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November 16th, 2010, 09:27 AM  #1 
Newbie Joined: Nov 2010 Posts: 10 Thanks: 0  Ring homomorphisms and kernel
Hi! In a book about number theory (Number fields  Daniel A.Marcus), he talk about the inertial degree. There is two steps here I don`t fully understand. He let K and L be numberfields where K? L and R=A?K og S=A?L (Where A is the set of algebraic integer in C) P and Q are non zero prime ideals, where P is in R and Q in S. Futhermore we know that Q lies over P. question 1: this is the point I wonder about : R? S => R > S/Q i.e. The containment of R in S induces a ringhomomorphism "R > S/Q". Why is that? Of course we know that the following is required for a ring homomorphism: i) f(a+b) = f(a) + f(b) and ii) f(ab) = f(a)f(b) As I understand, the inclusion "?" serves as our "f" over. We have that: i) f(a +R b) = f(a) +s f(b) ii) f(a *R b) = f(a) *s f(b) question 2: From the above the book concludes that the kernel must be R?Q Why is that? I would be really greatfull if somebody could explain to me more in detail about these questions. Thank you so much for your time and effort. Thomas 
November 16th, 2010, 11:05 PM  #2 
Senior Member Joined: Aug 2010 Posts: 195 Thanks: 5  Re: Ring homomorphisms and kernel
For your first question, note that since is an ideal of , there is a homomorphism which takes an element to its congruence class modulo . Furthermore, we have the inclusion map which is also a homomorphism. Composing these maps gives us a rather natural homomorphism . For the second question, note that has kernel exactly , and since the inclusion map is injective (and hence has kernel only the zero element), under the composition of maps, an element in will go to zero in if and only if the inclusion map sends an element of into , but since this map is just inclusion, these are exactly the elements in . 
November 17th, 2010, 12:28 AM  #3 
Newbie Joined: Nov 2010 Posts: 10 Thanks: 0  Re: Ring homomorphisms and kernel
Ah, ok I see. Thank you 

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homomorphisms, kernel, ring 
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