# When the Product of Non-Squares is a Square

1. Jan 15, 2012

### Kreizhn

1. The problem statement, all variables and given/known data
For some field F, if a and b are two elements which are not squares but whose product is a square, show that there is an element k such that $a = k^2 b$.

3. The attempt at a solution

This is a boiled down version of what I'm actually trying to show, so the statement above may not actually be correct. If it's not, please let me know!

I have a few ideas of how to hit this, but the following is the one I like the most:

If $ab = c^2$ for some element $c \in F$ then it is sufficient to show that either $a | c$ or $b | c$, since then (without loss of generality) if $a | c$ then there is some constant k such that ak = c in which case $ab = a^2 k^2$ which implies that $b = k^2 a$. Now the hard part (for me at least!) is showing that $ab = c^2$ implies that either a|c or b|c. It seems like this result should be true, but I've been having trouble showing it.

I start like this. If both a|c and b|c then we are done, hence assume without loss of generality that $b \not| c$. Here is where I get stuck. Any ideas?

2. Jan 15, 2012

### Bacle2

Sorry to change your strategy, but , in the factoring of a square, every factor should
appear an even number of times.

Then ab=∏xi2ki

3. Jan 15, 2012

### Dick

Are you sure you are working in a FIELD? If that's the case, then you can always solve ak=c for k if a is nonzero. k=c*a^(-1).

4. Jan 15, 2012

### Kreizhn

Bacle2, I agree entirely. In particular, if we apply this to the case when a and b are not squares it implies that a and b must share a prime factor, since otherwise they would be squares themselves. However, I don't quite see why that helps. Care to elaborate?

Yes, you are of course correct. I wonder if this would actually work for what I was originally trying to do. I'll have to think about this.

In any case, I believe that the statement

"If $ab = c^2$ then either a|c or b|c"

is still correct. I would be interested in seeing a proof of this if anyone can suggest one.

5. Jan 15, 2012

### Dick

The proof is trivial in a field. Everything divides everything. If you are working over the integers then 4*9=36=6^2. Neither 4 nor 9 divides 6.

6. Jan 15, 2012

### Kreizhn

Ah excellent. Thank you.

7. Jan 15, 2012

### Bacle2

While Dick gave you a nice answer, my idea was to use the fact that fields are UFD's, so that the product ab has a unique expression in which c must appear twice on the LH side.

8. Jan 15, 2012

### Dick

How many fields have you seen that are UFD's??

9. Jan 15, 2012

### Bacle2

Every field is a UFD, I'm almost 100%. Do you have a counter?

10. Jan 15, 2012

### Dick

The rationals, the reals, the complexes, integers modulo a prime p where p isn't 2, fields of order p^n where n>1, Galois extension fields. Do you want me to go on? Basically all of them. Think about it. What are the primes in the rationals, Q?

Last edited: Jan 15, 2012
11. Jan 15, 2012

### Bacle2

If F is a field, then F[X] is a UFD, e.g: http://planetmath.org/encyclopedia/PrimeFactor2.html [Broken]

This means that every poly. in F[X] has a unique factorization. Now,F embeds trivially

in F[X], right? Or isn't every f in F an element in F[X]?

Just use the embedding f--> f+ 0.x+0.x2+....+0xn

It then follows every f in F has a unique factorization.

Last edited by a moderator: May 5, 2017
12. Jan 16, 2012

### Dick

You seem to be googling instead of thinking. Unique factorization of polynomials is modulo the underlying field F. x^2-1=(x-1)(x+1)=(2x-2)(x/2+1/2). 2=3*(2/3)=5*(2/5). Does that look like unique factorization?

Last edited by a moderator: May 5, 2017
13. Jan 16, 2012

### Bacle2

I was referring to your statement:

So every field is a UFD, by the definition of UFD, since, trivially, every element of a field is a unit. I just used the link to avoid doing a proof of this fact. Maybe you were
using the term in relation to this problem/situation which I guess I missed, but, unless otherwise stated, every field is a UFD.

I just want to add that sometimes I go thru old posts, and I prefer to be maybe overly-finicky in using terminology to avoid picking up things that , however answer
a question, are not fully correct in the strict sense.

Last edited: Jan 16, 2012
14. Jan 16, 2012

### Dick

Yes, everything is a unit. So it's a trivial UFD. This post is what I've actually been going on about. What is the 'unique expression'? 1?

15. Jan 16, 2012

### Bacle2

My point is that once the expression is fixed, there is a unique, trivial solution, because every element divides every element, by the definition of field.

Then a|c is tautological for every non-zero element in the field, because every element is a unit, then a|c , inthe sense that a/c :=ac-1 , which is in F by closure properties of F.

Sorry, my connection is not good tonight.

16. Jan 16, 2012

### Bacle2

Anyway, this thread has become an exercise in futility and misunderstanding (misunderestimating?).

Yes, I guess I could have avoided bringing up the UFD issue, which does not really help, to say the least. My bad.

Let me delete my posts; they only seem to add confusion. Just wanted to ask.

Last edited: Jan 16, 2012
17. Jan 16, 2012

### Dick

You are right, of course. Not having any irreducible elements does NOT mean a field is not a UFD. Sorry, I seem to have gotten stuck in that impression. May have been too late.