# Any subfield of $\mathbb{C}$ must contain every rational number

I tried to prove that any subfield of $\mathbb{C}$ must contain every rational number by contradiction.

Proof:

Let $\mathbb{F}$ be any subfield of $\mathbb{C}$. Thus, $\mathbb{F}$ is itself a field under the usual operations of addition and multiplication of complex numbers. For $\mathbb{F}$ to be a field, $0$ and $1$ must belong to $\mathbb{F}$. We have to prove that any subfield of $\mathbb{C}$ must contain every rational number. Let us assume on the contrary that there exists at least one rational number $q \neq 0$ such that $q \notin \mathbb{F}$.

Let $-q$ and $\frac{1}{q}$ are in $\mathbb{F}$. Then since $\mathbb{F}$ is a field, it must contain additive inverse of $-q$ and multiplicative inverse of $\frac{1}{q}$; i.e. $q$ must be in $\mathbb{F}$. But, according to our assumption, $q \notin \mathbb{F}$; which implies that $\mathbb{F}$ is not a field. Thus we arrive at a contradiction.

Now, let $-q$ and $1/q$ are also not in $\mathbb{F}$ (edit - here I mean to say all elements of $\mathbb{Q}$, except $q$, $-q$ and $\frac{1}{q}$, are in $\mathbb{F}$). Then $\mathbb{F}$ does not satisfy closure under addition and multiplication, which again leads to the contradiction that $\mathbb{F}$ is not a field.

Hence any subfield of $\mathbb{C}$ must contain every rational number. Q.E.D.

I just need feedback on whether it is correct and how I can improve it (especially the last portion). Also, can we modify the statement of the result into - The set of rational numbers is the smallest subfield of $\mathbb{C}$ ?

Thanks.

Edit:

Thank you all of you for your valuable feedback. I think I did not write my arguments in the second part of the proof quite clearly. Here is how my chain of thoughts were - Since I assumed that at least one element $q$ of rational number is not in $\mathbb{F}$, so there may be more than one element of $\mathbb{Q}$ that are not in $\mathbb{F}$. This is why I assumed in the second part the non presence of $−q$ and $q^{−1}$. Now, consider for example all rational numbers, except $2$, $−2$ and $0.5$, are in $\mathbb{F}$ (as $q$, $-q$ and $q^{-1}$ in second part). Then $1−3=−2$ , such that $1,3 \in \mathbb{F}$ implies that closure does not hold. May be I should have added something like - Since $1, -(q+1)\in\mathbb{F}$, then by closure $1+{-(q+1)} = -q \in \mathbb{F}$. But $-q \notin \mathbb{F}$ which implies closures does not hold.

I hope my arguments are convincing enough. I look forward to more comments. Thanks..

Edit 2 - I realized where I made mistake in the above proof. Thank you all of you for your comments...:)

• This is not correct. Why does the non presence of $-q,q^{-1}$ imply non closure? You started by assuming $q\not\in\Bbb F$, so why should they be? – GPerez Apr 8 '15 at 10:28
• The way I'd do it: Every rational number is either $0$, or it can either be written in the form $\dfrac{1+1+\dotsb+1}{1+1+\dotsb+1}$ or in the form $-\dfrac{1+1+\dotsb+1}{1+1+\dotsb+1}$ (for some amount of $1$s in the numerator and denominator). Since $0$, $1$, sums, division, and their negatives exist in fields, all of these must be in the field. – Akiva Weinberger Apr 8 '15 at 10:50
• @Ritu You have no reason to assume though, that $1,3\in\Bbb F$ in your last example. By the same reasoning I could prove that $\Bbb C \subset\Bbb R$: suppose $1+i\not\in\Bbb R$. However, $i+1 = (-i) + (1+2i)$ and $\Bbb R$ would not be closed under addition. This is exactly the same as what you are saying. – GPerez Apr 8 '15 at 12:04
• Your argumentation is incomplete and very messy: you begin by assuming $\;0\neq q\notin\Bbb F\;,\;\;q\in\Bbb Q\;$, and then you first: suppose $\;-q\,,\,\,\frac1q\in\Bbb F\;$ , and then after that you suppose second : $\;-q\,,\,\,\frac1q\notin\Bbb F\;$ ....and what about $\;-q\in\Bbb F\;,\;\;\frac1q\notin\Bbb F\;$ , and $\;-q\notin\Bbb F\;,\;\;\frac1q\in\Bbb F\;$ ? You didn't check all the cases! Anyway, that way of reasoning is pretty cumbersome. – Timbuc Apr 8 '15 at 12:37
• Your new edit makes things even worse: what if exactly 124 elements of the rationals are not in $\;\Bbb F\;$ but all the rest are?! And what about 15 rational elements exactly aren't ? And... – Timbuc Apr 8 '15 at 12:39

Your last argument is not correct: why do you say "Then $\mathbb{F}$ does not satisfy closure under addition and multiplication"? This argument is not explained.

