# Answer to Question #16365 in Algebra for Irvin

Question #16365

Prove that a semilocal Dedekind ring is a PID

Expert's answer

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.

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.

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