Question

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

Answer 1

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.