March 22nd, 2018, 09:58 AM  #11 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
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. 
March 22nd, 2018, 02:04 PM  #12  
Senior Member Joined: Sep 2016 From: USA Posts: 376 Thanks: 205 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
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.  
March 22nd, 2018, 02:44 PM  #13  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 186 Thanks: 55 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
 
March 26th, 2018, 06:42 AM  #14 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,364 Thanks: 100 
If f(x)is reducible over F, so is f(x+c) f(x)=(xa)(xb)(xd) say. Then f(x+c)=(xa+c)(xb+c)(xd+c). ie, f(x+c) irr cannot imply f(x) red because that would imply f(x+c) red. 
March 30th, 2018, 08:34 AM  #15 
Member Joined: Sep 2011 Posts: 97 Thanks: 1 
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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