# If $A$ an integral domain contains a field $K$ and $A$ over $K$ is a finite-dimensional vector space, then $A$ is a field. [duplicate]

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?

• 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. May 19, 2012 at 0:30
• For an elaboration of Chris' hint see here, and see my various answers here. May 19, 2012 at 1:02

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.

• 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? May 18, 2012 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, 2012 at 23:15
• Shouldn't it be "pick a non-zero element from K" Nov 29, 2020 at 6:03

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 $c_0 + c_1 x +\cdots+c_mx^m = 0$. Note that $c_0 \neq 0$ (why?). Then what can you say about $x^{-1}$?

• 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. May 18, 2012 at 23:12
• Yes, you are correct! May 18, 2012 at 23:21
• Nice Answer.(+1) And exactly on the lines of how I was thinking :-) Aug 22, 2014 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).

• In what context do we consider extensions other than fields? In this case we would be assuming what we want to prove, namely that $A$ is a field extension of $K$. Nov 17, 2020 at 20:38
• @Junglemath By common definition $\,R\subseteq S\,$ is a ring extension if $\,R\,$ is a subring of $S\ \$ Nov 18, 2020 at 1:05