Proof by Contradiction: Irreducible Polynomials and Ideals

[Syrus]

Homework Statement

(see attachment)

Homework Equations

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)

 

[SammyS]

Homework Statement

(see attachment)

Homework Equations

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)

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

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

 

[micromass]

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

 

[Syrus]

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

 

[micromass]

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

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.

 

[Syrus]

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

 

[micromass]

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

 

[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.

 

[micromass]

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 implies that f(x) = g(x), a contradiction since we assumed that f and g differ in degree.

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.

 

[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.

 

[micromass]

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!

 

[Syrus]

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

 

[micromass]

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

 

[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 =)