This is a statement from Algebraic Geometry that I'm having some trouble parsing:
$\Bbb Z[X_1, X_2, ... , X_n]$ is the free commutative ring on $n$-generators.
I understand Z is a polynomial ring, meaning, when you add two polynomials and multiply two polynomials you still get a polynomial. Also, I understand commutative, but what does "free" mean? and what does it mean for it to have $n$-generators?
"Free," in this case, means that For any set $X$ (free on $X$), there exists an injection map $f: X\to F$ such that for any function $g: X\to A$, where $A$ is any other $\mathbb{Z}$-algebra, there is a unique $h: F\to A$ such that $ f\circ h=g$.
In our case $X=\{x_1,\dots,x_n\}$. The injection is just given by $f(x_i)=x_i$, now suppose that $g:\{x_1,\dots,x_n\}\to A$ is a function, let $a_i=g(x_i)$ the images of the $x_i$'s under $g$, we can define $h$ by setting $h(x_i)=a_i$, by linearity $h$ is extended to all $\mathbb{Z}[x_1,\dots,x_n]$ and it is easy to show that $f\circ h=g$.
In general, it is the idea, but in any case, the type of morphisms changes depending on the category we are talking about.
