May 19th, 2010, 11:28 AM  #1 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  factor ring
Let F be a field and f(x),g(x) in F[x]. Show that f(x) divides g(x) if and only if g(x) in <f(x)>. <= if g(x) in <f(x)> then f(x) divides g(x). we have that F[x]/<f(x)> and since g(x) in <f(x)> and f(x) is in F[x] we have that f(x)/g(x). => if f(x) divides g(X) then g(x) in <f(x)>. since f(x)/g(x) in F[x] and we have that F[x]/<f(x)> hence g(x) is in f(x). 
May 19th, 2010, 04:54 PM  #2 
Senior Member Joined: Apr 2008 Posts: 435 Thanks: 0  Re: factor ring
Whoa. What are you doing? This question looks like the definition of an ideal generated by f(x).

May 19th, 2010, 05:23 PM  #3 
Senior Member Joined: Nov 2009 Posts: 129 Thanks: 0  Re: factor ring
<=> if f(x) divides g(x) then g(x) in <f(x)> Proof: Suppose f(x) divides g(x)q(x). then g(x)q(x) in <f(x)>. which is maximal. Therefore <f(x)> is a prime ideal. Hence g(x)q(x) in <f(x)>. implies that either g(x) in <f(x)> giving f(x) divides g(x) or that q(x) in <f(x)> giving f(x) divides q(x). But we want that g(x) in <f(x)> giving f(x) divides g(x). can this prove go both way if it is right? 
May 21st, 2010, 04:39 AM  #4 
Senior Member Joined: Apr 2008 Posts: 435 Thanks: 0  Re: factor ring
Ah, those are supposed to mean 'divides.' You see, when I see A/B  I see A quotiented out by the ideal of B. When I see AB, I see that A divides B. This is good, as I didn't understand the first post at all.


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