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 November 16th, 2010, 08:27 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, 10:05 PM #2 Senior Member   Joined: Aug 2010 Posts: 195 Thanks: 5 Re: Ring homomorphisms and kernel For your first question, note that since $Q$ is an ideal of $S$, there is a homomorphism $\phi : S \rightarrow S/Q$ which takes an element to its congruence class modulo $Q$. Furthermore, we have the inclusion map $i : R \rightarrow S$ which is also a homomorphism. Composing these maps gives us a rather natural homomorphism $\phi \circ i : R \rightarrow S/Q$. For the second question, note that $\phi$ has kernel exactly $Q$, and since the inclusion map is injective (and hence has kernel only the zero element), under the composition of maps, an element in $R$ will go to zero in $S/Q$ if and only if the inclusion map sends an element of $R$ into $Q$, but since this map is just inclusion, these are exactly the elements in $R \cap Q$.
 November 16th, 2010, 11:28 PM #3 Newbie   Joined: Nov 2010 Posts: 10 Thanks: 0 Re: Ring homomorphisms and kernel Ah, ok I see. Thank you

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