$\mathbb Z[\sqrt{-7}]$ and $\,\Bbb Z[\sqrt{1+4k}]$ are not UFDs

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 $7$ ?

Solution 1:

Hint $ $ It's simpler: UFDs are integrally closed, but $\,\Bbb Z[\sqrt{-7}]\,$ 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}{d\equiv 1\pmod{\!4}}\ $

Proof $ $ Assume it is 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 $\,(2,w')\,\Rightarrow\, (2,w) = (2)\,\Rightarrow\, 2\mid w\,\Rightarrow\, w/2 = (1\!+\!\sqrt d)/2\in \Bbb Z[\sqrt d]\ \Rightarrow\!\Leftarrow$

Further $ $ if $\,w=\sqrt{c^2d}\,$ then $\,\Bbb Z[w]\,$ UFD $\,\Rightarrow\, c\mid 1,\,$ else $\,(w/c)^2 = d\,$ contra RRT (as above).
Alternatively by gcds $\,(c,w)^2 = (c^2,cw,c^2d) = c(c,w)\,\Rightarrow\, (c,w) = (c)\,\Rightarrow\, c\mid w\,$ in $\,\Bbb Z[w],\,$ i.e. $\,w/c\in \Bbb Z[w],\,$ so $\,c\mid 1\,$ in $\,\Bbb Z.$

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

Theorem $\,\ \Bbb Z[\sqrt d]\, $ a UFD $\,\Rightarrow\, d\,$ squarefree and $\,d\not\equiv 1\pmod{\!4}$

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).

Solution 2:

$$(1+\sqrt{-7})×(1-\sqrt{-7})=8=2×2×2.$$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 $e,f\in \Bbb Z$ and $f\not =0\implies e^2+7f^2\geq 7$. 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 $a,b,c,d\in \Bbb Z$. 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 $d\not=0\implies 8=a^2(c^2+7d^2)$ . Note that $d$ must be $+1,-1$ , otherwise $c^2+7d^2\geq 28$. Also $c\not=0$ , as $7\not |8$. Hence $c$ must be $+1,-1$ otherwise $c^2+7d^2\geq 11$. Therefore $a=1,-1$ i.e. $a+\sqrt{-7}b$ is unit.

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