Answer to Question #106439 in Abstract Algebra for Garima Ahlawat

Question #106439
Use the ring Z(underroot -2) to show that (1) the quotient ring of a ufd need not be a UFD. (2)an irreducible element of a UFD need not be a prime element
1
Expert's answer
2020-03-28T10:34:04-0400

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

Using this ring ,we have to show

1.The quotient ring of a UFD need not be UFD.

2.An irreducible elements of a UFD need not be a prime.

As we Know that ,"R" is a Euclidean domain.

Again, every Euclidean domain is PID and every PID is UFD.

Hence,"R" is a UFD.

Now consider the ideal ,

"I=<6>=\\{ 6(a+b\\sqrt{-2}):(a+b\\sqrt{-2})\\in R \\}" .

Then ,


"\\frac{R}{I}=\\{ (a+b\\sqrt{-2})+I:(a+b\\sqrt{-2})\\in R \\}"

clearly,

"2+I,3+I\\in \\frac{R}{I}"

but,"(2+I)(3+I)=6+I=I" .

Hence,"\\frac{R}{I}" is not a integral domain.

Therefore, it is not a unique factorization domain.

Since, To become a UFD ,It must be a integral domain .



2. As "R" is a PID ,so every irreducible elements is a prime.

Hence, the question is wrong in this case.



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