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March 18th, 2018, 06:00 PM   #1
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polynomial

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.
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Last edited by Alexis87; March 18th, 2018 at 06:04 PM.

 March 18th, 2018, 07:46 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 376 Thanks: 205 Math Focus: Dynamical systems, analytic function theory, numerics 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.
March 18th, 2018, 08:22 PM   #3
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Quote:
 Originally Posted by SDK 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.
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March 18th, 2018, 08:27 PM   #4
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Quote:
 Originally Posted by Alexis87 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.

 March 19th, 2018, 04:05 PM #5 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 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))
March 19th, 2018, 06:52 PM   #6
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Quote:
 Originally Posted by zylo 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.

Last edited by skipjack; March 22nd, 2018 at 10:42 AM.

March 20th, 2018, 06:34 AM   #7
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Quote:
 Originally Posted by SDK 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).

Last edited by skipjack; March 22nd, 2018 at 10:43 AM.

March 20th, 2018, 02:19 PM   #8
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Quote:
 Originally Posted by zylo 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 $x^2 + 1$?

Last edited by skipjack; March 22nd, 2018 at 10:44 AM.

March 22nd, 2018, 07:49 AM   #9
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Quote:
 Originally Posted by SDK 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.

March 22nd, 2018, 09:15 AM   #10
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Quote:
 Originally Posted by SDK 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 $g$.

Last edited by cjem; March 22nd, 2018 at 09:41 AM.

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