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Isomorphism homeworkby Icebreaker
Tags: isomorphism 
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#1
Nov1205, 01:30 PM

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"Let R be the ring Zp[x] of polynomials with coefficients in the finite field Zp, and let f:R>S be a surjective homomorphism from R to a ring S. Show that S is either isomorphic to R, or is a finite ring."
If S is isomorphic to R, then we're done. If S is not isomorphic to R, then by hypothesis, f is not injective. That is, ker(f) contains more than the zero element of R. ker(f) is an ideal of R, therefore there exists a surjective homomorphism p:R>R/ker(f), where p(x) = x + ker(f). My question is, if R is homomorphic to S and R is homomorphic to T, what is the relation between S and T? Are they the same? Isomorphic? In the question, R and S are given. In my answer, S is constructed. 


#2
Nov1205, 03:39 PM

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Did you finishing solving the problem? If Ker(f) has a nonzero element, then it has some polynomial of degree k. You know then that elements of R/Ker(f) have degree no more than k, and since there are only a finite number of possible coefficients for a given term in a polynomial, you've shown that R/Ker(f) is finite, hence so is S.
Certainly, S and T need not be the same or isomorphic. For every polynomial p in R, there is the ideal generated by p, (p), and there is always a surjective homomorphism from R to R/(p). But it's not true that for any polynomials p, q, that R/(p) is isomorphic to R/(q). 


#3
Nov1205, 05:14 PM

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I did finish solving the problem. I was just wondering whether the answer to my question was true. Thanks anyway.



#4
Nov1505, 06:48 PM

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Isomorphism homework
Well the correct answer to any question is true ;). Is your question the following: If there exists a surjective homomorphism f : R > S and a surjective homomorphism g : R > T, what is the relation between S and T? Well we know S = R/Ker(f), and T = R/Ker(g). We know Ker(f) and Ker(g) are ideals. Moreover, we know that for any ideal J, there is a surjective homomorphism j : R > R/J, which simply projects r in R to r + J (i.e. the coset of J containing r). Actually, I don't know all of this, this is what I would suspect is true from what I know of groups and what little I know of rings. Anyways, assuming the above is true, your question reduces to asking what the relationships are between two arbitary ideals of R. In general, R may have more than one ideal (in fact, it always has more than one ideal because it has R and {0}, but it may have more than one trivial ideal as well), and it may even have many nontrivial ideals which are not all isomorphic to one another. So if R is arbitrary, there is not much that can be said about the relationship between two ideals of R. Maybe in particular cases, R will only have nontrivial ideals that are isomorphic to one another. For an analogy, the group Z_{5} x Z_{5} has more than one nontrivial normal subgroup but any two nontrivial normal subgroups are isomorphic.



#5
Nov1505, 08:29 PM

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Exercise: Let R and S be commutative rings. Find a ring T and surjective homomorphisms T>R and T>S.
(commutativity isn't necessary) 


#6
Nov1505, 09:47 PM

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T = R x S, with the homomorphisms given by (r,s) > r and (r,s) > s respectively.



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