Primary Ideals, S_P0 and S_PH

S_P0 and S_PH

Verify that S_P0 is the saturation of 0. In the same way, S_PH is the saturation of H. This equals R iff H and S intersect, iff H is not contained in P. Keeping H inside P, S_PH also lies in P. This is because P is prime, and if ux lies in H and u is outside of P then x lies in P.

Let Q be a prime ideal that contains P. Select an ideal H, and the elements driven into H by R-Q are also driven into H by R-P. In other words, S_PH contains S_QH.

Quotient Ring

Map P forward to an ideal Q in the quotient ring R/H. Let x belong to S_PH, hence yx lies in H for some y outside of P. In the quotient ring, xy = 0, and y lies outside of Q. This is the definition of S_Q0 in R/H. Conversely, let the image of x lie in S_Q0 in R/H, as directed by an element outside of Q, which has a preimage y outside of P. Now xy lies in H, and x is in S_PH.

In summary, the image of S_PH, under the quotient map R/H, equals S_Q0 in the quotient ring. As a result, theorems about S_P0 often generalize to S_PH by pulling back through a quotient ring.

Minimal Prime and Radicals

Remember that S_P0 contains x if ux = 0. Taking the radical of this ideal, x could be an n^th power. In other words, x is in the radical of S_P0 if uxⁿ = 0 for some n and some u outside of P. Turn this around, and x is not in rad(S_P0) if the powers of x, times the elements of R-P, never reach 0. This happens iff x can be folded into R-P to create a larger multiplicatively closed set. The complement of this larger closed set is a smaller prime. There is a prime below P iff some x ∈ P is not in rad(S_P0). Equivalently, P is minimal iff P = rad(S_P0).

Given an ideal H in P, apply the above to the quotient ring R/H. Pull the result back to R, and P is a minimal prime containing H iff P = rad(S_PH).

d(R)

Suppose x is a nonzero element in the intersection. Place the annihilator [0:x] in a maximal ideal M. Pull this down to a minimal prime ideal P in d(R). Since x is in S_P0, ux = 0 for some u outside of P. Now u is in the annihilator of x, which lives in P. This is a contradiction, hence x cannot exist, and the intersection is empty.

S_PH is P Primary

To prove S_P0 is primary, start with xy in S_P0, even though x is not. Thus uxy = 0. With ux nonzero, y is a zero divisor. If y lies outside of P, x is in S_P0, which is a contradiction. Thus y is in P. If no power of y winds up in S_P0 then y can be added to R-P to give a larger multiplicatively closed set, which leads to a smaller prime. Since P is minimal, some power of y lies in S_P0, and S_P0 is primary.

With P minimal, rad(S_P0) = P, and S_P0 is P primary.

Now S_P0 is suppose to be the smallest P primary ideal. Suppose there is some other ideal H, not containing S_P0, that is P primary. Let x lie in S_P0-H. Now ux = 0, and some power of u lies in H. This places u in rad(H), inside P, yet u lies outside of P.

Conversely, if S_P0 is P primary, then P = rad(S_P0), and P is minimal.

In summary, P is minimal iff P = rad(S_P0), iff S_P0 is P primary - whence S_P0 is the smallest P primary ideal.

Let P be a prime ideal containing a smaller ideal J. Pass to the quotient ring R/J and apply the above. Pull the results back to R, and P is minimal over J iff P = rad(S_PJ), iff S_PJ is P primary - whence S_PJ becomes the smallest P primary ideal containing J.

Second Uniqueness

Let P be a prime ideal containing a smaller ideal J. Assume J has a primary decomposition. Pass to the quotient ring R/J and apply the above. Then pull the results back to R. If P is one of the isolated primes in σ(J), a minimal prime over J, and H is its P primary ideal, we can replace H with S_PJ.

I'm referencing a section that is yet to come, but the saturation of J through a minimal prime P containing J produces the P primary ideal that contributes to the intersection. The primary ideal has to be S_PJ; it can't be anything else. This will become clearer when we characterize the saturations of J.

If J has a primary decomposition, the prime radicals are fixed by first uniqueness, and the primary ideals below the isolated primes are fixed by second uniqueness. However, many different primary ideals are possible below an embedded prime. This is illustrated by the following example.

Example of an Isolated Prime and an Embedded Prime

Let x generate P, an ideal that is both prime and primary.

Let y-ax and x² generate H, for some fixed nonzero value a in K.

If f is a polynomial in P intersect H, part of it could be generated by x², but the other piece must be generated by xy-ax². Combine the x² pieces, and f is generated by x² and xy. In other words, J is the intersection of P and H.

Notice that P does not contain H, nor does H contain P.

Look at the quotient ring R/H. If a term contains x² it vanishes, and y can be replaced with ax wherever it appears. We are left with monomials in x. The nil radical consists of multiples of x, and if you mod out by this nil radical you get K, a field. Thus the nil radical is a maximal ideal. Pull this back to R and nil(H) is a maximal ideal, which I will call Q. With rad(H) = Q, H is Q primary.

The square of P is generated by x², which is part of H, hence P is contained in rad(H), which is Q. In this example, P is isolated and Q is embedded.

Second uniqueness fixes the primary ideal beneath the isolated prime P. Let's follow along as an exercise. We need to saturate J with respect to P. What elements of R are driven into J by elements outside of P? The elements outside of P are the polynomials in y. Take anything in P, generated by x, and multiply by y, and find something generated by xy, which lies in J. Therefore S_PJ = P. Sure enough, P is the primary ideal below P.

Second uniqueness does not constrain the primary ideal below Q. In fact each value of a in K leads to a different primary ideal, and none of these ideals contains the other.