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Given $F$ is a field, $R$ is a ring, and $\phi:F\to R$ is a surjective ring homomorphism. How do we prove that this makes $\phi$ is a bijection and $R$ is a field.

Simplest possible explanation is most appreciated! I am looking for an intuitive understanding.

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    Do you know what all of the ideals in $F$ are? Can you see why $R$ must be a field if $\phi$ is bijective? By the way, it is assumed that $R$ is not the zero ring for this to be true.2011-11-28

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The kernel of a ring homomorphism is a two sided ideal, so in this case $\ker\phi$ is an ideal of $F$. But $F$ is a field, and the only two possible ideals are $(0)$ or $F$, for any nonzero ideal must contain a unit, and thus is the whole field.

Assuming $R$ is not the zero ring, you must have $\ker\phi=(0)$, otherwise $\phi$ would not be surjective. So $\phi$ is injective with trivial kernel, and thus $\phi$ is a field isomorphism, so $R$ is a field also.

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The standard elegant ideal-theoretic proof is that given by yunone. But if you haven't yet learned about ideals you can instead proceed as follows. If $\phi$ is not $1$-$1$ then for some $\rm\:a\ne b$ we have $\rm\:\phi(a)=\phi(b)$ so $\rm\:\phi(a-b)=\phi(a)-\phi(b) = 0\:,\:$ so, for $\rm\ c = a-b\:,\:$ $\rm\phi(1) = \phi(c\:c^{-1}) = \phi(c)\:\phi(c^{-1}) = 0\:$ hence $\rm\:\forall x\!:\ \phi(x) = \phi(x\cdot 1) = \phi(x)\:\phi(1) = 0\:,\:$ so $\rm\:R = \phi(F) = \{0\}\:,\:$ contra (omitted) hypothesis.