March 18th, 2018, 06:00 PM  #1 
Member Joined: Sep 2011 Posts: 97 Thanks: 1  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.
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  
Member Joined: Sep 2011 Posts: 97 Thanks: 1  Quote:
Hi SDK, Thanks for the advice. I have done up as attached. Please see if it is correct. Thanks.  
March 18th, 2018, 08:27 PM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 376 Thanks: 205 Math Focus: Dynamical systems, analytic function theory, numerics  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 
Senior Member Joined: Sep 2016 From: USA Posts: 376 Thanks: 205 Math Focus: Dynamical systems, analytic function theory, numerics  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  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  Quote:
$\displaystyle g(x)=x^{n}+a_{1}x^{n1}+...... $ deg n $\displaystyle g(x+c)=(x+c)^{n}+a_{1}(x+c)^{n1}+......=x^{n}+........$ deg n So f(x+c) is reducible (can be factored into nonconstant 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  
Senior Member Joined: Sep 2016 From: USA Posts: 376 Thanks: 205 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
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 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100  
March 22nd, 2018, 09:15 AM  #10  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 186 Thanks: 55 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
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 $a_m {m\choose k} c^k x^{mk}$ (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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