Why is $\mathbb{Q}$ the minimal field that contains $0$ and $1$ ?

Question

Why is $\mathbb{Q}$ the minimal field that contains $0$ and $1$ ?

Why isn't it $\mathbb{Z}_2$ ?

Answer 1

It depends what you mean by "$0$" and "$1$" (and also by "minimal").

If "$0$" and "$1$" refer to the real numbers zero and one, then in essence you're asking, "Why is $\mathbb{Q}$ the minimal subfield of $\mathbb{R}$ containing $0$ and $1$?" To answer this, note that $\{0, 1\}$ is not a subfield of $\mathbb{R}$; it's not closed under addition ($1+1\not\in\{0, 1\}$) and it doesn't have additive inverses ($-1\not\in\{0, 1\}$). See also this question.

On the other hand, if "$0$" and "$1$" don't refer to the real numbers zero and one, but rather just the additive and multiplicative identity in any field - then yes, $\mathbb{Z}/2\mathbb{Z}$ is the smallest field in terms of cardinality. However, note that $\mathbb{Z}/2\mathbb{Z}$ does not embed into $\mathbb{Q}$! The map sending $0$ to $0$ and $1$ to $1$ is not a homomorphism. So in that sense, $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Q}$ are incomparable. Indeed, $\mathbb{Q}$ has no proper subfields, so I would say it is a minimal field.

Answer 2

Sometimes the "minimal field" is $\mathbb{Q}$, other times it is $\mathbb{Z}/p\mathbb{Z}$ for a prime number $p$. It depends on what the characteristic of your ambient field is.

By definition, any field $F$ must contain elements called $0_F$ and $1_F$ (or just $0$ and $1$), which are not equal and which satisfy $0+x=x$ and $1 \cdot x = x$ for all $x \in F$.

Let $F$ be a field. The characteristic of $F$ is the smallest number of times you have to add $1$ to itself to get $0$. If this never happens, then we say that $F$ has characteristic zero. Otherwise, you can show that the characteristic of $F$ has to be a prime number.

If the characteristic of $F$ is a prime number, say $p$, then you can show that the subset $\{0, 1, 2, ... , p-1\} \subseteq F$ is actually a field. Let's denote that field by $F_p$. If $K$ is any subfield of $F$ whatsoever, it is easy to see that $F_p \subseteq K$, because $K$ must contain $0$ and $1$, and hence $1+1, 1+1+1$ etc. Furthermore, there is an isomorphism $\mathbb{Z}/p\mathbb{Z} \rightarrow F_p$.

On the other hand, if the characteristic of $F$ is zero, then the set $R_0 =\{0,1,-1,2,-2, ...\}$ is a subring of $F$ which is obviously isomorphic to $\mathbb{Z}$. Since $F$ is a field, the inverses of the nonzero elements of $F$ exist, are unique, and are elements of $F$. Therefore,

$$F_0 := \{ \frac{x}{y} : x, y \in R_0, y \neq 0 \}$$ is a subfield of $F$. If $K$ is any subfield of $F$ whatsoever, it is easy to see that $F_0 \subseteq K$, because $K$ must contain $0$ and $1$, hence $-1,2,-2$ etc. as well as the inverses of these elements. Furthermore, there is an isomorphism $\mathbb{Q} \rightarrow F_0$. Let me know if you have any questions about this.