Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field

Question

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.

Answer 1

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 $R$.

Answer 2

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 $R$.

Answer 3

$\{1,2,4,8,\ldots\}$ 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?

Answer 4

This proof is "similar" to some of the above proofs, but not identical:

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: Let $0\neq x\in R$ and let $\phi: F[t]\rightarrow k\{1,x,x^2,..,\}$ be the map defined by $\phi(t):=x$. Since $R$ is a ring it follows $\phi$ is a ring homomorphism, and since $dim_F(R) < \infty$ it follows there is an integer $n$ with $x^{n+1}\in F\{1,x,x^2,..,x^n\}$. Hence the map $\phi$ is a surjective map of $F$-algebras with $ker(\phi)=(p(t))$, where $p(t)$ is a non-zero polynomial. Since $R$ is an integral domain and $Im(\phi) \subseteq R$ is a sub ring it follows $F[t]/(p(t))$ is an integral domain, hence $p(t)$ is an irreducible polynomial. Since $p(t) \neq 0$ it follows $(p(t)) \subseteq F[t]$ is a maximal ideal, hence $Im(\phi)\subseteq R$ is a field containing $x$. It follows there is an element $0\neq y\in Im(\phi)$ with $xy=yx=1$ and it follows $R$ is a field. QED

Question: "However the following proof looks wrong to me - am I misunderstanding something, or is it actually an incorrect proof? ........ It then goes on to show, given the above, that we can derive an inverse for r. ..... 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?"

Answer: The claim of the Lemma is correct but the proof you give above is incomplete. Above I give a(nother) compete proof of the Lemma.

PS: I do not have a copy of the book you mention, hence I cannot comment on the proof given in the book.