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Suppose $F$ is a field and $R$ is a ring. The function $f\colon F\to R$ is a surjective homomorphism. Prove that $R$ is either the trivial ring, or $R$ is isomorphic to $F$.

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Any thoughts of your own? – Alex B. Nov 16 '12 at 17:03
hint: the image of an invertible element is invertible (why?) – the L Nov 16 '12 at 17:04
Hint: by the first isomorphism theorem the image is characterized by the kernel, for which there are only two choices in a field, the ideals $(0)$ and $(1).\ \ $ – Bill Dubuque Nov 16 '12 at 17:05
Consider the kernel. – i. m. soloveichik Nov 16 '12 at 17:05
up vote 3 down vote accepted

We have a ring homomorphism $f : F \to R$ here are the key facts:

  • the inverse image of an ideal is an ideal.
  • the only ideals of a field $F$ are $(0)$ and $F$.
  • the kernel of a ring homomorphism is an ideal.

Take the inverse image of the kernel, if it's $(0)$ the rings are isomorphic. If it's $F$ the homomorphism is trivial.

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