I'm confused about this basic notion in algebraic geometry. If $u$ has an expression $u = p/q$, where $q(P) \neq 0$, then we call $u$ a regular function at $P$. Now we would like to know which rational functions (of one variable) are regular at the point at infinity in projective space. So if we have $u(x) = p(x)/q(x)$, we can homogenize this, and look at its representation in projective space. In particular, we want the denominator to be non-zero at the point $[1:0]$ of projective space.
Here's my problem. It is claimed in this question here: Which rational functions $\mathbb{P}^1\rightarrow k$ are regular at the point at infinity? that the rational functions regular at infinity have degree of $q$ at least that of degree of $p$. I don't understand why this is true.
Consider the rational function defined by $p(x) = x^2 + x + 1$, and $q(x) = x + 1$. We can homogenize this guy, and assuming I didn't goof any arithmetic, the result is $\frac{x^2 + xy + y}{x+y}$. My problem is that the denominator is not $0$ at infinity, although $\deg q < \deg p$.
What am I not understanding about this process?
Rational functions on projective space are ratios of homogeneous polynomials of the same degree. A ratio of polynomials of different degrees will not yield a well-defined function on projective space, since rescaling the coordinates by a constant will produce different values. For instance, let $F(x,y) = \frac{x^2 + xy + y}{x+y}$ and consider the point $P = [1 : 1] = [2 : 2]$. Then $$ F(P) = \frac{1 + 1 + 1}{1 + 1} = 3/2 $$ but also $$ F(P) = \frac{4 + 4 + 2}{2 + 2} = 10/4 = 5/2 \, . $$
To turn $g(x) := \frac{p(x)}{q(x)} = \frac{x^2 + x + 1}{x + 1}$ into a function on projective space, we have to homogenize the numerator and denominator so that they have the same degree. This yields $G(X,Y) = \frac{X^2 + XY + Y^2}{XY + Y^2}$, and you can check that $G$ has a pole at $\infty = [1:0]$ since $G([1:0]) = [1:0]$.
An alternative way of checking vanishing at infinity is to use the change of coordinate map $x \mapsto 1/x$ that exchanges $0$ and $\infty$. Then we can examine the behavior of $g(1/x)$ at $0$ to determine the behavior of $g(x)$ at $\infty$. We find \begin{align*} g(1/x) = \frac{\frac{1}{x^2} + \frac{1}{x} + 1}{\frac{1}{x} + 1} = \frac{1 + x + x^2}{x + x^2} = \frac{x^2 + x + 1}{x(x+1)} \end{align*} which indeed has a pole at $0$.