**Title:** Commutative Algebra
**Course web site:** `http://www.math.umd.edu/~jmr/603/`
**Meeting times:** MWF, 9:00am-9:50am (MTH 0411)
**Instructor:** Professor
Jonathan
Rosenberg. His office is room 2114 of the Math Building,
phone extension 55166, or you can contact him by
email.
Office hours are Mondays and Wednesdays 10-11, or by appointment.
Professor Rosenberg will be away in Japan from February 16 to
February 24. Professor Washington will teach the class on
February 16 and 18, and **there will be no class on
February 21 or 23**. Class will resume as usual on February 25.
Also, there will be no class on Monday,
April 25.

**Text:**
M. Atiyah and I. MacDonald, *Introduction to Commutative Algebra*,
Westview Press,
ISBN 978-0-201-40751-8. Other good references for certain topics are:

- N. Bourbaki,
*Commutative Algebra*, various editions in French and English. Reprinted, Springer, 1998. The bible on the subject, but probably much more than you really want to know. - D. Eisenbud,
*Commutative Algebra, with a view toward algebraic geometry*, Springer, 1995. A slightly different spin than A-M, slightly more advanced also, and covering more topics. - H. Matsumura,
*Commutative Algebra*, 2nd ed., Benjamin, 1980. A very good book, somewhat more advanced than A-M. - M. Reid,
*Undergraduate Commutative Algebra*, Cambridge University Press, Cambridge, 1995. Slightly lower level than A-M. - R. Y. Sharp,
*Steps in commutative algebra*, London Mathematical Society Student Texts, 19. Cambridge University Press, Cambridge, 1990. 2nd edition, 2000. Less terse than A-M, and at a slightly lower level. - O. Zariski and P. Samuel,
*Commutative Algebra*, 2 vols., Springer, 1958. The original classic text on the subject. Still valuable for some things.

**Prerequisite:** Math 600 or equivalent, or in other words,
one semester of graduate-level abstract algebra, getting up to basics
of rings, modules, and ideals. Math 601 can be taken concurrently.

**Catalog description:**
Ideal theory of Noetherian rings, valuations, localizations, complete local rings, Dedekind domains.

Commutative algebra means the study of commutative rings and modules over them. The subject is motivated by applications in algebraic geometry, number theory, and algebraic topology. This course will be a basic introduction to commutative algebra, starting from the basics, and covering what you need to know for basic applications to algebraic geometry, number theory, and algebraic topology. We will cover all of Atiyah-MacDonald, and then depending on time, we will cover some more modern topics, especially those involving homological and category-theoretic methods.

Homework will be assigned, collected and graded regularly, but there will be no exams. If you want to get an "A" in the course, you need to do most of the homework in a satisfactory way.

**Course evaluations.** Please
fill out the on-line course
evaluation questionnaire at CourseEvalUM before May 11.

**Conventions for these problems:** "ring" means commutative ring
with unit, unless stated otherwise. Integral domains are also
assumed commutative with unit. **Z** denotes the ordinary integers,
**N** denotes the non-negative integers,
**Q** denotes the rational numbers, and
**F**_{q} denotes the finite field of
*q* elements.

- Exercises on ideals (based on A-M, Ch. 1) (due Monday, January 31)
- Show that an integral domain with only finitely many elements
is a field. Deduce that if
*R*is a ring with only finitely many elements, then every prime ideal of*R*is maximal, and the nilradical and Jacobson radical of*R*coincide. - Show that if
*R*is a ring, the set of zero-divisors in*R*is a union of prime ideals of*R*. (This is A-M, Ch. 1, #14.) - If
*R*is a ring and*J*is an ideal in*R*, the*nilradical of*, denoted √*J**J*, is the set of*x*∈*R*with*x*∈^{n}*J*for some*n*. Show that √*J*is the intersection of the prime ideals containing*J*. In particular, it's an ideal containing*J*. - Let
*R*=**F**_{3}[*x*,*y*]/(*xy*,*x*^{2}). Give as explicit a list as possible of all the prime ideals in*R*. Which ones are maximal? What are the radical and nilradical of*R*?

