Field has Characteristic of Zero iff exists Monomorphism from Rationals

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Theorem

Let $F$ be a field.

Then:

there exists a field monomorphism $\phi: \Q \to F$ from the field of rational numbers $\Q$ and $F$.

if and only if:

$\Char F = 0$

where $\Char F$ denotes the characteristic of $F$.


Proof

Necessary Condition

Let $\Char F = 0$.

Then from Field of Characteristic Zero has Unique Prime Subfield, $F$ has a unique prime subfield $K$ such that:

$K \cong \Q$

where $\cong$ denotes isomorphism.


Thus there exists an isomorphism from $\Q$ to a subfield of $F$.

From Injection to Image is Bijection, it follows that $\phi$ is an injection into $F$.

Hence the result by definition of field monomorphism.

$\blacksquare$


Sufficient Condition

Let there exists a field monomorphism $\phi: \Q \to F$ from the field of rational numbers $\Q$ and $F$.

Let $K := \Img \phi$.

Then $K$ is isomorphic to $\Q$.

As $K$ is a subfield of $F$, both $0$ and $1$ are in $K$.

From Characteristic of Field is Zero or Prime, either $\Char F = 0$ or $\Char F = p$ for some prime number $p$.

Aiming for a contradiction, suppose $\Char F = p$.

Then:

$p \cdot 1 = 0$

But as $1 \in K$ it follows that $p \cdot 1 \in K$ as $K$ is closed under addition.

But as $K \cong \Q$ this cannot happen

Hence there is no prime number $p$ such that $\Char F = p$.

The result follows by Proof by Contradiction.

$\blacksquare$


Sources

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