# Polynomials and Ideals

1. Feb 9, 2012

### Syrus

1. The problem statement, all variables and given/known data

(see attachment)

2. Relevant equations

3. The attempt at a solution

I have been attempting a proof by contradiction (for the last statement) for a while now, but I can't seem to reach a contradiction from these premises:

1 ≤ deg(f) ≤ deg(g) (without loss)
N ≠ F[x]
both f and g are irreducible over F
I also know that since F is a field, N is principal (by a lemma)

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2. Feb 9, 2012

### SammyS

Staff Emeritus
https://www.physicsforums.com/attachment.php?attachmentid=43713&d=1328829169

Having the image appear directly in this thread, may help someone to answer this.

3. Feb 9, 2012

### micromass

Staff Emeritus
If f(x) is irreducible, what can you say about (f(x)) in relation to N?? Are they equal?? (use that N is principal)

4. Feb 9, 2012

### Syrus

So since f(x) is irreducible over F[x], <f(x)> is maximal. Is this what you're getting at micromass?

5. Feb 9, 2012

### micromass

Staff Emeritus
Yes. So <f(x)>=N, right?? But if g(x) is irreducible, then also <g(x)>=N.

So <f(x)>=<g(x)>!! Try to write out what that means.

6. Feb 9, 2012

### Syrus

I may be blanking, but how can we be sure that <f(x)> = N?

7. Feb 9, 2012

### micromass

Staff Emeritus
<f(x)> is maximal, N is an ideal that contains <f(x)>...

8. Feb 9, 2012

### Syrus

Foolish of me. Thank you.

So since <f(x)> = <g(x)>, {f(x)s(x) | s(x) in F[x]} = {g(x)r(x) | r(x) in F[x]}, which, for s(x), r(x) = 1, implies that f(x) = g(x), a contradiction since we assumed that f and g differ in degree.

9. Feb 9, 2012

### micromass

Staff Emeritus
No, that is not correct. You can't conclude that f(x)=g(x) because it is simply not true.

You now that f(x)=g(x)r(x) for some r. You also now that there is an s such that g(x)=f(x)s(x). Take the degrees of both equations.

10. Feb 9, 2012

### Syrus

Let deg(r) = j and let deg(s) = k.

deg(f) = m = deg(g) + deg(r) = n + j
deg(g) = n = deg(f) + deg(s) = m + k

So by substitution from above, n = (n + j) + k, which implies that j + k = 0. But the only way this is true is if j = k = 0, which is shows that m = n; a contradition.

11. Feb 9, 2012

### micromass

Staff Emeritus
Good!!

Another way to prove it is to manipulate the equations directly to prove that r(x)s(x)=1 and therefore r(x) and s(x) are invertible and thus be an element of F

So now you know that if (f(x))=(g(x)) then there must be a c in F such that f(x)=cg(x)!!

Well done!

12. Feb 9, 2012

### Syrus

One question remains: How do we know for sure that <f(x)> is not equal to {0}?

13. Feb 9, 2012

### micromass

Staff Emeritus
That would mean that f(x)=0. But f(x) is supposed to be irreducible, and 0 (by definition) isn't.

14. Feb 9, 2012

### Syrus

*embarrassed. I don't know why i keep wasting your time tonight micromass... I think too much problem solving for one day is drying me out. Much thanks =)