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In the link Is $\mathbb{R}$ a vector space over $\mathbb{C}$? it is proven that $\mathbb{R}$ can be given the structure of a $\mathbb{C}$-vector space. Since $\mathbb{C}$ is the simple extension $\mathbb{R}(i)$ of the field $\mathbb{R}$, I've been wondering if it's possible to do something similar with field extensions of $\mathbb{Q}$.

For example, it's not hard to prove that if $\mathbb{Q}$ is an $F$-vector space, then $F$ is a countable field (hence, $\mathbb{Q}$ can't be an $\mathbb{R}$-vector space). Therefore, since it is necessary for $F$ to be countable, two natural candidates are $\mathbb{Q}(\sqrt{2})$ or $\mathbb{Q}(\pi)$. Furthermore, since $\sqrt{2}$ is algebraic over $\mathbb{Q}$ and $\pi$ is transcendental over $\mathbb{Q}$, I believe it is worth considering both extensions.

Is it possible to show that $\mathbb{Q}$ is a vector space over $\mathbb{Q}(\sqrt{2})$ or over $\mathbb{Q}(\pi)$?

Peluso
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2 Answers2

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If $\mathbb{Q}$ is a vector space over some field $K$, then $K = \mathbb{Q}$. To show this, denote by $\cdot_K$ the action of $K$ on $\mathbb{Q}$.

First, we have, for any integer $k$: $$k = \overline{1}\cdot_K k = (\overline{1} \cdot_K 1) + \cdots + (\overline{1} \cdot_K 1) = (\overline{1} + \cdots + \overline{1}) \cdot_K 1 = \overline{k} \cdot_K 1$$ where $\overline{k}$ is $k$ viewed in $K$. This implies that integers $\overline{k}$ in $K$ act by multiplication by $k$ on $\mathbb{Q}$. In particular, we get for free that $K$ is of characteristic $0$ (otherwise there would be some integer $p >0$ acting as zero on $\mathbb{Q}$, which contradicts what we just showed).

Because $K$ is of characteristic zero, it contains a copy of $\mathbb{Q}$. This copy of $\mathbb{Q}$ acts by the usual multiplication on the $K$-vector space $\mathbb{Q}$: we have seen that this was the case for integers, I'll let you prove that this still holds for any rational number.

Finally, let $x \in K$, we will show that $x \in \mathbb{Q}$. Let $r = x \cdot_K 1$, this is a rational number. So we can consider: $$1 = \frac{1}{r}\cdot_K r = \frac{x}{r} \cdot_K 1.$$ Therefore $\frac{x}{r}$ acts as the identity, hence $\frac{x}{r} = 1$ i.e. $x = r \in \mathbb{Q}$, qed. Here we used the fact that $\overline{1}$ is the only element of $K$ acting as the identity: I'll let you prove this from the axioms of vector spaces.

François Gatine
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Let $(M,+)$ be an abelian group. Let $R$ be a unitary ring. Every $R$-module structure on $(M,+)$ is given by a ring homomorphism $R \to \text{End}(M,+) = \text{End}_\mathbb{Z}(M)$. If $R$ is a field then $R$-module is the same as $R$-vector space.

Let $\alpha \in L$ where $L \supseteq \mathbb{Q}$ is some field extension. Now here concretely $R= \mathbb{Q}(\alpha)$ and $(M,+) = (\mathbb{Q},+)$. It is $\text{End}(\mathbb{Q},+) = \text{End}_\mathbb{Z}(\mathbb{Q}) = \text{End}_\mathbb{Q}(\mathbb{Q}) \cong \mathbb{Q}$ as every $\mathbb{Z}$-linear map is automatically $\mathbb{Q}$-linear.

So now we are searching for a ring (field) homomorphism $\Phi: \mathbb{Q}(\alpha) \to \text{End}(\mathbb{Q},+)\cong \mathbb{Q}$. As field homomorphism, $\Phi$ is injective and $\mathbb{Z}$-linear, i.e. automatically $\mathbb{Q}$-linear. In particular $\dim_{\mathbb{Q}} \mathbb{Q}(\alpha) \leq \dim_\mathbb{Q} \mathbb{Q} = 1$. This means $\alpha \in \mathbb{Q}$ and $\mathbb{Q}(\alpha) = \mathbb{Q}$.

Thus $\mathbb{Q}$ is neither a $\mathbb{Q}(\sqrt{2})$-vector space nor a $\mathbb{Q}(\pi)$-vector space.

psl2Z
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