the preimage of F. It has not changed at all. E is the intersection of the corresponding prime ideals in R, giving a closed set. A set is closed in ox iff it is closed in spec S. The map is a homeomorphism, and the spaces are the same.

Spec S is compact, hence each base open set ox in spec R is compact.

It follows that an open set in spec R is compact iff it can be represented as a finite union of base open sets. This includes oE where E is finitely generated. The closed set of prime ideals containing g1 g2 g3 … gn is the intersection of the closed sets based on each generator gi. Take complements and oe is the union of the base open sets based on these generators.

Recall that a space is irreducible if every pair of nonempty open sets intersects. Equivalently, every pair of proper closed sets fails to cover the entire space. Is spec R irreducible?

Spec R is not irreducible iff there are two proper closed subspaces that cover all of spec R. this happens iff the ideals A and B are not in rad(0), so that vA and vB are proper, yet every prime ideal contains either A or B, contains A ∩B, contains AB, and contains either A or B. In other words, rad(0) contains AB, but neither A nor B.

In summary, spec R is irreducible iff rad(0) is prime.

Is the closed subspace vE irreducible, for a given ideal E? Again, this space is not irreducible iff we can find vA and vB that cover vE. We may as well assume A and B contain E, since our subspace is restricted to those prime ideals that contain E. Neither A nor B lies in rad(E), else one of the closed sets is the entire space. At the same time, AB lies in rad(E), else vA and vB would miss a prime ideal containing E. Once again rad(E) is not prime.

In summary, vE is irreducible iff rad(E) is prime, iff E is a prime ideal.

The dimension of a space is one less then the longest chain of irreducible closed sets, and the dimension of a ring is one less than the longest chain of prime ideals. Happily, these two definitions coincide. Prime ideals correspond with irreducible closed sets, an ascending chain of prime ideals becomes a descending chain of irreducible closed sets, and the dimension of R equals the dimension of spec R.

An irreducible closed set is called an irreducible component if it is maximal, i.e. at the top of its chain. This happens iff the corresponding ideal is a minimal prime. The irreducible components of spec R correspond to the minimal prime ideals of R.

Since every irreducible closed set lies in an irreducible component, every prime ideal descends to a minimal prime ideal. You can see a more direct proof of this fact here.

If R/lownil(R) is noetherian, spec R is a zariski space. In other words, spec R is noetherian, half hausdorff, and generic. We prove that spec R is noetherian in an earlier section; that leaves half hausdorff and generic.

Let P and Q be distinct prime ideals in R. Assume, without loss of generality, that P is not contained in Q. Thus vP is a closed set that includes P, but not Q. The complement is an open set that contains Q, but not P. This holds for any P and Q, hence spec R is T0.

As per the previous section, a set is closed and irreducible in spec R iff it is vP for some prime ideal P. Consider the closure of P in spec R. This is the intersection of all closed sets vE where P contains E, which is simply vP. Therefore P is the generic point for vP, and spec R is generic.

Note that spec R is rarely hausdorff. For instance, let R be an integral domain, and the 0 prime ideal is in every nonempty open set. There simply aren't disjoint open sets.

Assume R is the finite direct product of rings R1 R2 R3 … Rn. spec R is the disjoint union of the individual spaces spec Ri.

First look at the space of spec R as a set. Review the characterization of maximal and prime ideals in a direct product. An ideal in R is prime iff it is the direct product of a prime ideal Pi in Ri and all the other component rings. Sure enough, this is the union of all the points across spec Ri. In fact this defines a canonical injection. The point Pi in spec Ri maps into spec R by crossing Pi with all the other rings, thus producing the point P in spec R. This map is an embedding, and together, all the spectra from the component rings cover spec R.

Consider the closed set vE in spec R, where E is an ideal. Project E into each of the component rings, giving ideals Ei. A prime ideal P, with a proper projection Pi in Ri, contains E iff Pi contains Ei. Apply this criterion as i runs from 1 to n, and vE is the union of the individual closed sets defined by Ei. Conversely, closed sets, defined by Ei, combine to form a closed set in R, by setting E to the direct product of Ei.

Injection is a function from one space into another, and the preimage of a closed set is closed, according to Ei, hence the function is continuous. The image of a closed set is also closed, via Ei crossed with all the other rings, hence the function is bicontinuous. In fact it is an embedding of the space spec Ri into spec R.

The subspace determined by spec Ri inside spec R is both open and closed. Spec R includes an independent, disconnected subspace for each ring Ri.

The above reasoning remains valid if R is an infinite direct sum of component rings, but remember, the resulting ring does not contain 1. That's ok; 1 is not required to derive spec R.

How bout the converse?

Remember that all information about the nil radical is lost when viewing spec R, so assume rad(0) is 0, or, mod out by rad(0) to create a ring wherein rad(0) = 0. Now every nonzero x in R evades at least one prime ideal.

