I am a beginner in alegbraic geometry. I am trying to understand what the statement "function on a variety" means. Consider this extract taken from here:
Working in $\mathbb{C}[x]$ take for example $X = V( \langle x^2 \rangle) $ and the polynomial $x^2+1$ . What does it mean for that polynomial to "define a function on $X$" ?
Recall that in this setup, the points in $X$ are points living in $\mathbb{C}$. In other words, if I say $V(\langle x^2\rangle)$, then I mean the set of elements $c \in \mathbb{C}$ such that $c^2 = 0$. What I've just done is taken a polynomial $f(x) = x^2$, plugged in the value $c$, and tested whether I got zero. In this case, $f(x)$ defines a function on $\mathbb{C}$. In particular, the functions on $\mathbb{C}$ morally should just be polynomials, at least when we restrict our attention to algebraic geometry (if you do, say, differential geometry then you might care about holomorphic things, which are no longer polynomials in general).
So what about if you say "functions on $X = V(\langle x^2 \rangle)$"? Well, elements of $X$ still are still elements of $\mathbb{C}$, but it just so happens that the only element $c$ such that $c^2 = 0$ is $c = 0$. So $X = \{0\}$. Now, any polynomial, say $f(x) = x^2 + 1$ as in your example, is still a function on $X$, in the sense that I can pick any point of $X$ (which is some complex number) and plug it into $f$. But it just so happens that the only point of $X$ is $0$, and $f(0) = 1$. So even though $f(x) = x^2 + 1$ and the other constant polynomial $g(x) = 1$ are different polynomials, they still define the same function on $X$, because $f(0) = 1 = g(0)$. Formally, this means that really the $\mathbb{C}$-algebra of functions on $X$ (which also uniquely determines $X$ under some appropriate equivalence of categories) is just $\mathbb{C}[x]/\sqrt{(x^2)} = \mathbb{C}[x]/(x) = \mathbb{C}$: in other words, the only part of the polynomial that matters is the constant term, which is expressed by the fact that we took the quotient by the ideal generated by $x$.
So in this setting, a function should really just be a map $f: X \to \mathbb{C}$ which is the restriction of a polynomial $f(x) \in \mathbb{C}[x]$ (which gives a map $f: \mathbb{C} \to \mathbb{C}$ by plugging in complex numbers) to $X$. More generally, there's a notion of a "regular function" on a variety or scheme, which basically means "looks like a polynomial affine-locally" but where you're no longer working in affine space.
Of course from this perspective (this "classical" perspective of varieties), you do lose a bit of information: the $\mathbb{C}$-scheme $\mathrm{Spec}(\mathbb{C}[x]/(x))$ is only the reduced subscheme of $\mathrm{Spec}(\mathbb{C}[x]/(x^2))$, which means you've lost track of multiplicity at $0$.