What does it mean for elements to be algebraically independent?

Question

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

Answer 1

The evaluation map at a point $(b_1,\ldots,b_n)\in B^n$ takes a polynomial $f(x_1,\ldots,x_n)\in A[x_1,\ldots,x_n]$ to its value at the point: $f(b_1,\ldots,b_n)\in B$. Clearly the zero polynomial will map to zero, but if the $b_i$ are algebraically independent by the first definition, then only the zero polynomial will map to zero, making the evaluation map injective.

Answer 2

The two are equivalent. Suppose $b_1,\ldots,b_n$ are not algebraically independent according to your textbook's definition, so we have some nonzero polynomial $f(X_1,\ldots,X_n)\in A[X_1,\ldots,.X_n]$ which the evaluation map sends to $0$. This means $f(b_1,\ldots,b_n)=0$, i.e. the elements $b_1,\ldots,b_n$ satisfy some non-trivial polynomial equations with coefficients in $A$. The other direction is similar.

Answer 3

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 $\rm\:A.$