Isomorphism


by Icebreaker
Tags: isomorphism
Icebreaker
#1
Nov12-05, 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.
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AKG
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#2
Nov12-05, 03:39 PM
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Did you finishing solving the problem? If Ker(f) has a non-zero 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).
Icebreaker
#3
Nov12-05, 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.

AKG
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#4
Nov15-05, 06:48 PM
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Isomorphism


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 non-trivial 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 non-trivial ideals that are isomorphic to one another. For an analogy, the group Z5 x Z5 has more than one non-trivial normal subgroup but any two non-trivial normal subgroups are isomorphic.
Hurkyl
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#5
Nov15-05, 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)
AKG
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#6
Nov15-05, 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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