If a subspace is open and closed, then there are two closed sets vE and vF that are disjoint, and cover spec R. If x ∈ E times y ∈ F is nonzero then there is some prime ideal P that does not contain ef, and vE∪vF does not cover spec R. Therefore the ideals E and F are orthogonal.

If some nonzero x lies in both E and F, then some prime ideal P is not in vE or vF. Therefore E and F are disjoint orthogonal ideals. They act as independent rings within R.

If E and F do not span 1, then they lie in some maximal ideal P, which is common to both vE and vF. The individual rings span 1, and cover all of R. Therefore R is the direct product of the two subrings E and F.

If spec R separates into finitely many disjoint spaces, apply the above repeatedly, and R is the finite direct product of rings.

As a corollary, the spectrum of a local ring is connected. If it were disconnected, one could produce a maximal ideal for each component.

The spectrum of a field, or any simple ring, is a single point corresponding to the 0 ideal, which is the only proper ideal, maximal and prime. One step up from this is a pid, where 0 is still prime, and every nonzero prime ideal is maximal. The spectrum has a generic point at 0, and a row of primes above it, all of them closed points in the spectrum, and all of them brought in by the closure of 0. 0 is the generic point for spec R, and spec R is a zariski space with dimension 1. The finite union of closed points is closed, generated by the product of the corresponding ideals, or elements.

Polynomials over a field F form a pid, a ring of dimension 1 and a zariski space of dimension 1. 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 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 next generator g is either 0 or an irreducible polynomial over F. 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. This time F is the fraction ring of R, rather than a quotient ring of R. Once again F[x] is a pid. Let g be the generator of P. Multiply through by a common denominator to find g primitive in R[x].

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 F[x], hence {q,g(x)} is a closed point in spec R.

The closure of {q} or {g} brings in {q,g}. Finally, 0 is generic for the entire space, as with any integral domain.

To illustrate, consider **Z**[x].
Let h = x2+1.
This generates a prime ideal, which implies a point in spec R.
Take the closure of this point.
If p = 3 mod 4 then h remains irreducible mod p.
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.)

If you are familiar with category theory, then you know commutative rings and ring homomorphisms form a concrete category, that I will call CR. Topological spaces and continuous functions form another category CT. Spec R implements a contravariant functor from CR into CT.

We already know how to carry rings to topological spaces; let's see how a ring homomorphism can induce a continuous function between those spaces. The adjective contravariant means the functions run in the other direction. If a ring homomorphism h maps R into S, the induced function g carries spec S into spec R.

Every ideal in S has a preimage in R, which is called the contraction of that ideal. When rings are commutative, prime ideals pull back to prime ideals. this establishes the function g from spec S back to spec R.

Note that every prime in the image of g contains the kernel K of the ring homomorphism h. Thus the image of spec S lies in the closed set vK.

Let vE be closed in spec R. Assume E is a radical ideal, as large as it can be. If E does not contain K then it does not intersect the image of g at all. Its preimage under g is empty, which is closed. This isn't very interesting, so assume E contains K. E is an ideal, and it determines a well defined image set F in S. A prime ideal Q in S pulls back to a prime P in R containing E iff Q contains F. In other words, the preimage of vE under g is vF. This is a closed set, hence g is continuous.

Verify that the composition of ring homomorphisms upstairs induces the composition of continuous functions downstairs. This completes the characterization of spec R as a functor.

g need not be bicontinuous,
as when h embeds the integers into the rationals.
0 is closed in **Q**, which maps to 0 in **Z**, which is not closed.
But we can still examine the closure of g(vF).
There is no harm in pushing F up to a radical ideal.
F is now the intersection of various prime ideals Qi in S.
Let E be the preimage of F under h.
The primes Qi pull back to primes Pi in R,
and these all contain E.
Suppose their intersection includes more than E.
The additional element maps to something in S that is outside of F, and still in each Qi.
That is a contradiction, hence E is a radical ideal.

The image of vF is a subspace of vE. Any other closed set containing the image is based on an ideal inside of E, hence vE is the smallest closed set containing the image, and the closure of the image.

Let B be an ideal in S with rad(B) = F. B defines the same closed set in S as F. Let E be the preimage of F as before, hence E is a radical ideal. Let A be the preimage of B. Let x in E have image y in F. Some power of y lies in B, hence some power of x lies in A. E = rad(A), and both E and A give the same closed set in spec R. The closure of g(vB) = vA.

Apply the above to a single point in spec S. If g(Q) = P, then g applied to the closure of Q has closure equal to the closure of P.

Let e be an element of R, with image f in S. Every prime Q in S missing f pulls back to a prime P in R missing e. Conversely if Q contains f then P contains e. Combine these results and the preimage, under g, of the base open set oe, is of. Similarly, the preimage of ve is vf.