What you should do is the following:

$1 \in\mathbb{F}$ so all elements of the form $1+ \cdots +1$ belong to $\mathbb{F}$. This means that $\mathbb{Z}_{\geq 0} \subset \mathbb{F}$.

Clearly this implies that $\mathbb{Z} \subset\mathbb{F}$ (take $-(1+ \dots)$)

Clearly this implies that all inverses of natural numbers $\{ \frac{1}{n} \}_{n\geq 1}$ belong to $\mathbb{F}$, since $\mathbb{F}$ is a field.

And now, any rational number is of the form $m\cdot \frac{1}{n}$ with $m, n$ integers, $n \geq 1$, hence $\mathbb{Q} \subset\mathbb{F}$.

Finally, you can conclude that $\mathbb{Q}$ is the smallest subfield of $\mathbb{C}$, as you said. This can be generalized to any field: you can have a look at http://en.wikipedia.org/wiki/Characteristic_%28algebra%29#Case_of_fields

• Thank you. Please go through my edit, I have explained my chain of thoughts for that conclusion in second part. I hope they are good enough. – Ritu Apr 8 '15 at 12:24
• @Ritu When you try an argument like "If $x\notin \mathbb{F}$, then $y\notin\mathbb{F}$" you only now have to assume that $y\notin\mathbb{F}$ to avoid contradiction. Then you might argue that $y$ not being in the field makes some other element not in the field. Then to avoid contradiction you are forced to say the new element is not in the field. It creates a chain of assumptions of things not being in the field. But, it won't help to produce a contradiction. – Joe Johnson 126 Apr 8 '15 at 23:39
• Yes, I get your point. Thanks... – Ritu Apr 15 '15 at 5:54
• @Ritu In short, if yoou do not use induction somewhere in your proof, I'm sure its not correct. – Hagen von Eitzen Aug 29 '15 at 20:11

I agree with the first part, if $-q,\dfrac{1}{q}\in\mathbb{F}$, then there is a contradiction. I don't see why if $-q\notin\mathbb{F}$ or $\dfrac{1}{q}\notin\mathbb{F}$, closure implies a contradiction.

You may wish to show that $\mathbb{Z}\subseteq\mathbb{F}$, then use closure to show that $\mathbb{Q}\subseteq\mathbb{F}$.

• Thank you. Please go through my edit, I have explained my chain of thoughts for that conclusion in second part. I hope they are good enough. – Ritu Apr 8 '15 at 12:27

Unfortunately your proof is not correct as the comments have pointed out. Note however that, as you said, 1 must be contained in your field $F$. Therefore, since your field must be additively closed we conclude that $\mathbb{Z}\subseteq F$. Next, the smallest field containing $\mathbb{Z}$ is its field of fraction $\mathbb{Q}$. Note that a similar definition for this field is ''the field determined by the intersection of all fields containing $\mathbb{Z}$''. Therefore we conclude that $\mathbb{Q}\subseteq F$.

As other comments has pointed out, $\mathbb{Q}$ is the smallest subfield containing $\mathbb{Z}$. This is a special case of the more general fact that the fraction field $Frac(R)$ is the smallest field containing the ring $R$. To get $Frac(R)$, we take the following set (very intuitive): $\left \{\frac{m}{n} \mid m,n \in R \cap n \neq 0_R \right \}$. Just like in $\mathbb{Q}$ we have the concept of simplest terms, in $Frac(R)$ we have the representatives for equivalence classes, under the equivalence relation $\frac{m}{n}\sim \frac{p}{q}$ iff $mq=pn$, under multiplication in $R$. It is a worthwhile exercise to prove that this is in fact an equivalence relation, that addition and multiplication (and subtraction and division) are well-defined (i.e. don't depend on representative), and that $Frac(R)$ is in fact the smallest field containing $R$ (hint what would happen to $\alpha^{-1}$ if a nonzero, "non-$1_R$" element $\alpha$ were removed?).