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March 22nd, 2018, 10:58 AM   #11
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SDk

I think I see what's bothering you.

The operations which qualify F[x] as a ring treat x as a symbol, not a variable, so technically f(x+c) is undefined.

About the best you can say is, that if c is in F, then evaluating f(x+c) is a legitimate member of F[x].

As for the algebra of F[x] when x is a variable, I don't know. Is it still a Ring?
Another topic.

But it doesn't matter.You can interpret f(x+c) as a prescription for obtaining a legitimate polynomial with x as a symbol. That works.
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March 22nd, 2018, 03:04 PM   #12
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Quote:
Originally Posted by cjem View Post
Isn't this pretty clear?

Write $h(x) = a_0 + a_1 x + \dots + a_n x^n \in F[x]$. Then $h(x+c)$ can be expressed as a sum of monomials of the form $a_m {m\choose k} c^k x^{m-k}$ (where, here, ${m\choose k}$ really means the multiplicative identity of $F$ added to itself ${m\choose k}$ times). We know $c$ lies in $F$ and $a_m$ lies in $F$ for all $m$, so each such monomial is in $F[x]$. So any sum of these monomials (and in particular $h(x+c)$) lies in $F[x]$. Same idea for $g$.
I never claimed it was difficult to prove. I was simply pointing out that it was never even mentioned in either of the supposed proofs in this thread. In fact, each of these supposed proofs concluded by showing that $f(x+c)$ could be factored. However, without mentioning that the factors are in $F[x]$, this is true for EVERY polynomial, even irreducible polynomials (just find factors in the splitting field).

It appears this subtlety was (and still is) lost on some who have mistaken this exercise for a trivial calculus 1 problem about factoring polynomials.
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March 22nd, 2018, 03:44 PM   #13
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Originally Posted by SDK View Post
I never claimed it was difficult to prove. I was simply pointing out that it was never even mentioned in either of the supposed proofs in this thread. In fact, each of these supposed proofs concluded by showing that $f(x+c)$ could be factored. However, without mentioning that the factors are in $F[x]$, this is true for EVERY polynomial, even irreducible polynomials (just find factors in the splitting field).

It appears this subtlety was (and still is) lost on some who have mistaken this exercise for a trivial calculus 1 problem about factoring polynomials.
Fair enough. While I do appreciate your point and agree that this subtlety is something to be aware of, I'm not sure if it was actually being missed here. While it's slightly abusive to do so, it's pretty standard to say "$f(x)$ can be factored" to mean "$f(x)$ can be factored over $F[x]$" when the only field mentioned is $F$. There's not really any ambiguity in doing so - it wouldn't be used to mean "$f(x)$ can be factored over some extension" since, as you point out, this would apply to every polynomial and thus be a completely useless statement!
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March 26th, 2018, 07:42 AM   #14
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If f(x)is reducible over F, so is f(x+c)
f(x)=(x-a)(x-b)(x-d) say. Then
f(x+c)=(x-a+c)(x-b+c)(x-d+c).

ie, f(x+c) irr cannot imply f(x) red because that would imply f(x+c) red.
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March 30th, 2018, 09:34 AM   #15
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Thanks everyone for the help. I really appreciate it. I will share the outcome once the prof has reviewed his solution.
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