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I'm reading through Abstract Algebra by Hungerford and he makes the remark that the intersection of all subfields of the real numbers is the rational numbers.

Despite considerable deliberation, I'm unsure of the steps to take to show that the subfield is $\mathbb Q$.

Any insight?

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    ...Why do you edit that?2014-03-29

2 Answers 2

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First note that $\mathbb Q$ is itself a subfield of $\mathbb R$, so the intersection of all subfields must be a subset of the rationals.

Second note that $\mathbb Q$ is a prime field, that is, it has no proper subfields. This is true because if $F\subseteq\mathbb Q$ is a field then $1\in F$, deduce that $\mathbb N\subseteq F$, from this deduce that $\mathbb Z\subseteq F$ and then the conclusion.

Third, conclude the equality.

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    The second step is important on its own, and shows that any field of characteristics zero has a copy of the rational numbers inside.2012-06-08
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Any subfield of the reals must contain 0 and 1. Since the subfield is closed under addition and subtraction, it must contain all the integers. Since it's also closed under division (except division by zero), it must contain the rationals.

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    @H.Kabayakawa: But this answer is incomplete! It does not mention that "trivial" fact, and therefore cannot use it in proof. Many things in mathematics are "obvious" and "trivial" and the further you get, the more things become that way (and simultaneously, less obvious and less trivial in a significant way!) but you still **have to prove the things you wish to use in your arguments**. This is just how mathematics works. The fact that $\mathbb Q$ is a subfield of $\mathbb R$ is indeed true and trivial. It requires *at least* mentioning, if you wish to use it.2012-06-08