With regards to Theorem 3.2 in Hartshorne: Are regular functions on a variety simply polynomials? I am reading Hartshorne's Algebraic Geometry for the first time and I am having some trouble understanding Proposition 3.2.
The proposition implies that
$\begin{array}{ccccc}
\mathcal{O}(Y) &\hookrightarrow &\mathcal{O}_{p,Y}&\hookrightarrow & K(Y)\\
\cong & &\cong & & \cong\\
A(Y)&\hookrightarrow &A(Y)_{M_p}&\hookrightarrow &\text{field of fractions}(A(Y))
\end{array}$
Here $\mathcal{O}(Y)$ is the ring of regular functions over $Y$, $\mathcal{O}_p(Y)$ is the local ring of $p$ on $Y$, $K(Y)$ is the function field of $Y$, and $A(Y)$ is the coordinate ring. 
The reason I find this confusing is that all the rings in the top row are defined locally. Each definition implies that the functions have to be equal to fractions of polynomials in particular open sets. They do not, however, imply apriori that the functions have to be fractions everywhere on $Y$. Despite this,  I think the isomorphisms do imply that functions in these rings are indeed fractions of polynomials everywhere on $Y$, which seems very counter-intuitive. 
My question is: due to the above isomprphism, are the follwoing characterizations true? 
(1) The only regular functions in $\mathcal{O}(Y)$ are polynomials on $Y$.
(2)The only elements in $\mathcal{O}_{p,Y}$ can be represeted as $<Y,f/g>$ where $f$ and $g$ are polynomials in $A(Y)$ such that $g(p)\ne 0$.
(3)The only elements in $K(Y)$ can be represented as $<Y, f/g>$ where $f$ and $g$ are any non-zero elements in $A(Y)$.
On a similar note, I am having some difficulty with the proof of this proposition because I don't know much about transcendence degrees. Does anyone know of a proof that doesn't use this concept?
 A: I cannot say that there is an affirmative answer to all three of your questions, but the ideas that seem to be behind them are basically correct:  
This proposition simply says that in the case of an affine variety $Y$ (the word "affine" is key here) the ring $\mathcal{O}(Y)$ has a simple description, as follows. Say $Y$ is the zero locus of some polynomials in some $\mathbb{A}^n$. Then part (a) of the proposition states that all regular functions come from restricting polynomial functions on $\mathbb{A}^n$ to $Y$.  
Part (c) then implies that every element in $\mathcal{O}_p$ is represented by $\langle U_g, f/g\rangle$ where $f$ and $g$ come from polynomials functions on $\mathbb{A}^n$ restricted to $Y$, $g$ does not vanish at $p$, and $U_g$ is the complement in $Y$ of the zero locus of $g$. In other words, any element of $\mathcal{O}_p$ can be viewed as the quotient of two polynomials on $Y$, the denominator not vanishing at $p$ (which means exactly that elements of $\mathcal{O}_p$ are in the localization of $A(Y)$ at $\mathfrak{m}_p$; that's the idea behind the proof.)  
The first half of (d) says that every element of $K(Y)$ can be represented by  $\langle U_g, f^\prime/g^\prime\rangle$, where this time, the only requirement on $g^\prime$ is that is does not vanish everywhere on $Y$. This is similar to (c) except that there is no mention of $p$.
As for transcendence degrees, they can be avoided in this instance if you simply ignore the second half of part (d) in the proposition (which itself is about transcendence degrees.) But transcendence degrees are not too hard to deal with, and they are necessary to discuss the dimensions of varieties. I would recommend reading any basic treatment on them, and working out a couple exercises on them (especially ones that relate transcendence degrees to Krull dimension.)  
One last note: The assumption that $Y$ is affine is, as I mentioned, key; especially since $A(Y)$ does not even make sense otherwise. Also, for projective varieties and the ring $S(Y)$, things are considerably different. See Proposition 3.4 in the same chapter in Hartshorne.