- Show that an integral domain with only finitely many elements
is a field. Deduce that if
- Exercises on modules (based on A-M, Ch. 2) (due Friday, February 11)
- Show that the finite generation hypothesis in Nakayama's Lemma cannot
be removed, even over a local ring. (Hint: Let
*R*be the local ring of germs at 0 of continuous (real-valued) functions on small intervals (-ε, -ε) of the real line. Show that the maximal ideal*M*of*R*is equal to*M*^{2}. Why would this contradict Nakayama's Lemma if it weren't for the finite generation hypothesis?) - Show that if
*R*is a ring and*M*is an*R*-module with the property that every finitely generated submodule of*M*is flat, then*M*is flat. - Let
*R*be a ring. An*R*-module is called*torsion-free*if*x*in*M*,*a*in*R*, and*ax*= 0 implies*x*= 0 or*a*= 0. Show that if*R*is a PID and an*R*-module*M*is torsion-free, then*M*is flat. Use the previous problem plus the fact (which you can assume) that every finitely generated torsion-free module over a PID is a free module. - Let
*R*be a ring. Show that an*R*-module*M*is flat if and only if it has the property that every linear dependence relation in*M*comes from relations in*R*, i.e., that if Σ_{i}*a*_{i}*x*= 0 in_{i}*M*, with*a*in_{i}*R*and*x*in_{i}*M*, then there exist finitely many*b*in_{ij}*R*and*y*in_{j}*M*such that*x*= Σ_{i}_{j}*b*_{ij}*y*for each_{j}*i*and Σ_{i}*b*_{ij}*a*= 0 for each_{i}*j*.

- Show that the finite generation hypothesis in Nakayama's Lemma cannot
be removed, even over a local ring. (Hint: Let
- Exercises on localization (based on A-M, Ch. 3) (due Friday, February 25)
- Let
*R*be a local ring with maximal ideal*p*. Show that the localization map*R*→*R*is an isomorphism._{p} - Use Nakayama's Lemma to show that a finitely generated projective module
*P*over a local ring*R*is free (of finite rank). (Hint: reduce*P*modulo*m**P*,*m*the maximal ideal, to get a vector space over the field*R*/*m*, choose a basis and pull back. You want to show you get a free basis for*P*. Use the projectivity by doing this simultaneously for*P*and for a complement*Q*, whose direct sum is a free module.) Then show that if*R*is any ring and*P*is a finitely generated projective module, then for any prime ideal*p*of*R*,*P*is a finitely generated projective module over_{p}*R*, hence a free module of finite rank, and_{p}*p*→ rank*P*is a well-defined continuous function Spec_{p}*R*→**N**.

Solution to the hardest part: If*P*is a finitely generated projective*R*-module, and if*p*is a prime ideal of*R*, then*P*is a finitely generated projective module over the local ring_{p}*R*, hence it is free by the previous part, and so its rank is well defined. We need to show that the rank is a continuous function of_{p}*p*∈ Spec*R*. If*P*is a direct summand in*R*, then^{m}*P*=*R*(as a left^{m}A*R*-module) for some idempotent*m*×*m*matrix*A*with entries in*R*. Then rank*P*is the rank of_{p}*A*after localizing the entries at*p*. So rank*P*=_{p}*n*means there is an*n*×*n*submatrix of*A*with determinant not in*p*, and that any larger square submatrix of*A*does have determinant in*p*. But the minor determinants of*A*are elements of*R*, and for one of them to lie in*p*is a closed condition in the Zariski topology. (The prime ideals containing a given ideal are a closed set in Spec*R*.) Thus the condition of*P*having rank less than a specified value is a closed condition. But the condition of_{p}*P*having rank greater than a specified value is equivalent to the complement of_{p}*P*in*R*having rank less than a specified value, which is a closed condition on minor determinants of 1 −^{n}*A*, so rank*P*is both upper and lower semicontinuous and is thus continuous._{p} - Let
*R*=**Z**[*x*]/(*x*^{2}− 1). Show that*J*= (*x*− 1) is a prime ideal in*R*and show that*R*_{J}is isomorphic to**Q**. What is the kernel of the map*R*→*R*_{J}? - Suppose (
*a*) is an ideal in*R*,*R*a PID, and suppose the cyclic*R*-module*R*/(*a*) vanishes when localized at (*p*),*p*irreducible. What are the possibilities for*a*?

- Let
- Exercises on primary decomposition (based on A-M, Ch. 4) (due Monday,
March 7)
- Show that if
*R*is the ring of continuous real-valued functions on [0, 1], then the zero ideal in*R*is not decomposable.

