How to find equations of the graph of a rational map? Assume that $f:{\mathbb P}^n\dashrightarrow {\mathbb P}^n$ is a rational map. Then, how one can write down explicitly the equations of its graph?
 A: As 6006 says in the comments, we have to be careful what we mean by the terms used. A rational map is a class of actual morphisms $\varphi:U \mathbb \to \mathbb P^n$ where $U$ is an open set, where we consider $\varphi \sim \varphi'$ if they agree on a smaller open set. There is a canonical representative of $f: \mathbb P^n \dashrightarrow \mathbb P^n$ given by $f:U \to \mathbb P^n$, where $U$ is the largest possible open set on which $f$ is defined (abusing notation and giving them the same name).
Secondly, we have to be clear what we mean by the graph of a rational map. Do you mean the graph $\Gamma$ as a subset of $U \times \mathbb P^n$ (where $U$ is a maximal open set on which $f$ is defined)? Since you talk about the equations of the graph, you might mean the closure of $\Gamma$ inside $\mathbb P^n \times \mathbb P^n$. Let's decide for the latter.
Let me explain this with a concrete example.
Let $f:\mathbb P^2 \dashrightarrow \mathbb P^2$ be given by $(x:y:z) \mapsto (xy:xz:yz)$. This is a rational involution of $\mathbb P^2$ (i.e. it is its own inverse, which is easily seen because it can also be written as $(x:y:z) \mapsto \left(\frac{1}{z} : \frac{1}{y} : \frac{1}{x}\right)$. The map $f$ is undefined on the three points $(1:0:0),(0:1:0)$ and $(0:0:1)$. Then $U=\mathbb P^2 \backslash \{P_1,P_2,P_3\}$, where $P_i$ are these three points is the maximal open set on which $f$ becomes a morphism.
We thus have a well-defined morphism (not just a rational map) $f:U \to \mathbb P^2$.  
Then let $(a:b:c)$ be coordinates on the target. Then as in Mohan's comment, the equations of the graph are given by the $2\times 2$-minors of the matrix
$$
\begin{bmatrix}a & b & c \\ xy & xz & yz \end{bmatrix}.
$$
The resulting ideal is generated by the $2 \times 2$-minors of this matrix and represents a closed subset $Z \subset \mathbb P^2 \times \mathbb P^2$. But it is not the right answer!
Why? Because using the same equations in a larger space, we get more components than we asked for. If $(x:y:z)=(1:0:0)$, then the equations reduce to $0=0$, which gives us many points not on the graph.
To get the correct ideal, we have to remove the components corresponding to fibers over the undefined points of the map. In commutative algebra terms, this means saturating the ideal by the ideal of the indeterminacy locus. Here's how to do do this in Macaulay2:
i1 : R = QQ[x,y,z,a,b,c]

o1 = R

o1 : PolynomialRing

i2 : M = matrix{{a,b,c},{x*y,x*z,y*z}}

o2 = | a  b  c  |
     | xy xz yz |

             2       3
o2 : Matrix R  <--- R

i3 : I = minors(2,M) : ideal(x*y,x*z,y*z)

o3 = ideal (y*b - x*c, z*a - x*c)

o3 : Ideal of R

i4 : isPrime I

o4 = true

i5 : isPrime minors(2,M)

o5 = false

Hence the ideal of the closure of the graph is defined by two bilinear equations in $\mathbb P^2 \times \mathbb P^2$. This is the blowup of $\mathbb P^2$ in the three points $P_i$, and it is a del Pezzo surface of degree 6. 
