Transcendence Degree of Integral Domain over a Field This may be trivial, but I am confused on the following issues.
1)  If we have a finitely generated integral domain R over a field k, why is the transcendence degree of R over k (that is, the transcendence degree of its field of fractions over k) necessarily less than the number of generators of R as an algebra over k.
2) In what instances must the generators of R be algebraically independent?
 A: Here's an answer to your first question:
The transcendence degree of a field extension $L/K$ is the size of any transcendence basis for $L/K$, i.e. the size of any set of elements of $L$ that is maximal with respect to the property of being algebraically independent over $K$. The fact that you can use any maximal set is a really useful thing for computing transcendence degrees, akin to the fact that the dimension of a vector space is the dimension of any basis.
In your case, suppose that $R=k[a_1,\dots,a_n]$ is an integral domain. Then we have that $L = \operatorname{Frac}(R) = k(a_1,\dots,a_n)$. If all of the $a_i$ are algebraic over $k$, then $L$ is integral over $k$ so $tr.deg(L/k) = 0 \leq n$. Otherwise one of the $a_i$ is transcendental over $k$, and we can say $i = 1$ wlog. Then either all of the $a_i$ for $i \geq 2$ are integral over $k(a_1)$, or there is some $i$ with $\{a_1,a_i\}$ algebraically independent over $k$, again we can say $i = 2$ wlog. We can repeat the process until we get $m \leq n$ with $\{a_1,\dots,a_m\}$ algebraically independent over $k$ and $a_i$ integral over $k(a_1,\dots,a_m)$ for all $i > k$. But this means that $L$ is integral over $k(a_1,\dots,a_m)$ (since it is a finitely generated extension by integral elements), and so for any element $l$ of $L$,$l$ is integral over $k(a_1,\dots,a_m)$ and so $\{a_1,\dots,a_m,l\}$ is not an algebraically independent set. This is the same as saying $\{a_1,\dots,a_m\}$ is a maximal algebraically independent subset, which implies that $tr.deg(L/k) = m \leq n$, as required.
In fact, what we have actually proved is the the stronger statement that some subset of the $\{a_i\}$ is a transcendence basis for $L/k$. From this we can prove stronger statements. Firstly we can prove the second part of your question and see that the generators are algebraically independent iff $tr.deg(L/k) = n$. We can also prove that if any $r$ of the $a_i$ are algebraically dependent, then $tr.deg(L/k) \leq r$.
