Atiyah and Macdonald Exercise 1.27 Please do not ruin the fun by telling me why $\mu$ is surjective!
I am having trouble understanding the idea of the coordinate functions on the affine algebraic variety $X$. I am trying to understand that $P(X)$ is generated as a $k$-algebra by the coordinate functions. I understand what it means to be generated as a $k$-algebra, but the problem is I don't understand what the coordinate functions are!
Just some notation: $P(X) = k[t_1,t_2,\ldots, t_n]/I(X)$ where k is algebraically closed and $I(X)$ is the ideal of the variety $X$.
What do they explicitly mean when they say "Let $\xi_i$ be the image of $t_i$ in $P(X)$."? I can't really see what the image of an unknown is.
For example, if I have $k[t]/(t^2)$, then the image of the $t$ is still $t$ but with the relation that $f(t)t^2 = 0$ for $f \in k[t]$.
If I have $k[x,y]/(xy-1)$ then the images of $x$ and $y$ are still $x$ and $y$ with $f(x,y)(xy-1) = 0$? So there is no explicit way to show the image of a variable?
Then the problem states that if $x \in X$, then $\xi_i(x)$ is the $i$th coordinate of x. I guess this just means the usual "plugging in" of an element in an equation. Why doesn't this work for arbitrary elements in $k^n$?
 A: A lot of this repeats what Paul says in his comment above.
Since $k$ is algebraically closed it's infinite, so the map $A := k[x_1, \dots, x_n] \to \operatorname{Fun}(k^n, k)$ is injective. [This is not the important part, but it's comforting.] We can compose with restriction of functions to get a map into $\operatorname{Fun}(X, k)$. The ideal $I(X)$ is the kernel of this composition.
So, $A(X) := A/I(X)$ can be identified with a subring of the ring of functions on $X$ and this allows us to evaluate elements of $A(X)$ on points of $X$.
As with any quotient, there is a canonical map $\phi\colon A \to A(X)$ and we're setting $\xi_i = \phi(x_i)$. I would not say that $\xi_i$ and $x_i$ are the same thing: they live in different rings. $\xi_i$ is the coset $x_i + I(X)$. Of course it's common to say things like $x_1 = 2x_2$ in $A/I(X)$ but one should be aware of what's going on.
We can't evaluate outside of $X$ because if $x \notin X$ then there is some element $f \in I(X)$ such that $f(x) \neq 0$. To define $\bar g(x)$ for some $\bar g \in A(X)$ we would try to choose some representative $g \in A$ such that $\phi(g) = \bar g$ and then set $\bar g(x) = g(x)$. But $g + f$ is another "lift" of $\bar g$ and $g(x) \neq g(x) + f(x)$.
In your first example [note, however, that $(t^2)$ couldn't be of the form $I(X)$; if you want an associated geometric space, do the exercises on schemes!], $t$ is the same thing as $t + t^2$ in the quotient and these evaluate differently at $1$.
