# What does it mean for elements to be algebraically independent?

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?

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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.

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There is something else that confuses me (that's related to this). There is an example in the textbook that states "The null set forms a transcendence basis for an algebraic extension of K. In particular, algebraic extensions have finite transcendence degree. But a transcendence basis needs to be algebraically independent, right? But doesn't algebraic independnece imply that the null set is the identity element? (Since the evaluation map is injective). – user58289 Apr 17 '13 at 1:11
The null set is by definition algebraically independent. It doesn't make sense to consider the evaluation map at the null set. We need an $n$-tuple of elements to define our evaluation map. I'm not sure what you mean by "the null set is the identity element." – Jared Apr 17 '13 at 3:33

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.

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Equivalent in the case of fields, of course. The second definition is boarder. – Asaf Karagila Apr 16 '13 at 0:23
@AsafKaragila Yes, good point. Although the only reason it is broader is the use of the word "field" in the first. – Alex Becker Apr 16 '13 at 0:25

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 $\rm\:A.$

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