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[2]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 ,RR is a Euclidean domain.

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

Hence,RR is a UFD.

Now consider the ideal ,

I=<6>={6(a+b2):(a+b2)R}I=<6>=\{ 6(a+b\sqrt{-2}):(a+b\sqrt{-2})\in R \} .

Then ,


RI={(a+b2)+I:(a+b2)R}\frac{R}{I}=\{ (a+b\sqrt{-2})+I:(a+b\sqrt{-2})\in R \}

clearly,

2+I,3+IRI2+I,3+I\in \frac{R}{I}

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

Hence,RI\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 RR is a PID ,so every irreducible elements is a prime.

Hence, the question is wrong in this case.



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