abstract algebra - The definition of "algebraically independent"

In Lang's Algebra, he gives a definition that

Elements $x_1, \cdots, x_n\in B$ are called algebraically independent over $A$[a subring of $B$] if the evaluation map $$f\mapsto f(x)$$is injective. Equivalently, we could say that if $f\in A[X]$ is a polynomial and $f(x)=0$, then $f=0$.

I am confused about the "injective" here. Two possible interpretations in my mind:

1. Fix an $f$, for $x\neq y,f(x)\neq f(y).$


2. Fix an $x$, for $f_1\neq f_2,f_1(x)\neq f_2(x).$

I was wondering which one is correct and why. Could you give me some helpful examples?

Besides, why are these two definitions equivalent?

Thanks in advance.

Answer

The evaluation map is from $A[X_1,\dots,X_n]$ to $B$. I'll denote it $\text{ev}_x : A[X] \rightarrow B$. Writing it out very explicitly, $\text{ev}_x(f(X_1,\dots,X_n)) = f(x_1,\dots,x_n)$. To say that it is injective is to say $ev_x(f) = ev_x(g) \Rightarrow f = g$. This is your second interpretation. To see that the two definitions are equivalent, recall that for any ring homomorphism $\phi : R \rightarrow S$, $\phi(r) = \phi(r')$ $\Longleftrightarrow$ $r - r' \in \text{ker}(\phi)$. Then it is a little exercise to see that $\phi$ is injective if and only if $\text{ker}(\phi) = \{0\}$. Apply this result to $ev_x$ and it says that $ev_x$ is injective if and only if $ev_x(f) = 0 \Rightarrow f = 0$. But $ev_x(f) = f(x_1,\dots,x_n)$, so the condition becomes $f(x_1,\dots,x_n) = 0 \Rightarrow f(X_1,\dots,X_n) = 0$. It is important to understand throughout that $f(x_1,\dots,x_n)$ is always an element of $B$, while $f(X_1,\dots,X_n)$ is always a polynomial in several variables with coefficients in $A$ (that is, an element of $A[X]$).