Show that $α$ is a perfect square in the quadratic integers in $\mathbb Q[\sqrt{d}]$

Question:

Suppose that $\mathbb Q[\sqrt{d}]$ is a UFD, and $α$ is an integer in $\mathbb Q[\sqrt{d}]$ so that $α$ and $\barα$ have no common factor, but $N(α)$ is a perfect square in $\mathbb Z$. How can I show that $α$ is a perfect square in the quadratic integers in $\mathbb Q[\sqrt{d}]$?

What I have Done:

I'm not sure if I'm approaching this correctly but if $\alpha$ and $\bar{\alpha}$ have no common factor, then $\alpha=\pi_{1}\pi_{2}\cdots\pi_{k}$ and $\bar{\alpha}=\pi'_{1}\pi'_{2}\cdots\pi'_{j}$ where $\pi_{i}$ and $\pi'_{i}$ are prime in $\mathbb{Q}(\sqrt{d})$. But $N(\alpha) = \alpha\bar{\alpha}=\pi_{1}\pi_{2}\cdots\pi_{k}\pi' _{1}\pi'_{2}\cdots\pi'_{j}=n^{2}$ where $n \in \mathbb{Z}$. Somehow I need to show that $\alpha=\beta^{2}$ where $\beta$ is a quadratic integer in $\mathbb{Q}(\sqrt{d})$ (i.e., $\alpha$ is a perfect square in the quadratic integers in $\mathbb{Q}(\sqrt{d})$)

• @abiessu I added to my question. – Rachel May 4 '15 at 17:40
• It seems to me like $\gcd(\alpha, \overline\alpha) = 1$ is incompatible with $N(\alpha)$ being a perfect square. Can you give a concrete example, in, say, $\mathbb{Z}[\sqrt{-2}]$? – user153918 May 4 '15 at 18:09
• @AlonsodelArte, how about $3\pm4i$ in $\Bbb Z[i]$? – Lubin May 5 '15 at 13:54
• I think your conjecture is false, see my comment below. But I gave you an up-arrow, because I think the conjecture is pleasing. – Lubin May 5 '15 at 14:04
• @Lubin That's an excellent example, one I should've thought of more readily. – user153918 May 5 '15 at 14:12

[I guess you mean $\mathbb Z[\sqrt d]$ instead of $\mathbb Q[\sqrt d]$. The latter is a field, and it's not really meaningful to talk about unique factorisation in a field.]

You can complete your strategy by proving that (with your notation) the $\pi_i$ and $\pi_j'$ are different.

Lemma. Let $A$ be a UFD and $x$, $y$ be two coprime elements such that $xy$ is associated to a square. Then $x$ and $y$ are associated to squares.

Proof. If $x = \epsilon\, p_1^{e_1} \cdots p_r^{e_r}$ and $y = \eta \,q_1^{f_1}\cdots q_s^{f_s}$ are their decompositions in irreducible elements, (with $\epsilon, \eta \in A^\times$) then the $p_i$ and the $q_j$ are disjoint (because $x$ and $y$ are coprime) and the decomposition of $xy$ is then $xy = \epsilon\,\eta\, p_1^{e_1} \cdots p_r^{e_r}q_1^{f_1}\cdots q_s^{f_s}$.

Because $xy$ is a associated to a square, all the $e_i$ and all the $f_j$ are even integers, which proves that $$x = \epsilon\, \left( p_1^{e_1/2} \cdots p_r^{e_r/2}\right)^2 \qquad \text{and}\qquad y = \eta \,\left( q_1^{f_1/2} \cdots q_s^{f_s/2}\right)^2.$$

And to answer Alonso del Arte's concern: $\alpha = (1+2i)^2 \in \mathbb Q(i)$ works. Then $\alpha$ and $\overline{\alpha} = (1-2i)^2$ are coprime ($1 \pm 2i$ are irreducible elements and they aren't associated because $\pm 1$ and $\pm i$ are the only units) and $\alpha \overline{\alpha} = \left( (1+2i)(1-2i) \right)^2 = 25.$ You can do such examples every time a prime number $p$ (5 in this example) splits in $\mathbb Z[\sqrt d]$.
• But I’m a little worried about units. Even for $i\in\Bbb Z[i]$, you have an integer that’s (naturally) relatively prime to its conjugate, but is not a perfect square. This seems to me to apply to the integer $i(1+2i)^2$ just as well. – Lubin May 5 '15 at 14:03
• I think that the exercise has to be modified to mean: if $N(\alpha)$ is a square, then $\alpha$ is associated to a square. Otherwise, any time you have nonsquare units $\eta \in \mathbb Z[\sqrt d]^\times$, you will have exactly the same problem: $N(\alpha) = N(\eta \alpha)$ but at most one of $\alpha$ and $\eta \alpha$ can be a square). – PseudoNeo May 5 '15 at 14:13
• A second moral is that when you talk about divisibility, the natural things to consider are often elements up to association. Anyway, as soon as you advance in the field, you will think less and less about individual elements and more and more about ideals. As the principal ideal $(x)$ contains exactly the same information as $x$ up to association, this moral becomes more and more evident. – PseudoNeo May 6 '15 at 19:20