# How many elements does $R$ have?

Let $f: R\to S$ be a homomorphism of rings (with $R$ commutative) such that kernel of $f$ has $4$ elements and image of $f$ has $16$ elements. How many elements does $R$ have?

Would you simply use the first isomorphism theorem in the following way? $R/\ker(f)=R/4$, so $R$ must have $16\cdot 4= 64$ elements.

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The correct calculation looks like that. But you have details wrong. e.g. the assertion $R/\ker(f) = R/4$ is way off base. –  Hurkyl Mar 31 '12 at 10:24
In more proper notation $16=|{\rm Im}(f)|=|R/\ker(f)|=|R|/|\ker(f)|=|R|/4$ so that $|R|=16\cdot4=64$. –  Andrea Mori Mar 31 '12 at 10:29

By the fundamental isomorphism theorem for rings, the image $f({R})$ is isomorphic to $R/Ker(f)$, and hence their cardinalities are equal. Furthermore $|R|=|ker(f)||R/Ker(f)|$, which is a consequence of Lagrange's Theorem. You should be able to figure out the rest I think.