Jtdylktuy

I can prove this for $sqrt{-5}$. What changes for the above question? Is it a similar proof? What would be the irreducible elements? Are they $1$ and $7$ ?

Hint $ $ It's simpler: UFDs are integrally closed, but it is not since $,(1+sqrt{-7})/2 $ is a root of $,x^2-x+2.,$ Said more simply it fails RRT = Rational Root Test, which holds true in any UFD.

Remark $ $ The RRT proof fails because some gcd it employs fails to exist in $,Bbb Z[sqrt{-7}].,$ Since gcds always exist in a UFD, we can use such gcd failure to prove rings are not UFDs. Let's use this to prove a more general case (for variety vs. traditional proofs).

Lemma $ Bbb Z[sqrt d]$ is not a UFD if $,color{#0a0}{dequiv 1pmod{!4}} $

Proof $ $ Assume it's a UFD. $ color{#0a0}{ww'}=(1!+!sqrt d)(1!-!sqrt d) = 1-d = color{#0a0}{4k},, color{#c00}{w} = 2-w',$

therefore the gcd product $ ,(2,color{#0a0}w)(2,color{#0a0}w') = (4,2w',2w,color{#0a0}{4k}) = 2(2,w',color{#c00}{w}) = 2(2,w')$

Cancelling $,(2,w'),Rightarrow, (2,w) = (2),Rightarrow, 2mid w,Rightarrow, w/2 = (1!+!sqrt d)/2in Bbb Z[sqrt d] Rightarrow!Leftarrow$

Further $ $ if $,w=sqrt{c^2d},$ then $,Bbb Z[w],$ UFD $,Rightarrow, cmid 1,,$ else $,(w/c)^2 = d,$ contra RRT (as above).

Alternatively by gcds $,(c,w)^2 = (c^2,cw,c^2d) = c(c,w),Rightarrow, (c,w) = (c),Rightarrow, cmid w,Rightarrow,cmid 1$

Combining yields a purely gcd-based proof of the following well-known fact.

Theorem $, Bbb Z[sqrt d], $ a UFD $,Rightarrow, d,$ squarefree and $,dnotequiv 1pmod{!4}$

Note $ $ The proof of the theorem also works if we instead assume the ring is a PID and read the tuples as ideals vs. gcds [since the proof uses only laws common to both: $ $ distributive, associative, commutative, and $(a,b,bc) = (a,b),$]. Then it yields $,(2,w),$ is noninvertible (so nonprincipal).

$$(1+sqrt{-7})×(1-sqrt{-7})=8=2×2×2.$$We show $2,1+sqrt{-7},1-sqrt{-7}$ are irreducibles.

So let $2=(a+sqrt{-7}b)(c+sqrt{-7}d)$ where $a,b,c,d in Bbb Z$, then taking norms $4=(a^2+7b^2)(c^2+7d^2)implies b=d=0$ , since for $e,fin Bbb Z$ and $fnot =0implies e^2+7f^2geq 7$. Also either $a$ or $c$ is $+1,-1$ as $2^2=4=(ac)^2$ and $Bbb Z$ is a UFD i.e. either $a+sqrt{-7}b$ or $c+sqrt{-7}d$ unit.

Next, let $1+sqrt{-7}=(a+sqrt{-7}b)(c+sqrt{-7}d)$ with $a,b,c,din Bbb Z$. Again taking norms $8=(a^2+7b^2)(c^2+7d^2)implies bd=0$ and both $b,d$ are not zero as $8$ has no integral square root. So without loss of generality assume $b=0$ and $dnot=0implies 8=a^2(c^2+7d^2)$ . Note that $d$ must be $+1,-1$ , otherwise $c^2+7d^2geq 28$. Also $cnot=0$ , as $7not |8$. Hence $c$ must be $+1,-1$ otherwise $c^2+7d^2geq 11$. Therefore $a=1,-1$ i.e. $a+sqrt{-7}b$ is unit.

Similarly $1-sqrt{-7}$ is irreducible.