a question about the proof of theorem 5.3 of Hartshorne's Algebraic Geometry This is on page 33 of Hartshorne's Algebraic Geometry.
Let $Y$ be an affine variety, and Sing $Y$ be the set of singular points of $Y$. Suppose we already know that Sing $Y$ is a close subset of $Y$, now we need to show that $X := Y -$Sing $Y$, which is an open subset of $Y$, is not empty.
In order to do this, the author considers a hypersurface in $\mathbb P^n$ which is birationally equivalent to $Y$.

Since birational varieties have isomorphic open subsets, we reduce to the case of a hypersurface.

I have three questions about this.


*

*As in Corollary 4.5(i)(ii) of this same book, two varieties are birationally equivalent if and only if they each contains one nonempty open subsets isomorphic to each other. Does this mean the existence of an isomorphism correspondence between all the open subsets of the two varieties? 

*Another question is, let $Y,Y'$ be birationally equivalent varieties, and $X = Y -$Sing $Y$, $X'=Y'-$Sing $Y'$, then is it true that $X$ and $X'$ must be isomorphic? Combining with the first question, if the isomorphism correspondence between the open subsets of $Y$ and $Y'$ does exist, does it necessarily corresponds $X$ and $X'$?

*I think the answer to both of my questions must be negative, since the open subset might be the variety itself. But how can I understand the proof? Why can the author just consider $Y$ as a hypersurface?
Thanks.
 A: For question 1, the answer is no. Consider $\mathbb{A}^1$ and $\mathbb{A}^1-\{pt\}$. $\mathbb{A}^1$ is not isomorphic to any proper open subset of itself (this is an exercise somewhere in Chapter 1 of Hartshorne).
The answer to problem 2 is also no. If $Y=\mathbb{A}^1$ and $Y'=\mathbb{A}^1-\{pt\}$, then $X=Y$ and $X'=Y'$, so they're obviously non-isomorphic. On the other hand something close to the statement of 2 is true- $X$ and $X'$ are birational. Since $\operatorname{Sing}(Z)\subset Z$ is a proper closed subset for all varieties $Z$, we know that $X$ and $Y$ are birational, and similarly for $X'$ and $Y'$, so $X$ is birational to $X'$.
In reading the proof of I.5.3, note that the first sentence of paragraph 2 of the proof is "To show $\operatorname{Sing} Y$ is a proper closed subset of $Y$, we first apply I.4.9 to get $Y$ birational to a hypersurface in $\mathbb{P}^n$". So we go and look up I.4.9, which tells us that any variety $X$ of dimension $r$ is birational to a hypersurface $Y\subset \mathbb{P}^{r+1}$. The proof is as follows (all of this is essentially Hartshorne's proof with some extra comments from me):
The function field $K$ of $X$ is a finitely generated extension field of $k$. Since $k$ is algebraically closed, it is perfect, and by a theorem in commutative algebra, $K/k$ is separably generated, so we can find a transcendence basis $x_1,\cdots,x_n\in K$ so that $K$ is a finite separable extension of $k(x_1,\cdots,x_n)$. By the theorem of the primitive element, we can find one further element $y\in K$, algebraic over $k(x_1,\cdots,x_n)$ such that $K=k(x_1,\cdots,x_n,y)$. Since $y$ is algebraic over $k(x_1,\cdots,x_n)$, it satisfies a polynomial equation with coefficients which are rational functions in the $x_i$s. Clearing denominators, we get an irreducible polynomial $f(x_1,\cdots,x_n,y)=0$. This gives a hypersurface in $\mathbb{A}^{r+1}$ with function field $K$, which exactly means a hypersurface birational to $X$. In order to get this hypersurface in $\mathbb{P}^{r+1}$, we use the operation of projective closure, from exercise 2.9 of chapter I.
