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 $\Z[\sqrt{-2}]$ $=\{ a+b\sqrt{-2} :a,b\in \Z \}$ .

Claim: The Quotient ring of a UFD need not be a UFD.

Since $R$ is an Euclidean domain ,so $R$ is PID.

Again , we known that every PID is UFD.

Hence, $R$ is a UFD.

Now ,consider the ideal $<6>=\{$ $6(a+(\sqrt{-2})b \ ):a,b\in \Z \}$

Then ,$\frac{R}{I}=\{ (a+b(\sqrt{-2}))+<6> :a,b\in \Z \}$

Clearly,$2\notin <6>,$ and $3\notin <6>$ but $6\in<6>$

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

But we known that ,to became a UFD ,it must be first integral domain.

Hence , it is not a unique factorization domain.

(Proved).

Since $R$ is a PID , Every irreducible elements is prime .

Hence, 2nd part of the question is wrong. That is ,we can't find a irreducible element in the given ring R which is not a prime.

But in place of $R$ if you take $R'=$ $\Z [\sqrt{-5}]$ then there are beautiful results,

For example:

The ring $\Z[x]$ is a UFD but when we quotient by the ideal$I=<x^2+5>$ then,

$\frac{R'}{I}=\frac{ \Z[x]}{I}\approx\Z[\sqrt{-5}]$

Which is not a UFD.

Since ,$46=2.23$

$=(1+3\sqrt{-5})(1-3\sqrt{-5})$

We can easily verify that $2,23,(1+3\sqrt{-5})(1-3\sqrt{-5})$ are irreducible.

Define a norm on $\Z[\sqrt{-5}]$ by $N(a+b\sqrt{-5})=a^2+5b^2$

Then $x\in R' \ is \ a \ unit \ iff \ N(x)=1$

Suppose $x=a+b\sqrt{-5}$is a unit $\iff$ there exit $y=a'+b'\sqrt{-5}$ such that xy=1 \

Suppose $x \ is \ a \ unit \implies\exists y \in R' s.t . \ xy=1$

So,$N(xy)=N(1)\implies \ N(x)N(y)=1$

Hence ,$N(x)=1. \ since$ norm gives non negative integer.

Now ,assume $N(x)=1\implies\ a^2+5b^2=1$

Since,$a^2 \ and \ 5b^2 are \ positive \ number\implies a=+1 \ or \ -1$

Hence,$x=a$

Which is a unit .

Also it follows that the only unit in $R'$ are +1 and -1.

If possible let $2=xy$ $for \ some \ x,y\in R'$ ,where neither x nor y is a unit.

Then $N(2)=N(x)N(y)\implies N(x)N(y)=4$

$\therefore N(x)=N(y)=2$ ,Since $N(x) \ and \ N(y) are$ non negative integer.

Hence,$2=4$

Which is not possible .Hence 2 is irreducible.

Similarly 23 is not irreducible.

If possible let $1+3\sqrt{-5}=xy$ ,for some $x,y\in R'$ where neither x nor y is a units.

Then $N(xy)=N(1+3\sqrt{-5})=46=N(x)N(y)$

$N(x)N(y)=2×23\implies \ either N(x)=2 \ or \ N(x)=23$

Therefore, $N(x)=a^2+5b^2=2 \ or \ 23$

Since,$a^2,5b^2 \ are \ positive \ number \ so N(x)\neq2$

If N(x)=23 then $5b^2=23-a^2\geq0$

Which is not possible because a is an integer .

Hence, $1+3\sqrt{-5}$ is irreducible.

Similarly,$1-3\sqrt{-5}$ is irreducible.

Also ,we can prove $(1+3\sqrt{-5})$ is irreducible but not a prime

just using the definition of prime in an integral domain.

If possible ,let this number is a prime.

Then,

$(1+3\sqrt{-5})(1-3\sqrt{-5})=46=2.23$

$\implies \ (1+3\sqrt{-5}) |2.23$

But neither $(1+3\sqrt{-5})|2 \ nor \ (1+3\sqrt{-5})|23$

Suppose,$(1+3\sqrt{-5})\mid 2$ then there exit a number $(a+b\sqrt{-5} ) \ such \ that \ (a+b\sqrt{-5})(1+3\sqrt{-5})=2$

$\implies a+15b=2 \ and \ a+b=0$

Which is impossible.

Hence$(1+3\sqrt{-5})\nmid 2.$

Similarly,$(1+3\sqrt{-5})\nmid23$

Therefore,$(1+3\sqrt{-5})$ is not a prime.