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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:

Based on the comments I received:

$\mathbb{Z}$ 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?

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  • $\begingroup$ 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. $\endgroup$ – rschwieb Mar 30 '20 at 17:49
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    $\begingroup$ 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 $R$? $\endgroup$ – Berci Mar 30 '20 at 18:00
  • $\begingroup$ @rschwieb I don’t quite follow this. What are your $F$ and $K$ in this case? $\endgroup$ – user489116 Mar 30 '20 at 18:09
  • $\begingroup$ @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. $\endgroup$ – rschwieb Mar 30 '20 at 18:26
  • $\begingroup$ 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 $\endgroup$ – user489116 Mar 30 '20 at 18:46
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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 $S$.

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

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

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