Is g(spec S) dense in spec R? It is, iff the kernel K of h lies in nil(R). If K is not in the nil radical then some prime ideal does not contain K, and cannot be the image of anything under g. Therefore oK is a nontrivial open set disjoint from the image of g.

Conversely, assume nil(R) contains K. let oE be an open set in R, and ratchet E up to a radical ideal, whence E contains K. To be nontrivial there are prime ideals in R that do not contain E. Suppose all prime ideals in S contain h(E), whence every element in h(E) is nilpotent. This means every element of E, raised to some power, winds up in K, which is in the nil radical, hence everything in E is nilpotent. This contradicts the fact that some prime ideals do not contain E. Therefore there must be some prime ideal in S not containing h(E), that maps to a prime ideal in R not containing E, and the image of g intersects every open set in spec R.

For the last theorem of this section, assume h is surjective, and maps R onto S. Primes in S correspond 1 for 1 with primes of R containing K. The image of spec S is the closed subspace vK. Since ideals in R and S correspond, closed sets correspond, and g embeds spec S into spec R.

This is an important theorem in algebraic geometry: the dimension of the polynomial ring R[x] is 1 more than the dimension of R. By induction, you can adjoin l indeterminants, and the dimension increases by l.

Let P be a prime ideal in R and consider P[x], the polynomials with coefficients in P. This is an ideal Q in the ring R[x], and it happens to be a prime ideal. This is familiar territory is it not? You saw the proof for commutative rings here, and even for noncommutative rings here, but they were part of larger theorems about radical ideals, so let's run through the proof one more time. In this formulation, R and x commute, but R need not be commutative.

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 still 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]. In addition, you 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.

If P is maximal and R is commutative, 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, 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 mod g. The extension gives another division ring, hence 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. Then push a 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, adjoin x to give the maximal ideal, thus increasing the dimension by 1.

Don't worry, I'm done with division rings. For the rest of this section, rings are assumed to be commutative.

Let R be an integral domain, so that R[x] is an integral domain. Thus 0 is a prime ideal in both rings. I'm looking for prime ideals lying over 0. These are prime ideals that yield 0 when intersected with R. These are prime ideals wherein none of the polynomials are constants. Everything has degree 1 or higher. 0 is of course such a prime ideal; are there any others?

The nonzero elements of R form a multiplicatively closed set. The prime ideals that do not contain constant polynomials are the prime ideals that miss this multiplicative set. Since prime ideals correspond under localization, carry these prime ideals into the fraction ring F[x]. Since F[x] is a pid, every nonzero prime ideal is maximal. We cannot have a chain of three prime ideals. Start with 0, then jump up to the maximal ideal generated by g(x) irreducible, and you're done.

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 with all other polynomials, divided by all elements of R, provided the result lies in R[x]. 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. The two prime ideals forming a chain over 0 are 0 and g(x).

Just as 0 has two primes over it, 0 and g(x), so P has two primes over it. The first is the least prime containing P, namely P[x]. If S = R/P, then mod out by the prime ideal P[x] and get S[x]. These are polynomials in x with coefficients in S. Remember that S and S[x] are integral domains. By the above, 0 is one prime ideal in S[x] , and there is one more above it, defined by an irreducible polynomial g. These pull back to two prime ideals in R[x] lying over P. The lower prime ideal is of course P[x]. The latter is generated by p and g. 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, or noetherian, which implies a factorization domain, 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, just as cosets of P are represented by elements of R. Once again g is primitive, with coefficients in R. g, in concert with P, fractionally generates the prime ideal, or it generates the prime ideal directly if S is a ufd.

For the rest of this section, R is a ufd. The dimension of R[x] is precisely 1 more than the dimension of R. 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. Let S = R/P, whence the quotient ring S is a ufd. 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. S being a ufd, J is directly generated by p and g.

Step up to the next prime Q containing P. Within R, Q is minimal over P. The image of Q is minimal over S, and by the quotient lemma, 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. It is generated by q and h. 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 you're curious about the step from g to h in the last section, it's not hard. Let S = R/P, and let T = R/Q. Since g remains nonzero in T[x], the prime ideal over Q, containing g, is generated by q and h, where h is primitive and irreducible in T[x]. In fact h generates g, and is a factor of g. Now I know we said g is irreducible, but that was relative to S[x]. Map everything in S onto T, including the coefficients of g. The new polynomial may be reducible in T[x]. Either g remains irreducible, or it has an irreducible factor h, which does the trick.

If the higher prime ideal is not given, and you want to construct it, let h be any irreducible factor of g in T[x]. There may be several factors to choose from, but you must pick one, then move on to the next prime, and so on. This is the chain lifting property that is sometimes called "going up". The going up theorem usually refers to an integral extension of R, but it applies here as well. An ascending chain of prime ideals in R lifts to a chain of prime ideals in R[x], with the option of one more prime ideal somewhere along the way.