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I have seen complicated arguments to establish whether $\mathbb Z$ extended by this or that radical is a UFD. They usually suppose that $u$ is a unit and then compute the reciprocal and then make arguments about each component being an integer. Some of them go by defining a norm and proving things about that. I'm a little confused about why these arguments are so elaborate when it seems to me that there is a very straight-forward argument in many cases. My guess is that I'm overlooking some important caveat in one of my steps but I'm not sure where.

Take for example $\mathbb Z[\sqrt{7}]$. To prove it's not a UFD I look at two factorizations of 6:

$$ 6=3\cdot 2 $$

$$ 6 = (\sqrt 7 - 1)(\sqrt 7 + 1) $$

I then argue that there is no unit (in fact no element at all) such that $3u=\sqrt 7-1$. Since $u=a+b\sqrt 7$ then we would need

$$ 3a = -1 $$

but this is not true for any $a\in\mathbb Z$. Of course the same argument could be used on any of $\pm 3,\pm 2$ on the left and $\pm (\sqrt 7-1),\pm (\sqrt 7+1)$ on the right.

Is there some logical error in saying that if $w+x\sqrt 7 = y+z\sqrt 7$ for integers $w,x,y,z$ then we must have $w=y$ and $x=z$?

user26857
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Addem
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    You left out an important point: why are those factorizations of $6$ actually irreducible factorizations? For comparison, in $\mathbf Z[i]$, if we write $10 = 2 \cdot 5 = (3+i)(3-i)$, that does not show $\mathbf Z[i]$ isn't a UFD since those factors of $10$ are all reducible in $\mathbf Z[i]$. And in fact $\mathbf Z[i]$ is a UFD. The "complicated" arguments you have seen are for showing $\mathbf Z[\sqrt{d}]$ is a UFD, not that it isn't: to show such a ring is not a UFD, a single example of nonunique irrreducible factorization is enough. A proof of being a UFD can be a lot harder. – KCd Nov 20 '20 at 05:16
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    In a word, as @KCd points out, $6=3\cdot2$ is not a factorization in primes. – Lubin Nov 20 '20 at 05:28
  • For what it's worth, ${\bf Z}[\sqrt{13}]$ is not the ring of integers in ${\bf Q}[\sqrt{13}]$, since it doesn't contain $(1+\sqrt{13})/2$, which is an algebraic integer in ${\bf Q}[\sqrt{13}]$. – Gerry Myerson May 22 '23 at 06:00
  • "While 2 and 3 are primes in Z, but they are not primes in Z[7–√] and hence not irreducible". Is this statement true? – Maged Zakaria May 22 '23 at 05:35
  • $2=(3+\sqrt7)(3-\sqrt7)$. – Gerry Myerson May 22 '23 at 05:43
  • This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review – cs89 May 22 '23 at 06:06
  • Note is is easy to show that $\Bbb Z[\sqrt d]$ is not a UFD if $d\equiv 1\pmod{!4},$ since it is not integrally closed (i.e. it fails rational root test),e.g. here the fraction $, x = (1+\sqrt{13})/2,$ is a root of $,x^2-x-3.\ \ $ – Bill Dubuque May 22 '23 at 18:39

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There is a counter example. Since $4=2 \cdot 2 = (\sqrt{13}+3)(\sqrt{13}-3)$, it suffices to show those factors are irreducible. Suppose to the contrary that $2$ is not irreducible. It follows that there exists $a,b \in \mathbb{Z}[\sqrt{13}]$ such that $2=ab$, $N(a) \ne 1$ and $N(b) \ne 1$. It boils down to show that there are no solutions to $u^2=13v^2 \pm 2$ in $\mathbb{Z}$. Observe that $2 \nmid \sqrt{13} \pm 3$ and we are done.

Someone asked the same about $\mathbb{Z}[\sqrt{8}]$. You might find Is $\mathbb{Z}[{ \sqrt 8 } ] $ a Euclidean domain? relevant.

user26857
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Chan Tai Man
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To repeat what KCd and Lubin had already said in a verbose way. While $2$ and $3$ are primes in $\mathbb{Z}$, but they are not primes in $\mathbb{Z}[\sqrt{7}]$ and hence not irreducible.

\begin{equation} 2 = (3+\sqrt{7})(3-\sqrt{7}) \\ 3 = (\sqrt{7}+2)(\sqrt{7}-2) \\ \sqrt{7}+1 = (3-\sqrt{7})(5+2\sqrt{7}) \\ \sqrt{7}-1 = -(3+\sqrt{7})(5-2\sqrt{7}) \end{equation}

Definitions: Let $R$ be an integral domain. Then an element $a \in R$ is said to be prime in $R$ if:

(a) $a \ne 0$ and $a$ is not a unit, and

(b) For all $b,c \in R$, $a \mid bc$ implies $a \mid b$ or $a \mid c$.

The element $a \in R$ is said to be irreducible (a weaker condition than primeness) if:

(a) $a \ne 0$ and $a$ is not a unit, and

(b) If $a = bc$ for some $b,c \in R$, then $b$ is a unit or $c$ is a unit.

If you are still unsure, you might want to look up the definition of unit (not unity). Very briefly, $a$ is a unit if and only if norm $N(a)=1$. BTW, $\mathbb{Z}[\sqrt{7}]$ is a Euclidean domain and hence a UFD.

Chan Tai Man
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Picking up on the answer of Chan Tai Man, from $$ 3\times2=(\sqrt7-1)(\sqrt7+1) $$ we get $$ (\sqrt7+2)(\sqrt7-2)(3+\sqrt7)(3-\sqrt7)=-(3+\sqrt7)(5-2\sqrt7)(3-\sqrt7)(5+2\sqrt7) $$ We can cancel $-(3+\sqrt7)(3-\sqrt7)$ from both sides, leaving $$ (2+\sqrt7)(2-\sqrt7)=(5+2\sqrt7)(5-2\sqrt7) $$ Then a simple calculation shows $$ {5+2\sqrt7\over2-\sqrt7}=-(8+3\sqrt7) $$ which is a unit. So in fact, up to units, we don't have two different factorizations of $6$ into irreducibles.

Gerry Myerson
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