Solution to this problem: Suppose we had primary ideals*Q*_{1}, ...,*Q*, with intersection equal to the zero ideal. The associated prime ideals_{n}*P*_{1}, ...,*P*would have to arise as radicals of annihilators of elements_{n}*f*of_{j}*R*. But the annihilator*I*of a function*f*∈*R*consists of all functions*g*∈*R*that vanish on the closure of the set*E*of all*x*∈ [0, 1] where*f*(*x*) ≠ 0. This ideal*I*is determined by*E*, which can be any closed subset of [0, 1] with dense interior. Furthermore,*I*is always equal to its radical, since*R*/*I*≅*C*(*E*) has no nilpotents. But*I*can never be a prime ideal, since if*I*≠*R*,*f*(*x*) ≠ 0 on some open interval (*a*,*b*), and*C*(*E*) has zero-divisors (take a function vanishing on [*a*, (*a*+*b*)/2] and another one vanishing on [(*a*+*b*)/2,*b*]). So there can't be any associated prime ideals. - Let
*R*=**C**[*x*,*y*]/(*x*−*y*^{2}). Find a minimal primary decomposition of the ideal*I*in*R*generated by [the image of]*x*− 1, resp., (*x*− 1)^{2}. - (A-M, #16, p. 56) Assume that every ideal in
*R*is decomposable. Show that the same property holds for any localization*S*^{−1}*R*of*R*.

- Show that if
- Exercises on integral closure (based on A-M, Ch. 5) (due Monday,
March 14)
- Show that if
*k*is a field and*R*=*k*[*x*,*y*]/(*x*^{2}−*y*^{3}), then*R*is an integral domain but is*not*integrally closed (in its field of fractions). (From the algebraic geometry point of view, that means the curve*x*^{2}=*y*^{3}is not a normal variety.) What is the integral closure*S*? Show explicitly that*S*is finitely generated as an*R*-module. - Do A-M problem 1 from Ch. 5, that an integral homomorphism of
rings induces a
*closed*mapping of spectra. - Let
*d*be a square-free integer (so*d*is not a perfect square and is not divisible by any perfect square > 1), and let*R*=**Z**[√*d*]. Show that*R*is integrally closed in its field of fractions*unless**d*≡ 1 mod 4, in which case it is*not*integrally closed. - Do A-M problem 2 from Ch. 5, that given a homomorphism
*A*→ Ω with*A*a ring and Ω an algebraically closed field, and given*B*⊇*A*integral over*A*, the homomorphism extends to one*B*→ Ω. Exhibit a case where*B*is an extension of*A*which is*not*integral and no such extension exists. Is integrality*necessary*for the extension to exist? Why or why not?

- Show that if
- Exercises on going down, valuations, Noether
normalization, and the Nullstellensatz (based
on A-M, Ch. 5) (due Monday, March 28)
- Show that the hypothesis that
*B*be an integral domain is not superfluous in the Going Down Theorem (A-M Theorem 5.16), as follows. Let*A*=**Z**,*B*=**Z**[*x*] / (2*x*,*x*^{2}−*x*). Check that*A*is integrally closed in its field of fractions and that*B*is integral over*A*, but that*B*is not an integral domain. Then check that ``going down'' fails for (0) ⊂ (2) in**Z**, (2,*x*− 1) over (2) in*B*. (Here we are using the slight abuse of notation of letting*x*also stand for its image in*B*.) - One version of the Nullstellensatz asserts that if
*K*is an algebraically closed field and*A*is a finitely generated*K*-algebra ("finitely generated" here means*as an algebra*, not as a*K*-vector space), then the inclusion*K*→*A*splits (i.e., there is a homomorphism of*K*-algebras in the other direction, such that the composite*K*→*A*→*K*is the identity). Prove this assertion as follows. Choose a maximal ideal*I*in*A*. Then show that*A*/*I*has to be algebraic over*K*, hence equal to*K*. (Hint: Show that a transcendental extension of*K*cannot be finitely generated as a*K*-algebra.) Show also that the theorem is false if*K*=**R**, or if*A*is only assumed countably generated, but not finitely generated, over*K*. - An
*ordered abelian group*means an abelian group (written additively) with a translation-invariant transitive and reflexive relation ≥, such that for every*x*, either*x*≥ 0 or −*x*≥ 0, and if both*x*≥ 0 and −*x*≥ 0, then*x*= 0. Let*K*be a field. A*valuation*on*K*is defined to be a homomorphism ν:*K*^{×}→*G*, where*G*is an ordered abelian group (caution: we write*G*additively but*K*^{×}multiplicatively) with the property that ν(*x*+*y*) ≥ min(ν(*x*), ν(*y*)) whenever this makes sense (i.e., none of*x*,*y*, or*x*+*y*are zero).- Show that any ordered abelian group is torsion-free. (Hint:
first show that
*x*≥ 0 implies*nx*≥*x*for*n*positive.) - Let ν:
*K*^{×}→*G*be a valuation. Show that it must send any root of unity to 0. (Use #1.) - Give an example of a valuation ν on
**Q**, with*G*=**Z**with the usual ordering and with ν(2) = 1. (By #2, you know that ν(−1) = 0.) - Show that if ν is a valuation on a field
*K*, then {*x*: ν(*x*) ≥ 0} ∪ {0} is a valuation ring in*K*(in the sense of A-M, Ch. 5). - What is the valuation ring in your example from #3?

