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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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    $\begingroup$ ...Why do you edit that? $\endgroup$
    – Najib Idrissi
    Mar 29, 2014 at 10:30

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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    $\begingroup$ The second step is important on its own, and shows that any field of characteristics zero has a copy of the rational numbers inside. $\endgroup$
    – Asaf Karagila ♦
    Jun 8, 2012 at 21:21
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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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    $\begingroup$ This is only enough to show that $\mathbb Q$ is contained in the intersection, but why is it equal? $\endgroup$
    – Asaf Karagila ♦
    Jun 8, 2012 at 21:22
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    $\begingroup$ @Asaf Karagila:Perhaps because $\mathbb{Q}$ is itself subfield of $\mathbb{R}$. $\endgroup$
    – H. Kabayakawa
    Jun 8, 2012 at 22:43
  • $\begingroup$ @H.Kabayakawa: Yes, precisely because of that. However equality of sets is defined as a statement about inclusion of sets ($A=B\iff A\subseteq B\land B\subseteq A$), and this answer only gives $\subseteq$. $\endgroup$
    – Asaf Karagila ♦
    Jun 8, 2012 at 22:45
  • $\begingroup$ @Asaf Karagila: But Syd proved $\mathbb{Q}$ is contained in the intersection, and the intersection is in $\mathbb{Q}$ because $\mathbb{Q}$ is itself subfield. Then ... $\endgroup$
    – H. Kabayakawa
    Jun 8, 2012 at 22:57
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    $\begingroup$ @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. $\endgroup$
    – Asaf Karagila ♦
    Jun 8, 2012 at 23:12

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