# $n$ algebraically independent elements in a field of fractions implies $n$ algebraically independent elements in the $k$-algebra

Let $A$ be a $k$-alegbra and $B$ be its field of fractions. Suppose $\{\frac{f_{1}}{g_{1}},...,\frac{f_{n}}{g_{n}}\}$ is an algebraically independent set in B. Question: Are there $n$ algebraically independent elements in A?

I can show that either $f_{i}$ or $g_{i}$ is algebraically independent for each of the fractions, using facts about field extensions. So suppose each $f_{i}$ is algebraically independent in $A$. Does this imply that $\{f_{1},...,f_{n}\}$ is an algebraically independent set in $A$? If not, how can I show that there are at least $n$ algebraically independent elements in A? My two ideas are (1) use linear combinations of the $f_{i}$, and (2) successively choose different fractions in B that guarantee that $\{f_{1},...,f_{n}\}$ is algebraically independent. I am not sure how to go about proving that either of these two ways actually works.

I believe some of the comments of the following are relevant: Extension of residue fields and algebraic independence

Let $T=\{f_1,\dots,f_n,g_1,\dots,g_n\}$. Replacing $A$ by $k[T]$ and $B$ by $k(T)$, we may assume $A$ is generated by $T$.
Now let $S\subseteq T$ be a maximal algebraically independent subset of $T$. By maximality of $S$, every element of $T$ is algebraic over $k(S)$, and hence so is every element of $B$, since $B$ is generated by $T$ as a field extension of $k(S)$. So $S$ is a transcendence basis for $B$ over $k$. But since $\{\frac{f_{1}}{g_{1}},...,\frac{f_{n}}{g_{n}}\}$ is algebraically independent, $B$ has transcendence degree at least $n$ over $k$. It follows that $S$ has at least $n$ elements.
• Take any element of $F$ which is not algebraic over $k(T)$, and add it to $T$. – Eric Wofsey May 20 '18 at 21:43