# Perfect Square in an UFD

Let $R$ be an UFD with quotient field $F$. Show that an element $d\in R$ is a square in $R$ if and only if $d$ is a square in $F$.

And then get a counterexample that above statement is not true if $R$ is not UFD.

-

If $d$ is the square of an element in $R$, then $d$ is certainly the square of an element in $F$.

Now suppose $d$ is the square of an element in $F$. A typical nonzero element in $F$ is a quotient of elements in $R$. And since every element of $R$ is a product of irreducibles, every element in $F$ is a quotient of products of irreducibles. In other words, a nonzero element in $F$ is just a product of the form $A^aB^bC^c$ etc. where $A, B, C$ etc. are irreducible members of $R$ and $a, b, c$ etc. are integers. These latter integers may, of course, be negative.

So what happens if $d$ is the square of an element in $F$? We have that $d = (A^aB^bC^c$ etc.)$^2$, or $d = A^{2a}B^{2b}C^{2c}$ etc. Since $d$ is a member of $R$, every power belonging to the irreducibles which compose $d$ (that is, $2a, 2b, 2c$ etc.) must be positive. But this means that $a, b, c$ etc. must also be positive, meaning that $A^{a}B^{b}C^{c}$ etc. has to be an element of $R$. Thus $d$ is the square of an element in $R$.

-
Can you give me a counterexample – Muniain Dec 25 '12 at 15:39
lol maybe. can you think of any non-UFDs, first of all? – D_S Dec 25 '12 at 15:52
@Firmino, see my answer. I have a counterexample. – Amr Dec 25 '12 at 16:16

Let me give a counterexample for the case when $R$ is not an UFD: set $R=K[X^2,X^3]$, the ring of polynomials over a field $K$ whose monomial of degree one is missing. Then the field of fractions of $R$ is $F=K(X)$. Now take $X^2\in R$. This is obviously a square in $F$, but there is no $a\in R$ such that $a^2=X^2$ (otherwise $X\in R$, a contradiction).

-
Dear @Firmino, now you have also a counterexample. – user26857 Feb 17 '13 at 11:18

Forward direction: If $d$ is a squre in $R$, then $d=c^2$ for some $c\in R$. Thus, $\frac{d}{1}=[\frac{c}{1}]^2$.

Backward direction: Let $\frac{d}{1}=[\frac{r}{s}]^2$. Thus, $s^2d=r^2$. By writing $d,s,r$ as products of irreducibles one can see that the exponents of the irreducibles that appear in the prime factorization of $d$ are even.

-
How is that a counterexample? A counterexample would entail giving an example of an integral domain $R$ with quotient field $K$ such that the square of something in $F$ is in $R$, but is not the square of anything in $R$. $\mathbb{Z_6}$ isn't an integral domain. – D_S Dec 25 '12 at 17:05
i was not concentrating – Amr Dec 25 '12 at 17:26

Just as for irrationality proofs in $\,\Bbb Z,\,$ this is an immediate consequence of the monic case of the Rational Root Test (RRT), which is true in any UFD (or any GCD domain), since the proof uses only Euclid's Lemma $\rm\:(a,b)=1,\ a\mid bc\:\Rightarrow\:a\mid c.\:$ In particular, if $\rm\:x^2\! -c,\ c\in R,\:$ has a "rational" root $\rm\:x\in F\:$ then it must be "integral" $\rm\:x\in R\:$ by RRT.

This fails in non-UFD domains, e.g. there are very simple quadratic integer counterexamples:

$\rm\ \ x^2\! = (1\!+\!\sqrt{d})^2\!\in \Bbb Z[2\sqrt{d}]\$ has root $\rm\ x = 1\!+\!\sqrt{d} = \dfrac{2\!+\!2\sqrt{d}}2\:$ a proper fraction over $\rm\:\Bbb Z[2\sqrt{d}]$

Remark $\$ Rings satisfying the monic case of the Rational Root Test are called integrally closed. Thus the remark above translates as: the usual proof of RRT immediately generalizes to show that UFDs and GCD domains are integrally closed.

-