Question #17277

Show that if f : R → S is a surjective ring homomorphism, then f(rad R) ⊆ rad S.
Give an example to show that f(rad R) may be smaller than rad S.

Expert's answer

The inclusion *f*(rad *R*)*⊆** *rad *S *is clear since, for any maximal left ideal m of *S*,the inverse image *f*^{−}^{1}(m) is a maximal left idealof *R*. To give an example of strict inclusion, let *R *= Z and let *f*be the natural projection of *R *onto *S *= Z*/*4Z. Here, *f*(rad*R*) = *f*(0) = 0*, *but rad *S *= 2Z*/*4Z is nonzero.

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