Answer to Question #16365 in Algebra for Irvin

Question #16365
Prove that a semilocal Dedekind ring is a PID
1
Expert's answer
2012-10-15T11:48:44-0400
If R is (commutative) semilocal, we know that Pic(R) = {1}. If R is also Dedekind, then
next argument shows that R is a PID:
If R is a PID, it is well-known that R is a UFD. If R is a UFD, then gives Pic(R) = {1}. Finally,
if Pic(R) = {1},
then every invertible ideal in R is principal. Since R is a Dedekind ring, every nonzero
ideal is invertible, and hence principal. This shows that R is a PID.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS