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 Alexis87 March 18th, 2018 06:00 PM

polynomial

1 Attachment(s)
I would like to have verification if the following attached proof is correct. If it is not correct, what can be done to make it correct? Thanks.

 SDK March 18th, 2018 07:46 PM

You have only shown that a factorization of $f(x)$ gives rise to a factorization of $f(x + c)$ which is trivial. The problem is that EVERY polynomial over a field splits linearly over some extension field. What you haven't shown is that the $h(x+c)$ and $g(x+c)$ lie in $F[x]$ which is required to establish a contradiction.

I think the way to prove this is to note that an ideal of $F[x]$ is maximal if and only if it is principally generated by an irreducible (over $F$) polynomial in $F[x]$. This means that $f \in F[x]$ is irreducible is equivalent to the statement that $F[x]/(f)$ is a field.

 Alexis87 March 18th, 2018 08:22 PM

1 Attachment(s)
Quote:
 Originally Posted by SDK (Post 590227) You have only shown that a factorization of $f(x)$ gives rise to a factorization of $f(x + c)$ which is trivial. The problem is that EVERY polynomial over a field splits linearly over some extension field. What you haven't shown is that the $h(x+c)$ and $g(x+c)$ lie in $F[x]$ which is required to establish a contradiction. I think the way to prove this is to note that an ideal of $F[x]$ is maximal if and only if it is principally generated by an irreducible (over $F$) polynomial in $F[x]$. This means that $f \in F[x]$ is irreducible is equivalent to the statement that $F[x]/(f)$ is a field.

Hi SDK, Thanks for the advice. I have done up as attached. Please see if it is correct. Thanks.

 SDK March 18th, 2018 08:27 PM

Quote:
 Originally Posted by Alexis87 (Post 590231) Hi SDK, Thanks for the advice. I have done up as attached. Please see if it is correct. Thanks.
You haven't changed anything in your "proof". I agree that $f(x + c) = g(x + c)h(x+c)$. Now, prove to me that $h(x+c),g(x+c) \in F[x]$. If you can't, then there is no contradiction.

 zylo March 19th, 2018 04:05 PM

I agree with OP.

Given f(x+c) is irreducible.
Assume f(x) = g(x)h(x)
Then f(x+c) =g(x+c)h(x+c), contradiction

Note: f(x+c) has same degree as f(x). (ditto g(x+c))

 SDK March 19th, 2018 06:52 PM

Quote:
 Originally Posted by zylo (Post 590282) I agree with OP. Given f(x+c) is irreducible. Assume f(x) = g(x)h(x) Then f(x+c) =g(x+c)h(x+c), contradiction Note: f(x+c) has same degree as f(x). (ditto g(x+c))
There is a difference between a true statement and a proof of that statement. Your "proof" has proved nothing since you need to show that $g(x+c),h(x+c) \in F[x]$. If this contradiction is so clear, please explain why it's a contradiction.

 zylo March 20th, 2018 06:34 AM

Quote:
 Originally Posted by SDK (Post 590319) There is a difference between a true statement and a proof of that statement. Your "proof" has proved nothing since you need to show that $g(x+c),h(x+c) \in F[x]$. If this contradiction is so clear, please explain why it's a contradiction.
IF you assume f(x) is reducible, then it can be factored into non-constant polynomials, f(x)=g(x)h(x). Then f(x+c)=g(x+c)h(x+c) is also a product of non constant polynomials because if g(x), h(x) are non-constant polynomials, so are g(x+c) and h(x+c).

$\displaystyle g(x)=x^{n}+a_{1}x^{n-1}+......$ deg n
$\displaystyle g(x+c)=(x+c)^{n}+a_{1}(x+c)^{n-1}+......=x^{n}+........$ deg n

So f(x+c) is reducible (can be factored into non-constant polynomials) if f(x) is.
But f(x+c) is irreducible. Therefore so is f(x).

 SDK March 20th, 2018 02:19 PM

Quote:
 Originally Posted by zylo (Post 590369) IF you assume f(x) is reducible, then it can be factored into non-constant polynomials, f(x)=g(x)h(x). Then f(x+c)=g(x+c)h(x+c) is also a product of non-constant polynomials because if g(x), h(x) are non constant polynomials, so are g(x+c) and h(x+c). $\displaystyle g(x)=x^{n}+a_{1}x^{n-1}+......$ deg n $\displaystyle g(x+c)=(x+c)^{n}+a_{1}(x+c)^{n-1}+......=x^{n}+........$ deg n So f(x+c) is reducible (can be factored into non-constant polynomials) if f(x) is. But f(x+c) is irreducible. Therefore so is f(x).

It is more subtle than that. If you aren't mentioning fields or polynomials rings then you have missed the point.

As an example: Is $x^2 - 2$ irreducible? How about $x^2 + 1$?

 zylo March 22nd, 2018 07:49 AM

Quote:
 Originally Posted by SDK (Post 590414) It is more subtle than that. If you aren't mentioning fields or polynomials rings then you have missed the point.
The conditions were given in OP Thumbnail: F[x] reducible over F.

 cjem March 22nd, 2018 09:15 AM

Quote:
 Originally Posted by SDK (Post 590227) What you haven't shown is that the $h(x+c)$ and $g(x+c)$ lie in $F[x]$
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 $g$.

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