# Extension of Rational Roots Test to UFDs

1. Jul 21, 2011

### Kreizhn

1. The problem statement, all variables and given/known data
Show that $p(x) = x^2 - \sqrt2$ is irreducible in $\mathbb Z[\sqrt 2]$.

3. The attempt at a solution

I think I have this, but I just want to make sure my reasoning is correct. I'm sure there are other ways.

Firstly, it is sufficient to show that p(x) is irreducible in the field of fractions, which is $\mathbb Q[\sqrt 2]$. Next, since it has degree 2, it is sufficient to show that p(x) has no root in $\mathbb Q[\sqrt 2]$. Finally, we apply the Field of Fractions Root Test (my name for the rational roots test extended to general UFD's) to say that the only possible roots of p(x) are $\pm \sqrt2$. Since neither is a root, we're done.

I think the only thing I'm not really sure about here is the "rational roots test," and perhaps the fact that $\mathbb Q[\sqrt 2]$ is a UFD. As a matter of fact, a good question might be, if R is a UFD, is $R[\alpha]$ a UFD?

I think this would be true, though I'm not certain. In particular, if we move to field theory and S is a field extension of R with $\alpha$ transcendental over R, then $R[\alpha] \cong R[x]$ and since R is a UFD, then R[x] is a UFD, so $R[\alpha]$ is a UFD. (In fact we can so more right? Since then $R[\alpha]$ is actually a Euclidean domain). But what if alpha is algebraic? It seems to me that this would actually be weaker and should still hold.

What then do we do when we move out of fields and back into UFDs? Can a similar argument be applied?

Edit: Certainly in this case, $\mathbb Q$ is a field so we needn't consider UFD's exclusively, but it's interesting to me as a general question.

2. Jul 21, 2011

### micromass

Staff Emeritus
This is true and it is called Gauss' lemma. However, I really don't see the point in moving to the field of fractions. It only makes the question harder!

In general, if we have R a UFD and if we have a solution b to

$$a_nX^n+...+a_1X+a_0=0[/itex], then $$b(a_nb^{n-1}+...+a_1=-a_0$$ so any solution in R must divide a0. The only problem with your method is how you can be certain that $\pm\sqrt{2}$ are the only numbers dividing $\pm\sqrt{2}$$? This is not true. For example [itex]\mathbb{Z}[\sqrt{-5}]$ is not a UFD. So you'll still have some more things to check to see if $\mathbb{Z}[\sqrt{2}]$ is actually a UFD...

Also, you should consider Eisenstein's criterion, which is a really handy tool to decide divisibility. You'll only need to know the primes in $\mathbb{Z}[\sqrt{2}]$ to be able to apply it.

3. Jul 21, 2011

### Kreizhn

I guess that I was perhaps a little sloppy here. Indeed, $\pm 1$ should also divide it, but they clearly don't work. Anyway, to be more complete, if $\alpha = a + b\sqrt 2$ and we give the norm $N(\alpha) = a^2 - 2b^2$ then if $p \big| \sqrt{2}$ then $N(p) \big| N(\sqrt2) = -2$, so $N(p) = \pm 1, \pm 2$. But then I need to go into Pell's equation and so maybe that's not the best way of doing this.

Ah yes, I actually knew this. Don't know why that must have slipped my mind.

How can one determine these primes? The easiest case here would be to use $\sqrt 2$ assuming it's prime. Assuming that I use the fact that $\mathbb Z[\sqrt2]$ is a UFD, this amounts to showing that $\sqrt2$ is irreducible. But then arne't we back to Pell's equation?

4. Jul 21, 2011

### micromass

Staff Emeritus
So it by brute force. Say that

$$(a+b\sqrt{2})(c+d\sqrt{2})=\sqrt{2}$$

We can assume gcd(a,b)=1 and gcd(c,d)=1. Then we have

$$ac+2bd=0~\text{and}~bc+ad=1$$

The first equation gives us ac=2bd. Since gdc(c,d)=1, we obtain that c divides 2b.
However, when multiplying the second equation by c, we get that

$$c=bc^2+acd=b(c^2+2d)$$

so b divides c. Thus $c=\pm b$ or $c=\pm 2b$. In the same fashion $a=\pm d$ or $a=\pm 2d$. Use this information to finally obtain that only +/- 1 or $\pm \sqrt{2}$ can be the divisors.