# Showing the quotient is not a UFD

1. Nov 28, 2011

### Kreizhn

1. The problem statement, all variables and given/known data
Show that $\mathbb Q[x,y]/(x^2+y^2-1)$ is not a unique factorization domain.

3. The attempt at a solution

We have tried a few approaches. Using [] to denote equivalence classes, we note that we can write $[x]^2 = [1-y][1+y]$. Our goal was to show that this is a non-unique prime decomposition, but this doesn't work since neither of these elements are prime. Similarly, $[x+y-1][x+y+1] = [2x][y]$ but the same problem applies.

Next we tried playing around with evaluations of polynomials. In particular, in the quotient ring we can write any element as $p(x) + y q(x)$. Taking a product of such polynomials and evaluating at zero gives a structure similar to $\mathbb C$.

Another suggestion was to use Laurent polynomial. Any ideas?

2. Nov 28, 2011

### Office_Shredder

Staff Emeritus
Does your definition of UFD require a unique decomposition into primes or a unique decomposition into irreducibles?

3. Nov 28, 2011

### Kreizhn

Both. They are equivalent in a UFD no?

4. Nov 28, 2011

### Office_Shredder

Staff Emeritus
If it's not a UFD then it's not the same thing (namely, elements can be irreducible without being prime). So your first attempt at x2=(1-y)(1+y) is a good place to look... it's not a prime factorization, but they are irreducible factorizations

5. Nov 28, 2011

### Kreizhn

Yeah, that had occurred to me, though I'm not too sure how to show that $1 \pm y$ is irreducible. It seems obvious since they're monic first order polynomials, but I don't think I can just claim that (can I?). In any case, I would like to use Eisenstein's criterion but then I run into the problem of dealing with the prime ideals of $\mathbb Q$, of which there are no non-trivial proper ideals. Should I perhaps be looking at the prime ideals of in $\mathbb Q[x]$ instead?

6. Nov 28, 2011

### micromass

Staff Emeritus
I fear you need to start with the very definition of irreducible. You ring is pretty strange, so you can't use Eisenstein.

Assume that y+1=b*c and obtain a contradiction.

7. Nov 28, 2011

### Kreizhn

I think what you're hinting at is a degree argument, in that the only way that product could be a first order polynomial is precisely when one is a constant and the other is a first order polynomial. Since $\mathbb Q$ is a field, the constant is a unit, so we can conclude that 1-y is irreducible. However, does this necessarily apply in the quotient? Degree of equivalence classes is definitely not preserved in the quotient so I don't see how that argument would work.

Edit: The degree of the product of equivalence classes of polynomials in the quotient is not well behaved, is what I meant to say. That's a mouthful!

8. Nov 28, 2011

### micromass

Staff Emeritus
A direct attack would indeed not work. But let's look at it this way:

Assume

$$[y+1]=[P(x,y)][Q(x,y)]$$

Then

$$y+1-P(x,y)Q(x,y)\in (x^2+y^2-1)$$

So

$$y+1-P(x,y)Q(x,y)=R(x,y)(x^2+y^2-1)$$

and now try to obtain a contradiction.

9. Nov 28, 2011

### Kreizhn

Sorry. I've been trying to play around with this but I don't see it.

Okay. So I don't think this is a degree argument. I say this because there is no restriction on the degree of x in this case. The degree of y could have lead somewhere, but the presence of the -1 really means that y can take on any degree.

So I'm guessing this is a divisibility argument. I tried arguing that by re-arranging, we could get
$$(y+1)(R(x,y)(y-1) - 1) = -P(x,y)Q(x,y) - R x^2$$
in which case since y+1 divides the left hand side, it must also divide the right hand side. Though I don't see anything coming from this.

10. Nov 29, 2011

### Kreizhn

If I could show that [x] is irreducible and that the only the image of constants in $\mathbb Q[x,y]/(x^2+y^2-1)$ are units, could I show that it is not a UFD?