_rqy's Blog

0%

Solutions to the exercises of Introduction to Commutative Algebra, Atiyah, Chapter 8.

## Artin Rings

Exercise 1. Let be a minimal primary decomposition of the zero ideal in a Noetherian ring, and let be -primary. Let be the th symbolic power of (Chapter 4, Exercise 13). Show that for each there exists an integer such that .

Proof. Since is Noetherian and , it follows that there exists such that . Taking on both side, we have .

Suppose is an isolated primary component. Then is an Artin local ring, hence if is its maximal ideal we have for all sufficiently large , hence for all large .

Proof. If is an isolated primary component, then by definition is a minimal prime ideal. Hence is an Artin local ring. By (8.6), for all sufficiently large , hence for all large .

If is an embedded primary component, then is not Artinian, hence the powers are all distinct, and so the are all distinct. Hence in the given primary decomposition we can replace by any of the infinite set of -primary ideals where , and so there are infinitely many minimal primary decompositions of which differ only in the -component.

Proof. If is an embedded primary component, then is not minimal. Hence is not Artinian. By (8.6) again, the powers are all distinct, so are all distinct. Hence replacing by where gives another primary decomposition of , which is also minimal by the 1st uniqueness theorem of primary decompositions.

Exercise 2. Let be a Noetherian ring. Prove that the following are equivalent:

 1 is Artinian; 2 is discrete and finite; 3 is discrete.

Proof. i) ii) iii) is clear.

Suppose is discrete, then particularly is Hausdorff. Hence every prime ideal of is maximal. So is Artinian by (8.5).

Exercise 3. Let be a field and a finitely generated -algebra. Prove that the following are equivalent:

 1 is Artinian; 2 is a finite -algebra.

Proof.

 i) ii): If is Artinian, then is a finite direct product of Artin local rings by (8.7). Clearly each multiplicand is also a finitely generated -algebra. So it reduces to the case that is local. Let be the maximal ideal of , then is a finite extension of by the Hilbert Nullstellensatz.If , then ; each is a -vector space of finite dimension since is Noetherian. Hence is finite. Therefore is also finite. ii) i): Every ideals of is also a -vector subspace of . Since is a finite-dimensional -vector space, it satisfies d.c.c on ideals.

Exercise 4. Let be a ring homomorphism of finite type. Consider the following statements:

 1 is finite; 2 the fibres of are discrete subspaces of ; 3 for each prime ideal of , the ring is a finite -algebra ( is the residue field of ); 4 the fibres of are finite.

Prove that i) ii) iii) iv).

Proof.

 i) iii): If is finite, then also is . Therefore is a finite -algebra. ii) iii): Let denote . The fibres of are . Since is of finite type, there is a surjective homomorphism . Tensoring with gives a surjective homomorphism . Therefore is a finitely generated -algebra, so Noetherian.By Exercise 2, discrete if and only if is Artinian. By Exercise 3 it’s equivalent to that is a finite -algebra. iii) iv): Under the same notation above, iii) implies that is Artinian. Hence is finite.

If is integral and the fibres of are finite, is necessarily finite?

Proof. No. If is a field and is an algebraic extension, and the inclusion map, then is integral (algebraic), the fibres of are finite (since and are one-point spaces). But is not necessarily finite, for example if and .

Exercise 5. In Chapter 5, Exercise 16, show that is a finite covering of (i.e. the number of points of lying over a given point of is finite and bounded).

Proof. Let and denote the coordinate ring of and respectively. Then can be considered as the subring of . By the Nullstellensatz, and . And if denotes the inclusion map, then by Chapter 5, Exercise 16, is the induced mapping of . Since is integral over , is finite. By Exercise 8.4, the fibres of are finite.

But we want to bound the size of the fibres of . Suppose is generated by elements . Then for each maximal ideal of (also the points in ), . Let denote , then is generated by ; so . Let be the prime ideals of , then they are coprime since is Artinian; hence , so . Hence we have for all .

Exercise 6. Let be a Noetherian ring and a -primary ideal in . Consider chains of primary ideals from to . Show that all such chains are of finite bounded length, and that all maximal chains have the same length.

Proof. Let . Then primary ideals lying between and correspond to ideals of . Clearly is an Artinian local ring. Then the result follows immediately from (6.8).