13
$\begingroup$

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 $\mathbb Q$.

$\endgroup$
  • 3
    $\begingroup$ $\;\underbrace{1+1+...+1}_{n\;\text{times}}=n\;$ 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 '14 at 14:26
  • 1
    $\begingroup$ This is not a question... $\endgroup$ – 5xum Dec 1 '14 at 14:26
  • $\begingroup$ Sorry, I should have added the "proof-verification" tag to make my question more explicit. $\endgroup$ – Vizuna Dec 1 '14 at 14:28
  • 1
    $\begingroup$ 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 '14 at 15:30
  • $\begingroup$ Can we prove it by contradiction also? $\endgroup$ – Ritu Feb 20 '15 at 13:41
5
$\begingroup$

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).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 '14 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 '14 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 '14 at 14:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.