More thoughts: On $\mathbb{R}P^1$ functions that are locally polynomial must be constant (the standard proof still applies since it doesn't use the field being algebraically closed). Here is a map that identifies $\mathbb{R}P^1$ with $S^2$: $f:[x:y]\mapsto \left(\frac{x^2-y^2}{x^2+y^2},\frac{2xy}{x^2+y^2}\right)$. Regular functions on $S^2$ pull back to rational functions (homogeneous of degree 0) on $\mathbb{R}P^1$ whose denominators never vanish. If we regard such functions as regular, we get plenty of nonconstant regular functions on $\mathbb{R}P^1$. So the key is how we define regular functions: if they must be local polynomials, they are constants. If they are nonsingular rational functions, it is easy to find regular functions that are not constant. When the field is algebraically closed, the Nullstellensatz implies that nowhere-singular rational functions are polynomials. (I may be opening myself up to flames for asking stupid questions, but I'm writing a book on algebraic geometry and want these little issues beaten to death). Thanks for your answers and comments!

More thoughts: On $\mathbb{R}P^1$ functions that are locally polynomial must be constant (the standard proof still applies since it doesn't use the field being algebraically closed). Here is a map that identifies $\mathbb{R}P^1$ with $S^1$: $f:[x:y]\mapsto \left(\frac{x^2-y^2}{x^2+y^2},\frac{2xy}{x^2+y^2}\right)$. Regular functions on $S^1$ pull back to rational functions (homogeneous of degree 0) on $\mathbb{R}P^1$ whose denominators never vanish. If we regard such functions as regular, we get plenty of nonconstant regular functions on $\mathbb{R}P^1$. So the key is how we define regular functions: if they must be local polynomials, they are constants. If they are nonsingular rational functions, it is easy to find regular functions that are not constant. When the field is algebraically closed, the Nullstellensatz implies that nowhere-singular rational functions are polynomials. (I may be opening myself up to flames for asking stupid questions, but I'm writing a book on algebraic geometry and want these little issues beaten to death). Thanks for your answers and comments!

More thoughts: On $\mathbb{R}P^2$ functions that are locally polynomial must be constant (the standard proof still applies since it doesn't use the field being algebraically closed). Here is a map that identifies $\mathbb{R}P^2$ with $S^2$: $f:[x:y]\mapsto \left(\frac{x^2-y^2}{x^2+y^2},\frac{2xy}{x^2+y^2}\right)$. Regular functions on $S^2$ pull back to rational functions (homogeneous of degree 0) on $\mathbb{R}P^2$ whose denominators never vanish. If we regard such functions as regular, we get plenty of nonconstant regular functions on $\mathbb{R}P^2$. So the key is how we define regular functions: if they must be local polynomials, they are constants. If they are nonsingular rational functions, it is easy to find regular functions that are not constant. The glue between these is the Nullstellensatz, which implies that nowhere-singular rational functions are polynomials. (I may be opening myself up to flames for asking stupid questions, but I'm writing a book on algebraic geometry and want these little issues beaten to death). Thanks for your answers!

More thoughts: On $\mathbb{R}P^1$ functions that are locally polynomial must be constant (the standard proof still applies since it doesn't use the field being algebraically closed). Here is a map that identifies $\mathbb{R}P^1$ with $S^2$: $f:[x:y]\mapsto \left(\frac{x^2-y^2}{x^2+y^2},\frac{2xy}{x^2+y^2}\right)$. Regular functions on $S^2$ pull back to rational functions (homogeneous of degree 0) on $\mathbb{R}P^1$ whose denominators never vanish. If we regard such functions as regular, we get plenty of nonconstant regular functions on $\mathbb{R}P^1$. So the key is how we define regular functions: if they must be local polynomials, they are constants. If they are nonsingular rational functions, it is easy to find regular functions that are not constant. When the field is algebraically closed, the Nullstellensatz implies that nowhere-singular rational functions are polynomials. (I may be opening myself up to flames for asking stupid questions, but I'm writing a book on algebraic geometry and want these little issues beaten to death). Thanks for your answers and comments!