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I just started reading Linear Algebra by Hoffman and Kunze, and I came across the following line:

The interested reader should verify that any subfield of $\mathbb C$ must contain every rational number.

Of course, I am interested, so I tried to come up with a proof. I just need verification on the correctness of my proof, and any feedback on how I can improve it. So here it goes.

Proof

A subfield is simply a subset of a field. A field $\mathbb F$ must obey the following axioms:
1. Addition is commutative, i.e. $x + y = y + x$ for all $x, y$ in $ \mathbb F$.
2. Addition is associative, i.e. $x + (y + z) = (x + y) + z$ for all $x, y, z$ in $\mathbb F$.
3. There exists a unique additive identity $0$ such that $x + 0 = x$ for all $x$ in $\mathbb F$.
4. There exists a unique additive inverse $-x$ such that $x + (-x) = 0$ for all $x$ in $\mathbb F$.
5. Multiplication is commutative, i.e. $x \cdot y = y \cdot x$ for all $x, y$ in $\mathbb F$.
6. Multiplication is associative, i.e. $x \cdot (y \cdot z) = (x \cdot y) \cdot z$ for all $x, y, z$ in $\mathbb F$.
7. There exists a unique multiplicative identity $1$ such that $x \cdot 1 = x$ for all $x$ in $\mathbb F$.
8. For every nonzero $x$ in $\mathbb F$, there exists a unique multiplicative inverse $x^{-1}$ such that $x \cdot x^{-1} = 1$.
9. Multiplication is distributive over addition, i.e. $x \cdot (y + z) = x \cdot y + x \cdot z$ for all $x, y, z$ in $\mathbb F$.
10. Closure under addition, i.e. for all $x, y$ in $\mathbb F$, $x + y$ must also be in $\mathbb F$.
11. Closure under multiplication, i.e. for all $x, y$ in $\mathbb F$, $x \cdot y$ must also be in $\mathbb F$.

From the above it follows that any subfield $\mathbb F$ of $\mathbb C$ must contain the elements $0$ and $1$ (as the additive and multiplicative identity respectively). This satisfies axioms $3$ and $7$. By our axiomatic definition of fields, any subfield of $\mathbb C$ must also satisfy axioms $1, 2, 5, 6, 9$. To satisfy axiom $10$, we see that $\mathbb F$ must contain $\mathbb Z^+$ by induction (I do not know how to prove this). To satisfy axiom $4$, it follows that $\mathbb F$ must contain $\mathbb Z^-$ as well. Hence $\mathbb F$ contains $\mathbb Z$. To satisfy axiom $8$, we need to include all scalars of the form $\frac 1x$, where $x \in \mathbb Z \setminus\{0\}$. To accomodate axiom $11$, we must include all scalars of the form $\frac 1x \cdot y$, where $x \in \mathbb Z \setminus \{0\}, y \in \mathbb Z$. Hence $\mathbb F$ must contain $\mathbb Q$. Now every axiom is satisfied and we see that any subfield of $\mathbb C$ must at least contain $\mathbb Q$.

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    $\begingroup$ $\;\underbrace{1+1+...+1}_{n\;\text{times}}=n\;$ must be in $\;\Bbb C\;$ , and this is true for any $\;n\in\Bbb N\;$ . Taking now inverses you get that all $\;\Bbb Z\subset \Bbb C\;$ and etc. $\endgroup$
    – Timbuc
    Dec 1, 2014 at 14:26
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    $\begingroup$ This is not a question... $\endgroup$
    – 5xum
    Dec 1, 2014 at 14:26
  • $\begingroup$ Sorry, I should have added the "proof-verification" tag to make my question more explicit. $\endgroup$
    – Vizuna
    Dec 1, 2014 at 14:28
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    $\begingroup$ As a side note to this problem (Which is a top down approach of the minimality of $\mathbb Q$), it can also be shown that $\mathbb Q$ is the smallest field that contains $\mathbb Z$ (It is in fact the 'field of quotients' for the integral domain of the integers $\endgroup$
    – Alan
    Dec 1, 2014 at 15:30
  • $\begingroup$ Can we prove it by contradiction also? $\endgroup$
    – Ritu
    Feb 20, 2015 at 13:41

1 Answer 1

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What you have done so far is ok.

To prove that $\Bbb Z^+$ is in $\Bbb F$, just note that $1\in\Bbb F$ and that $n+1$ is in $\Bbb F$ whenever $n$ is in $\Bbb F$ (axioms 7 and 10).

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  • $\begingroup$ It makes sense that if $1 \in \mathbb F$, and $n \in \mathbb F$, then $n + 1 \in \mathbb F$ (by axiom 10). But how do we prove that it covers the more general case "for all $x,y$ in $\mathbb F$, $x + y$ must also be in $\mathbb F$"? Is such a proof even necessary? $\endgroup$
    – Vizuna
    Dec 1, 2014 at 14:36
  • $\begingroup$ @Vizuna If you prove that every element of $\mathbb Z$ is in $\mathbb F$, then if $x,y\in\mathbb Z$, $x+y$ is also in $\mathbb Z$, therefore in $\mathbb F$. $\endgroup$
    – 5xum
    Dec 1, 2014 at 14:40
  • $\begingroup$ That's convincing enough for me, thanks. It uses the axiomatic definition that $\mathbb Z$ is closed under addition, right? If it's an axiom then it doesn't require proving. $\endgroup$
    – Vizuna
    Dec 1, 2014 at 14:47

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