# Field with characteristic zero is vector space over $\mathbb{Q}$

Here’s my problem:

Prove that a field R of characteristic $$0$$ is a vector space over $$\mathbb{Q}$$.

I am unsure how to proceed here, as checking the vector space axioms seems wrong.

Here are my thoughts:

Since $$\operatorname{char}(R) =0$$, $$\mathbb{Z}$$ is isomorphic to some subdomain of $$R$$. Now $$\mathbb{Z}$$ is itself not a field but has field of fractions $$\mathbb{Q}$$...

Can anyone help me out here?

Edit:

$$\mathbb{Z}$$ is isomorphic to some subdomain of $$R$$. Extend this to a field of fractions inside $$R$$. This extension will be isomorphic to $$\mathbb{Q}$$ as this is the field of fractions of $$\mathbb{Z}$$. Then verifying the vector space axioms (is this necessary or is there shortcut?) gives the desired result.

Is this correct?

• If $D$ is a domain with field of fractions $F$, part of the definition of "field of fractions" is "if $D\subseteq K$ for some field, then $F\subseteq K$ too (all this is up to isomorphism, of course.) You'd be done if you applied that. – rschwieb Mar 30 '20 at 17:49
• But. Isn't $\Bbb Z$ itself an integral domain? It doesn't carry any vector space structure over $\Bbb Q$. Maybe you mean the field of fractions of $R$? – Berci Mar 30 '20 at 18:00
• @rschwieb I don’t quite follow this. What are your $F$ and $K$ in this case? – user489116 Mar 30 '20 at 18:09
• @user Sorry, what I wrote is correct but doesn't apply to your question, since I apparently misread your question. I now have the same incredulity about the way it is stated that Berci has. $\mathbb Z$ is not a $\mathbb Q$ vector space. So it looks like something is wrong with your statement. – rschwieb Mar 30 '20 at 18:26
• Whoops! Just realized that as stated my question is indeed incorrect/problematic. This should be a field rather than integral domain! In which case I believe I see how to use your hint @rschwieb – user489116 Mar 30 '20 at 18:46

Hint: Because $$R$$ is a field of characteristic $$0$$, every nonzero element of $$\Bbb{Z}\subset R$$ is invertible.

Once you have shown that $$\Bbb{Q}\subset R$$ all the vector space axioms are easily verified because $$R$$ is a field that contains $$\Bbb{Q}$$ as a subfield. Note that half of the axioms are already satisfied a priori because $$R$$ is a field. It might even be worth proving that in general:

If $$R$$ is a field and $$S\subset R$$ is a subfield then $$R$$ is a vector space over $$S$$.

• That last sentence is VERY helpful for me, as is your answer in general. Thanks! – user489116 Mar 30 '20 at 18:59
• I'm very glad it's helpful :) – Servaes Mar 30 '20 at 19:01

Consider $$Z \subseteq R$$ defined by $$Z= \{\dots,-1-1,-1,0,1, 1+1, \dots\}$$

and put $$Q := \{ab^{-1}: a,b \in Z, b \neq 0\}$$

Then $$\mathbb{Q}\cong Q$$. Let $$f: \mathbb{Q} \to Q$$ be this isomorphism. We can view $$R$$ as a $$Q$$-vector space by letting $$Q$$ act on $$R$$ by ring multiplication.

Then $$q. r = \phi(q)r; \quad q \in \mathbb{Q}, r \in R$$

defines a $$\mathbb{Q}$$-scalar multiplication on $$R$$ and we get that $$R$$ is a $$\mathbb{Q}$$-vector space.

So basically the idea is to find an isomorphic copy of $$\mathbb{Q}$$ inside $$R$$ and use this to get the vector space structure you like.

• Very nice! I think I am getting the picture now. Thanks! – user489116 Mar 30 '20 at 19:03