Bhaskaracharya Pratishthana, Pune
Instructional School for Teachers on Commutative Algebra
Tutorial-6 (3 - Dec - 2016)
R denotes a commutative ring containing 1.
1. Suppose that the polynomial ring R[X] is a PID. Show that R is a field.
2. Let k be a field. Show that the power series ring k[[X]] is a local ring with the unique
maximal ideal (X).
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3. If R is an integral domain, then show that R = P ∈Spec(R) RP = m∈Spm(R) Rm .
4. Let M be an R-module. Show that the following are equivalent:
(i) M = 0,
(ii) MP = 0 for all prime ideals P of R, and
(iii) Mm = 0 for all maximal ideals m of R.
5. Let S be a multiplicatively closed subset of R and M be a finitely generated R-module.
Prove that S −1 M = 0 if and only if sM = 0 for some s ∈ S.
6. Let I be an ideal of R and S = 1 + I. Show that
(i) S is a multiplicatively closed subset of R.
(ii) S −1 I is contained in the Jacobson radical of S −1 R.
7. If RP is an integral domain for each prime ideal P of R, is R necessarily an integral
domain ?
8. Suppose R is Noetherian. Let S be a multiplicatively closed subset of R. Is S −1 R
Noetherian ?
9. Let R = C[X, Y, Z]/(X 2 − Y Z) and P = (X, Y )/(X 2 − Y Z). ShowTthat P is a prime
ideal of R and find generator/s of the ideal P (2) := (P 2 )ec = P 2 RP R.
10. Let R be a ring such that
(i) for each maximal ideal m of R, the local ring Rm is Noetherian, and
(ii) for each x 6= 0 in R, the set of maximal ideals of R which contain x is finite.
Show that R is Noetherian.
11. Suppose that R is a UFD. Let S be a multiplicatively closed subset of R. Show that
S −1 R is a UFD.
12. Suppose that R is Noetherian. For a prime ideal P of R, the height of P , denoted by
htP is defined by
htP = sup {r | there exists a chain P0 ( P1 ( · · · ( Pr = P in Spec(R)}.
Let R = k[X, Y, Z]/(XY, XZ), where k is a field. Let P = (Y, Z)/(XY, XZ) ∈
Spec(R). Find htP .
13. Let R be a Noetherian ring. Recall that the Krull dimension of R is defined by
dimR := sup {ht P | P ∈ Spec(R)}.
Find dim Z. If R is Artinian, show that its dimension is zero.
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