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

Question

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...:)

Answer 1

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

Answer 2

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 $\mathbb{Q}\subseteq\mathbb{F}$.

Answer 3

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 $\mathbb{Q}\subseteq F$.

Answer 4

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?).