Integral extensions and going up
Use monic equations to control algebraic extensions and distinguish the direction and hypotheses of prime-chain lifting theorems.
On this page
The question for today
What you’ll understand
- Verify integrality and distinguish finite modules from finite-type algebras.
- Apply lying over and going up with the correct given prime.
- State the normal-domain hypothesis of a sufficient going-down theorem without confusing it with going up.
An element is integral when it satisfies a monic polynomial over the base ring. The monic condition is what keeps its powers inside a finite module span; an arbitrary polynomial relation does not provide the same control. Integral extensions also organize the prime spectrum: primes exist above every base prime, and a chosen prime can be extended upward along a base chain. Going down asks for a different lift with a different prescribed endpoint. It is not another name for going up and requires additional hypotheses in the theorem we will state.
Monic equations bound the necessary powers
For an inclusion A⊆B of commutative unital rings, b is integral over A if it satisfies b^n+a_{n−1}b^{n−1}+⋯+a₀=0 with coefficients in A. Higher powers can then be reduced to the first n powers, so A[b] is finite as an A-module. Conversely, a finite A-algebra is integral: multiplication by an element acts on a finite generating list, and the determinant trick yields a monic annihilating polynomial. Applying that polynomial to the algebra’s identity turns the module relation into the desired element relation.
Finitely many integral elements generate a finite A-module algebra by adjoining them one at a time. Thus an integral algebra of finite type is finite as a module. A finite-type algebra need not be integral: A[x] has one algebra generator but arbitrarily many independent powers as an A-module when A is a nonzero ring. Integrality is also transitive and behaves well under localization and quotient. These properties allow prime arguments to reduce to local or residue situations without discarding the monic equations that supplied the original finiteness control.
Lying over and going up choose different data
For an integral inclusion A⊆B, lying over says every prime p of A is the contraction of some prime q of B. Going up says that if p⊆p′ and q over p is already chosen, there exists q′ over p′ containing that chosen q. The direction matters: we keep the lower prime upstairs and extend the chain upward. These are theorem statements; a standard proof localizes and passes to suitable quotients, using the fact that an algebra integral over a field has maximal prime ideals. No Noetherian hypothesis is needed for these integral-extension results.
Incomparability supplies a further constraint: two comparable distinct primes upstairs cannot contract to the same prime downstairs in an integral extension. It prevents an integral map from hiding a nontrivial upstairs chain entirely over one base point. These facts do not imply each prime has only one lift, or that the lift is unramified. For example, arithmetic extensions can split a base prime into several primes or give a repeated factor. Lying over asserts existence, not uniqueness, and going up preserves a chosen chain rather than selecting a canonical lift.
Going down needs its own hypotheses
Going down starts with p⊆p′ and a chosen q′ over the upper prime p′, then seeks q⊆q′ over p. One sufficient theorem assumes A and B are domains, B integral over A, and A integrally closed in its fraction field. Under those hypotheses going down holds. The normality requirement here means integrally closed domain; it is not a claim that every base ring in every going-down result must be normal. Flat maps provide another route to going down, introduced later.
An integral extension without this extra condition still has going up, but one cannot simply invoke the normal-domain going-down theorem. Failure of that sufficient hypothesis does not by itself prove going down fails in the particular example. This distinction is especially relevant for cusp subrings, whose normalization adjoins a missing integral parameter. Tracking which ring is the base, which ring is integrally closed and which prime endpoint is prescribed avoids a large class of incorrect chain-lifting arguments in dimension theory.
Work it through
The cusp acquires its missing parameter
Let A=k[t²,t³]⊂B=k[t]. Show B is finite integral over A and lift (0)⊂(t²,t³) to B.
The element t satisfies X²−t²=0 with coefficient t² in A. Also every polynomial in t is in A+At, so B is generated by 1,t as an A-module.
The prime (t) in B contracts to the polynomials in A with zero constant coefficient, namely (t²,t³). The zero prime contracts to zero because both rings are domains.
Thus (0)⊂(t) is an explicit going-up lift. The base A is not integrally closed: t lies in its fraction field as t³/t² and is integral, but t is not in A.
The example verifies finite integrality and an upward chain. Its nonnormal base means the stated normal-domain going-down theorem cannot be invoked automatically, not that this specific chain proves a going-down failure.
The living notebook
Explore the model
Interactive lab · function
A monic relation does not determine a unique branch
Pause and Reset, then set a=1,x=1. Predict y=1 on the quadratic y=x². For a fresh comparison Reset and set a=1,x=−1; predict the same y=1. This illustrates the two possible parameters satisfying the monic equation T²−1=0 over fields of characteristic not two. The real graph is not a prime-spectrum map and does not prove lying over or distinguish arithmetic ramification.
- y at chosen x
- 0.608
- Tangent slope · 2ax
- 1.35
- Signed area · ax³/3
- 0.182
This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA monic equation controls powers but need not make the element or the prime above a point unique.
Pause & reason
Check your understanding
Your turn
Try it yourself
For Z⊂Z[i], verify integrality and show that (1+i) lies over (2). Explain why this does not make (2) prime in Z[i].
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use the monic equation for i and the quotient map sending i to 1 in F₂.
Compare with the solution
The element i satisfies X²+1=0 and Z[i]=Z+Zi is module-finite, hence integral. The quotient by (1+i) is F₂, so that ideal is prime and its contraction to Z is (2). But (1+i)²=2i, hence (2)=(1+i)² as ideals because i is a unit. The ideal (2) is not prime: the nonzero class of 1+i in Z[i]/(2) has square zero. A prime above a base prime need not equal the extended base ideal.
Keep the idea
Take it with you
- Monic relations give bounded power spans and finite-type integral algebras are module-finite.
- Integral inclusions have lying over, incomparability and going up.
- Going down is a different endpoint problem with its own sufficient hypotheses.
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