# Ring of total fractions of finite ring extension only has denominators of the ground ring

Let $B \supseteq A$ be a finite ring extension where $A$ is a domain with field of fractions $k = \operatorname{Frac}(A)$ and $B$ need not be a domain. Further, let $B$ be torsion-free over $A$.

Now I want to show that the ring of total fractions $\operatorname{Frac}(B)$ of $B$ is equal to $B \otimes_A k$.

By assumption, the RHS is contained in the LHS (the elements of $A$ are regular elements of $B$). But why can we write every fraction with denominator contained in $B$ as a fraction where the denominator is an element of $A$?

The case where $B$ is also integral is not that hard, but what if we cannot use that $\operatorname{Frac}(B)$ is a field?

If this is not true in the generality mentioned above, then is it true if we assume $A$ to be a PID (from which we then deduce that $B$ is free over $A$)?

$B\otimes_Ak$ is equal to its own total ring of fractions, because it is a finite dimensional $k$-vector space, so that, if $x$ is a non-zero divisor in $B\otimes_Ak$, the linear map of multiplication by $x$ is injective by definition, hence it is also bijective, and $x$ is invertible.
• This looks great and easy! Thank you very much! But the rank-nullity theorem also holds for non-finite dimensional vector spaces. So do we really need that $B$ is finite over $A$ here? – windsheaf Aug 29 '17 at 9:59
• Injective iff surjective iff bijective is valid only in finite dimension. You hve a counter-example with the ring of polynomials: multiplication by $X$ is injective in $K[X]$, but certainly not bijective. – Bernard Aug 29 '17 at 10:02