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Show that for a commutative ring $R$, there always exists a ring homomorphism from $R$ to some field.

I try to extend $R$ to a division ring by throwing in multiplicative inverse for each element, but then I don't know whether I can finally construct a division ring.

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  • $\begingroup$ You cannot necessarily extend R to a division ring because R may contain non-zero a,b such that ab=0. For example, let r be arithmetic on the integers,modulo 12. We cannot extend R to include a solution to 3x=1 as this implies 0=12=12x=4.3x=4.1=4 (mod 12). $\endgroup$ – DanielWainfleet Jan 25 '16 at 10:24
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Rings may contain zero divisors which cannot logically have inverses. Fortunately the homomorphism to the field does not need to be injective, so we can 'quotient out' these zero divisors. Note that $R$ can be injected into a field if and only if it is an integral domain. Taking the natural quotient and localization maps, $R$ maps into a field $(R/ \mathfrak{p})_{(0)}$ (the quotient field of $R/\mathfrak{p}$) for each prime ideal $\mathfrak{p}$. Because the 'only if' part of the consideration above, these are really the only cases we have a map from $R$ to a field: if $f:R \to k$ is such a map then $k$ is an extension of some $(R/\mathfrak{p})_{(0)}$.

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    $\begingroup$ We might as well take the quotient of $R$ with a maximal ideal, so that we immediately have a homomorphism onto a field. $\endgroup$ – hardmath Jan 25 '16 at 5:20
  • $\begingroup$ that provides one kind of example but it does not describe all cases of maps $\endgroup$ – basket Jan 25 '16 at 5:22
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I wanted to add a comment but are not allowed to. Your ring has to be nonzero, otherwise there is no morphism from $R$ into a field, supposing that you consider unitary morphisms i.e. morphisms that map $1$ to $1$.

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  • $\begingroup$ What do you mean by "nonzero"? $\endgroup$ – DanielWainfleet Jan 25 '16 at 10:28
  • $\begingroup$ A ring is zero if it contains only one element, namely $0$. In such a ring one has $1=0$. Hence, such a ring does not admit a unitary morphism into a field. Remember that by definition a field is a nonzero ring, and hence $1\neq0$ in a field. $\endgroup$ – Johannes Huisman Jan 26 '16 at 12:16
  • $\begingroup$ Thanks. What is or isn't "zero" depends on the subject. I just needed to be sure, because of the issue here of divisors of 0. If a set-theorist says zero, he may mean the empty set. Of course a ring is not empty. $\endgroup$ – DanielWainfleet Jan 26 '16 at 20:53

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