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

I need to prove this result, but the only starting point I think of is to argue by contradiction, by assuming $x$ is non-invertible and extending it to a basis over $K$ but have no idea how to go further than that. Any ideas?

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marked as duplicate by Dylan Moreland, Pedro Tamaroff, The Chaz 2.0, Martin Sleziak, Jyrki Lahtonen May 19 '12 at 5:57

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It's probably a good idea to first prove that a finite integral domain is a field. This case is then a slight extension of the argument. –  Chris Eagle May 19 '12 at 0:30
For an elaboration of Chris' hint see here, and see my various answers here. –  Bill Dubuque May 19 '12 at 1:02

3 Answers 3

up vote 11 down vote accepted

We will prove this using the Rank - Nullity Theorem applied to $A$ viewing it as a finite dimensional vector space over $K$. Pick a non-zero element $y \in A$ and consider the $K$ - linear transformation $$\begin{eqnarray*} T&\colon&A \longrightarrow A \\ && x\mapsto yx.\end{eqnarray*} $$

Now the kernel of this linear transformation is trivial because $A$ is an integral domain, and hence by rank-nullity we have that

$$\begin{eqnarray*} \dim_K A &=& \dim \ker T + \dim \operatorname{im} T \\ &=& 0 + \dim \operatorname{im} T\\ &=& \dim \operatorname{im} T \end{eqnarray*}$$

from which it follows that $T$ is surjective. In particular there exists $z \in A$ such that $zy = 1$, so that since $y$ was arbitrary we have shown that every element in $A$ is invertible, i.e. $A$ is a field.

Edit: I should say it is important to include the hypothesis that $A$ is an integral domain otherwise the result is not necessarily true. Consider the $\Bbb{C}$ - algebra

$$\Bbb{C} \oplus \Bbb{C} \oplus \Bbb{C} $$

as a finite dimensional vector space over $\Bbb{C}$. This is not an integral domain because $(1,0,0) \cdot (0,1,0) = (0,0,0)$. Now we also have

$$\dim_{\Bbb{C}} \big( \Bbb{C} \oplus \Bbb{C} \oplus \Bbb{C} \big) = 3 < \infty$$

but $\Bbb{C} \oplus \Bbb{C} \oplus \Bbb{C}$ cannot be a field because elements like $(1,0,0)$ don't have an inverse.

If we also drop the hypothesis that $\dim_K A < \infty$ the result is not true as well. For example consider the polynomial ring $\Bbb{C}[T]$ in the indeterminate $T$. This is an integral domain because $\Bbb{C}$ is. Then if we view this as a vector space over $\Bbb{C}$ it is infinite dimensional, but it is already a well known fact that $\Bbb{C}[T]$ is not a field.

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Thank you. The proof is simple and powerful--am I right that your proof does not assume $K$ is contaned in $A$ and applies to any general field? –  Montez May 18 '12 at 23:14
@user26655 It assumes that $K \subset A$ and recall that we must have $\dim_K A < \infty$ in order to apply rank nullity. –  user38268 May 18 '12 at 23:15

How about this: say $A$ has dimension $n$ over $K$. Let $x \in A, x\neq 0$. Consider elements $1,x,x^2,x^3...$ They can't be linearly independent (why?). So we may choose $m$ so that $1,x,..,x^m$ is linearly dependent over $K$ and $m$ is as small as possible. This means that we may find $c_0,c_1,..,c_m \in K$, not all equal to $0$, such that $c_0 + c_1 x +..+c_mx^m = 0$. Note that $c_0 \neq 0$ (why?) Hence what can you say about $x^{-1}$?

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Thank you for your reply. $1, x, x^2, \cdots$ cannot be independent because of the finite dimension. $c_0=0$, $x\neq 0$ would contradict minimality of $m$. Then $x(c_1+\cdots+c_mx^{m-1}$ is invertible so $x$ is invertible. –  Montez May 18 '12 at 23:12
Yes, you are correct! –  algebra_fan May 18 '12 at 23:21
Nice Answer.(+1) And exactly on the lines of how I was thinking :-) –  Wanderer Aug 22 '14 at 19:40

Hint $\ $ Being finite dimensional, the extension is algebraic. But a domain algebraic over a field is a field, since one obtains an inverse of an element $\ne 0$ from a minimal polynomial. See my post here for much further discussion of this and related methods of lifting the existence of inverses (e.g. the well-known method of rationalizing denominators).

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