I'm reading Galois Theory by Steven H. Weintraub (second edition), and finding that I'm at least somewhat short on the prerequisites. However the following proof looks wrong to me - am I misunderstanding something, or is it actually an incorrect proof?

Lemma 2.2.3. Let $F$ be a field and $R$ an integral domain that is a finite-dimensional $F$-vector space. Then $R$ is a field.

Proof. We need to show that any nonzero $r \in R$ has an inverse. Consider $\{1, r, r^2, \cdots\}$. This is an infinite set of elements of $R$, and by hypothesis $R$ is finite dimensional as an $F$-vector space, so this set is linearly dependent. Hence $\sum_{i=0}^n{c_i r^i} = 0$ for some $n$ and some $c_i \in F$ not all zero.

It then goes on to show, given the above, that we can derive an inverse for $r$.

However, if I consider examples like $r = 2 \in Q[\sqrt{2}]$, $r = \sqrt{2} \in Q[\sqrt{2}]$ or $r = 2 \in Q[X]/{<X^2>}$, the set $\{1, r, r^2, ...\}$ doesn't look linearly dependent to me.

I do believe the lemma is true (and might even be able to prove it), but this does not look like a correct proof to me. Am I missing something?

[Edit] Well yes, I am. Somehow I had managed to discount the possibility of any $c_i$ being negative, despite repeatedly looking at each fragment of the quoted text in an attempt to find what I might be misunderstanding.

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    $\begingroup$ You should check if the set is linearly dependent or not... For example, if $r=\sqrt2$ in $\mathbb Q[\sqrt2]$, is the set $\{1,r,r^2,\dots\}$ linearly independent or not? Notice I am not asking if you believe it is, or if it looks so, but if it is :) $\endgroup$ – Mariano Suárez-Álvarez Sep 12 '11 at 18:23
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    $\begingroup$ Note that $\mathbb Q[X]/\langle X^2\rangle$ is not an integral domain (since $XX=0$ there), so the lemma is not supposed to hold there. $\endgroup$ – Henning Makholm Sep 12 '11 at 18:28
  • $\begingroup$ You probably meant $Q[X]/\langle X^2-2\rangle$... still true that $\{1,2,4,8,\ldots\}$ ( or perhaps $\{1+(X^2-2), 2+(X^2-2), 4+(X^2-2),\ldots\}$) is $\mathbb{Q}$-linearly dependent. $\endgroup$ – Arturo Magidin Sep 12 '11 at 18:34
  • $\begingroup$ @Henning: ah sorry, clearly I need to look again at precisely what $\mathbb{Q}[X]/<X^2>$ means. But it turns out my question was deeply misguided in any case (or at least, revolved solely around the "what have I misunderstood" part). $\endgroup$ – Hugo van der Sanden Sep 12 '11 at 18:47
  • $\begingroup$ For some extended discussion see my answer here. See also this duplicate question. $\endgroup$ – Bill Dubuque May 19 '12 at 0:40

$\{1,2,4,8,\ldots\}$ is certainly $\mathbb{Q}$-linearly dependent in $\mathbb{Q}[\sqrt{2}]$; in fact, it is linearly dependent in $\mathbb{Q}$ already! $0 = 2(1) -1(2)$, with the elements in parentheses being the vectors. So this is a nontrivial linear combination of the vectors in the set which is equal to $0$.

For $\sqrt{2}$, the set is $\{1,\sqrt{2},2,2\sqrt{2},4,\ldots\}$. Again, this is $\mathbb{Q}$-linearly dependent, since $0 = 2(1) + 0(\sqrt{2}) -1(2)$. Again, this is a nontrivial linear combination of the vectors in the set which is equal to $0$.

What is it that makes it look "not dependent" to you?

  • $\begingroup$ @Henning: Sigh; thanks. $\endgroup$ – Arturo Magidin Sep 12 '11 at 18:32
  • $\begingroup$ Stupidity, mainly, perhaps with a bit of "out of practice" thrown in. Somehow I had it fixed in my head that the sum could not be zero without the powers somehow "wrapping around" past infinity, such as they might in modular arithmetic. Thanks. :) $\endgroup$ – Hugo van der Sanden Sep 12 '11 at 18:39

Weintraub's 15-line proof is correct but clumsy. Here is a 2-line proof:

Given $0\neq r\in R$ the $F$-linear map $R\to R:x\mapsto rx$ is injective ($R$ is an integral domain!), hence surjective ($R$ is finite-dimensional!). So $1$ is the image of some $s\in R$, i.e. $sr=1$ and so $s=r^{-1}$ belongs to $R$.

  • $\begingroup$ Nice simple proof indeed but does it answer the question as asked? $\endgroup$ – lhf Sep 12 '11 at 21:46
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    $\begingroup$ @lhf: The OP asks twice (line 3 and last line before the edit) if Weintraub's proof is correct and I have answered that it is. As for his other questions, there is no need to repeat what has been well explained by Arturo and the commentators. On the other hand I thought it could be psychologically useful for the OP to know that his difficulties are due in great part to the suboptimal quality of the proposed proof: students tend to think that it is always their fault if they fail to understand something. That is false. $\endgroup$ – Georges Elencwajg Sep 12 '11 at 22:04
  • $\begingroup$ @GeorgesElencwajg :Is $r$ taken from $R$ ? $\endgroup$ – user217921 Mar 13 '15 at 15:29
  • $\begingroup$ Dear @Saun: yes. I have edited my answer to make that explicit. $\endgroup$ – Georges Elencwajg Mar 13 '15 at 18:08
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    $\begingroup$ @GeorgesElencwajg : But then can you please explain why $x \to rx$ is $F$-linear ? I am having trouble with the homogeneous part ; let $\alpha \in F$ , then $f(\alpha .x)=r(\alpha . x)=(\alpha.x)r$ (as $R$ is assumed to be commutative) and $\alpha.f(x)=\alpha .(rx)=\alpha.(xr)$ , I am unable to see then how $f(\alpha .x)$ and $\alpha.f(x)$ are equal . Please help $\endgroup$ – user217921 Mar 14 '15 at 13:31

Another easy solution. I will explicitly construct the inverse.

Since $R$ is finite dimensional over $F$, we have $\{1,r,r^{2},...,r^{n}\}$ is a linear dependent set for some finite $n$ over $F$. In particular, if $r \neq 0$ and $r \in R$, then $a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots+a_{0} =0$ has a nontrivial solution where each $a_{i} \in F$. If $a_{0}=0$ then $$a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots+a_{1}r=0 \implies r(a_{n}r^{n-1}+a_{n-1}r^{n-2}+\cdots+a_{1})=0 \implies a_{n}r^{n-1}+a_{n-1}r^{n-2}+\cdots+a_{1}=0$$ since $R$ is an integral domain. If $a_{1}=0$ repeat the previous step. Clearly this process will terminate once we get to some nonzero $a_{i}$. Therefore we may assume WLOG that $a_{0} \neq 0$. But then $$a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots+a_{0} =0 \implies a_{n}r^{n}+a_{n-1}r^{n-1}+\cdots+a_{1}r=-a_{0} \implies b_{n}r^{n}+b_{n-1}r^{n-1}+\cdots+b_{1}r=r(b_{n}r^{n-1}+b_{n-1}r^{n-2}+\cdots+b_{1}) =1$$ where $b_{i}=-a^{-1}_{0}a_{i}$, showing that $r$ has an inverse in $R$.


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