TITLE: What does it mean for elements to be algebraically independent?
QUESTION [6 upvotes]: Wikipedia's definition:
"In abstract algebra, a subset S of a field L is algebraically independent over a subfield K if the elements of S do not satisfy any non-trivial polynomial equation with coefficients in K."
My textbook's definition:
"Let A be a subring of the commutative ring B. We say that the
elements $b_1, . . . , b_n$ of B are algebraically independent over A if the evaluation map
$ε_{b_1,...,b_n} : A[X_1, . . . , X_n] → B$ that evaluates each $X_i$ at $b_i$ is injective."

I'm a bit confused about how these two definitions are related. I think wikipedia's definiton is clearer but I'm not sure how this relates to the definition in the textbook. In other words, why are we looking at this evaluation map in order to understand the definition of algebraic independence?
Thanks in advance

REPLY [3 votes]: The kernel of the evaluation homomorphism is the ideal $\rm\,I\,$ of polynomial relations  (over $\rm\:A)\:$ among the $\rm\:b_i,\:$ i.e. the set of polynomials with coefficients in $\rm\:A\:$ such that $\rm\:f(b_1,\ldots,b_n)\, =\, 0.\:$ Therefore the evaluation map is injective iff the kernel $\rm\:I = 0,\:$ i.e. iff there exist no nonzero polynomial relations ($\rm A$-algebraic dependencies) among the $\rm\:b_i.$ When this holds true, then $\rm\:A[b_1,\ldots,b_n]\cong A[x_1,\ldots,x_n],\:$ so $\rm\:b_i\:$ are independent indeterminates (transcendentals) over $\rm\:A.$