Page 180 - DMTH403_ABSTRACT_ALGEBRA

Unit 17: Ring Homomorphisms

Secondly, take any f(x)  f(1) and r  R . Since f is surjective,  s  R such that f(s) = r. Notes
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Then

rf(x) = f(s) f(x) = f(sx)  f(I), since sx  I.
Thus, f(1) is an ideal of R . 2
Now, consider an epimorphism f : R  S and an ideal I in R. By Theorem 3 you know that f(1) is
an ideal of S and f (f(I)) is an ideal of R. How are I and f (f(I)) related? Clearly, I  f (f(1)).
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Can f (f(T)) contain elements of R\I? Remember that Ker f  f (f(1)) also. Thus,
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I + Ker f  f (f(1)). In fact, I + Ker f = f (f(1)). Let us see why.
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Let x  f (f(l)). Then f(x)  f(1). Therefore, f(x) = f(y) for sdme y  I. Then
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f(x – y) = 0.
 x – y  Ker f, i.e., x y + Ker f  I + Ker f.
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 f (f(I))  I + Ker f.
Thus, f (f(I)) = I + Ker f.
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This tells us that if Ker f  I, then
f (f(I)) = I (since Kerf  I + Ker f = I).
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Theorem 4: Let f : R  S be an onto ring homomorphism. Then

(a) if I is an ideal in R containing Ker f, I = f (f(I))
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(b) the mapping 1 – f(I) defines a one-to-one correspondence between the set of ideals of R
containing Ker f and the set of ideals of S.

Proof: We have proved (a) in the discussion above. Let us prove (b) now.
Let A be the set of ideals of R containing Ker f, and B be the set of ideals of S.
Define  : A  B : 4(I) = f(I).
We want to show that  is one-one and onto.
 is onto : If J  B then f (J) A and Ker f  f (J) by Theorem 3.
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Now  (f (J)) = f(f (J)) = J,
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 is one-one : If I and I are ideals in R containing Ker f, then
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(I ) = (I )  f(I ) = f(I )
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1
1
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 f (f(I )) = f (f(I ))
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2
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 I = I , by (a).
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Thus,  is bijective.
And now let us look closely at the sets Ker f and Im f, where f is a ring homomorphism. In Unit
6 we proved that iff : G – G is a group homomorphism then Ker f is a normal subgroup of G 1
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and Im f is a subgroup of G . We have an analogous result for ring homomorphisms, which you
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may have already realised from the examples you have studied so far.
Theorem 5: Let f : R – R be a ring homomorphism. Then
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(a) Ker f is an ideal of R .
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(b) Im f is a subring of R .
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