Answer to Question #109069 in Abstract Algebra for Durgesh singh

Question #109069
Show that z(sqrt-2) is non ufd
1
Expert's answer
2020-04-13T16:49:36-0400

The given ring is R=Z[2]R=\Z[\sqrt{-2}] .

Your question is wrong , Since we known that RR is a Euclidean domain by defining a norm on RR ,i,e,N(a+b2)=a2+2b2N(a+b\sqrt{-2})=a^2+2b^2 and every Euclidean domain is PID .

Again every PID is UFD.

Hence RR is a UFD.


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