- Show that any ordered abelian group is torsion-free. (Hint:
first show that

- Show that the hypothesis that
- Exercises on chain conditions and Noetherian
rings (based on A-M, Chs. 6-7) (due Wednesday, April 6)
- Do A-M exercises 5, 6, 8, 12 on page 79. These are all about Noetherian topological spaces and applications to Spec of a ring. Terminology: "quasi-compact" means "compact but not necessarily Hausdorff", i.e., each open covering has a finite subcovering, but distinct points need not have disjoint neighborhoods.
- Let
*R*be a Noetherian ring and let*S*be the set of non-zero divisors in*R*. Show that*S*^{−1}*R*has only finitely many maximal ideals. Also show by example that this is not necessarily the case if*R*is not Noetherian.

Solution to this problem

- Exercises on Artinian
rings and discrete valuation rings
(based on A-M, Chs. 8-9) (due Wednesday, April 13)
- Let
*G*be a finite abelian group, let*K*be a field, and let*R*=*K**G*be the group ring of*G*over*K*. Show that*R*is Artinian. What can the local factors of*R*look like? Show that*R*is semisimple (has trivial radical) exactly when all the local factors of*R*are fields, which happens if and only if the characteristic of*K*is relatively prime to the order of*G*. (Hint: if*x*∈*G*has order*q*and*q*is a power of the characteristic of*K*, then (*x*− 1)^{q}= 0.) - (A-M Ch. 9, exercise 3) Show that any Noetherian valuation ring is automatically a discrete valuation ring.
- Deduce from exercise 2 above that if
*R*is a PID (for example,**Z**or a polynomial ring in one variable over a field) and if*P*is a non-zero prime ideal of*R*, then the localization of*R*at*P*is a discrete valuation ring.

- Let
- Exercises on Dedekind domains
and completions
(based on A-M, Chs. 9-10) (due Friday, April 29)
- Let
*R*=**R**[*x*,*y*]/(*x*^{2}+*y*^{2}−1) be the ring of polynomial functions on the unit circle. - Show that
*R*is a Dedekind domain. - If
*P*is a non-zero prime ideal in*R*, show that*R*/*P*is isomorphic to either**R**or**C**. Show that both possibilities can occur, and that in the first case,*P*= (*x*−α,*y*−β), where α, β ∈**R**and α^{2}+ β^{2}= 1. Show that in the second case,*P*is a principal ideal generated by a linear polynomial. - Deduce from #2 that the class group of
*R*is generated by the ideals of the form (*x*−α,*y*−β). Show that all these ideals define the same element of the class group, which is of order 2, by showing that if*P*is a prime ideal of this form, then*P*is not principal but*P*^{2}is principal, and that if*P*_{1}and*P*_{2}are distinct prime ideals of this form, then*P*_{1}*P*_{2}is also principal. - For each of the following
**Z**-modules*M*, compute the*P*-adic completion*M*^{^}, where*P*= (2). Does*M*^{^}coincide with**Z**^{^}⊗_{Z}*M*?*M*=**Z**/(3).*M*=**Z**/(2).*M*=**Q**/**Z**.*M*=**Q**.

- Let
- Exercises on dimension theory
(based on A-M, Ch. 11) (due Monday, May 9)
- Is there a purely topological proof of the fact that
Noetherian rings have the descending chain condition on prime ideals?
In other words, answer the following questions:
- What property of a Noetherian topological space
*X*corresponds to the descending chain condition on prime ideals of*R*when*X*= Spec*R*? - Does every Noetherian topological space
*X*have the above property? Give a proof or a counterexample.

- What property of a Noetherian topological space
- Compute the Hilbert function of the graded ring
*R*=*K*[*x*,*y*,*z*]/(*x*^{2}*z*−*y*^{3}−*y**z*^{2}), where*K*is a field and*x*,*y*,*z*each have degree 1. - Show that a regular Noetherian local ring of dimension
*d*is integrally closed (see comments in A-M, bottom of p. 123).

- Is there a purely topological proof of the fact that
Noetherian rings have the descending chain condition on prime ideals?
In other words, answer the following questions: