Mathematics G4261xCommutative Algebra

Assignment #8Due Oct. 29, 2014

1. A short exact sequence 0→ M1φ1→ M2

φ2→ M3 → 0 of R-modules is said to be split if φ2 hasa right inverse which is a module homomorphism.

(a) Show that this is equivalent to φ1 having a right inverse which is a module homomorphism.

(b) Show it is also equivalent to M2∼= M1 ⊕M3, with φ1 and φ2 being the obvious maps.

(c) Show that the corresponding statement is false if M1, M2, M3 are nonabelian groups. Inthis case, what can be said about M2 in a split sequence?

2. Show that any exact sequence

· · · →Mi →Mi+1 →Mi+2 →Mi+3 →Mi+4 → · · ·

can be “spliced” into two exact sequences

· · · →Mi →Mi+1 → N → 0 and 0→Mi+2/N →Mi+3 →Mi+4 → · · ·

for some N .

*3. Let V be a vector space of dimension n over the complex numbers. For m ≤ n, let Gr(m,V )be the Grassmannian of m-dimensional subspaces, given the structure of an abstract varietyas stated in class. Show carefully that the Plucker map P : Gr(m,V ) → PΛmV taking Wto w1 ∧ · · · ∧ wm := 1

m!

∑σ∈Σm

(sgn σ)wσ(1) ⊗ · · · ⊗ wσ(m), where w1, . . . , wm is any basis forW , is well-defined and is an embedding, that is, an isomorphism onto its image. The hardpart is to do this concisely. You may use without proof facts from multilinear algebra, suchas that ΛmKm is spanned by the determinant.

*4. (a) If M and N are flat R-modules, then so is M ⊗R N ;

(b) If S is a flat R-algebra and M is a flat S-module, then M is flat as an R-module.

*5. For a module M over a ring R, prove that the following are equivalent:

(i) M is flat;

(ii) for all R-modules N and for all i > 0, Tori(M,N) = 0;

(iii) for all R-modules N , Tor1(M,N) = 0.

You may use the definition of Tor in terms of a free resolution, as well as its elementaryproperties such as the long exact sequence.

*6. Let 0 → M1 → M2 → M3 → 0 be a short exact sequence of R-modules with M3 flat. ThenM1 is flat if and only if M2 is flat.

CONTINUED OVERLEAF. . .

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*7. For a finitely generated module M over a Noetherian local ring R with maximal ideal m andresidue field k = R/m, show that the following are equivalent:

(i) M is free;

(ii) M is flat;

(iii) the map m⊗M → R×M is injective;

(iv) Tor1(k,M) = 0.

Hint for (iv)⇒ (i): Let x1, . . . , xn be elements of M whose images in M/mM are a basis. UseNakayama to show they generate M . Then consider the module homomorphism Rn → Mdetermined by the xi and use Nakayama again to show it is an isomorphism.

*8. A free resolution of an R-module M is an exact sequence

· · ·M2 →M1 →M0 →M → 0

where each Mi is free. The computation of Ext and Tor does not depend on the choice of afree resolution. Use this to compute the Ext and Tor groups of the Z-modules Z/(m) andZ/(n).

*9. Flat or not? Give a brief proof and sketch the corresponding varieties in each case.

(a) K[x, x−1] over K[x]

(b) K[x] over K[xn]

(c) K[x, y]/(y2 − (x+ 1)(x− 1)) over K[x]

(d) K[x, x−1, y]/(y2 − (x+ 1)(x− 1)) over K[x]

(e) K[x, y]/(xy) over K[x]

10. Read about transcendence bases and transcendence degree in §9 of Milne’s lecture notesFields and Galois Theory, available from http://www.jmilne.org/math/CourseNotes/

*11. By reading in Artin’s Algebra, §13, you may learn that the ideal (2, 1 +√−5) in Z[

√−5] is

not principal (see 13.3.3) and that every maximal ideal m ⊂ Z[√−5] satisfies either m = (p),

where p is a prime number such that −5 is not a square in Fp, or mm = (p), where −5 is asquare in Fp, and m is the complex conjugate of m (see 13.5.4(d) and 13.6.1). Note: m mayor may not equal m.

Prove that the ideal M = (2, 1+√−5) is not free as a module over R = Z[

√−5], but that for

every m, M/mM is 1-dimensional over the field R/m. This furnishes another counterexampleto Ma’s question. Chapter 13 is well worth reading in full.

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