Abstract Algebra: ED implies PID implies UFD, some number theory, 11-29-17

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Doesn't  <2, x> = { a + f(x) | a in 2Z, f(x) in Z[x], f(0)=0 }  ?

mgaryhickman
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sir is it sufficient to prove that euclidean ring is PID for proving ED is ufd IN 4 marks

narenderbhardwaj