I have been reading about Riemann surfaces from Rick Miranda's Book and now the term transcendent degree has come in to play, but I have no knowledge of field theory so I dont quite understand it I have some intuition behind it but I might be wrong.
So when proving that $\mathcal{M}(X)$ has transcendent degree one over $\mathbb{C}$ when $X$ is an algebraic curve there a few steps that I don't completely understand, when he says let $f$ and $g$ be two algebraically independent elements of $\mathcal{M}(X)$ do we mean that there is no combination of $g$ with elements of $\mathcal{M}(X)$ that can give us $f$? Is that why latter we will have that $\dim(L(nD))\geq (n^2+3n+2)/2$? New edit : Ok I think I already understood this part because we consider $L(D)$ as vector space over $\mathbb{C}$ and that will give us that number.
Now this is still something I dont understand: Also another thing thats is making me confusion due to my lack of knoweledge of field theory , but the author says that in order to prove that $\mathcal{M}(X)$ is finitetely generated extension field of $\mathbb{C}$ it suffices to see that $\mathcal{M}(X)$ is a finite algebraic extension of $\mathbb{C}(f)$, now this is making me a bit of cofusion because if $\mathbb{C}(f)$ is isomorphic to $\mathbb{C}(t)$, wouldn't this have infinite dimension as a vector space over $\mathbb{C}$? I am probably confusing definitions because maybe the algebraic extension dimension is not the same one as this but I really don't know, so any help would be appreciated.
Thanks in advance.
I assume you are asking about Prop. 1.17 on pp.174-176.
The notion of transcendence degree is analogous to dimension. The dimension of a vector space $E$ over a field $F$ is the maximum number of linearly independent elements, where $v_1,\ldots,v_n$ are linearly independent if there is no non-trivial linear relation $$L(v_1,\ldots,v_n)=c_1v_1+\cdots+c_nv_n=0$$ among them. (If $E$ is an extension field of $F$, then $E$ is a fortiori a vector space over $F$.) Analogously, the transcendence degree of an extension field $E$ of $F$ is the maximum number of algebraically independent elements, where $v_1,\ldots,v_n$ are algebraically independent if there is no non-trivial polynomial relation $$P(v_1,\ldots,v_n)=0$$ where $P$ is a non-zero polynomial in $n$ variables. For example, the field $\mathbb{Q}(\sqrt{e},\pi,\sqrt[3]{e\pi})$ has transcendence degree at most 2 over $\mathbb{Q}$, since we have the polynomial relation $x^2y-z^3=0$ when $x=\sqrt{e}$, $y=\pi$, and $z=\sqrt[3]{e\pi}$. (Also just about everybody believes it has transcendence degree exactly 2, but nobody's proven that yet.)
If $f$ and $g$ are algebraically independent, then the set of all monomials $f^ig^j$, $i,j\geq0$ is linearly independent, since a non-trivial linear relation among these monomials (with coefficients in $F$) is the same thing as a non-zero polynomial relation between $f$ and $g$. So the monomials belonging to $L(nD)$ are also linearly independent, and since there are more of them (for large $n$) than the bound on the dimension of $L(nD)$, we have a contradiction.
As for your second question, $\mathbb{C}(t)$ indeed has infinite dimension over $\mathbb{C}$, and so does $\mathcal{M}(X)$. But for finite dimension, we need to be able to express all the elements of the extension field linearly in terms of a finite number of basis elements. For finite generation, we allow rational expressions. Precisely, $E=F(v_1,\ldots,v_n)$ if for every element $z\in E$, there are polynomials (in $n$ variables) $P$, $Q$, with coefficients in $F$, such that $$z=\frac{P(v_1,\ldots,v_n)}{Q(v_1,\ldots,v_n)}$$ As a famous example, an elliptic function field $\mathcal{M}(X)$ is generated by its Weierstrass function and its derivative, $\wp$ and $\wp'$. There is a cubic equation relating $\wp$ and $\wp'$, which is why the transcendence degree isn't 2. Nonetheless, you can't express either one as a rational function of the other.
Hope this helps.