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I just learned about the concept of an algebraic independent set. Here is the definition I started working with.

Let $E/K$ be a field extension. A subset $M \subset E$ is said to be algebraically independent over $K$ if every element $\alpha \in M$ is transcendent over $K(M-\{\alpha\})$. Otherwise, we say that $M$ is algebraically dependent over $K$.

I saw an alternative definition that goes like this

$M$ is algebraically independent over $K$ if for any $n \geq 1$ and $\alpha_1,\dots, \alpha_n \in M$, pairwise disjoint and for all $f \in K[X_1,\dots,X_n]$, $$f(\alpha_1,\dots,\alpha_n) =0 \Rightarrow f=0$$

I now want to show that the first definition implies the second. Here is how I went about it but I am unsure about my reasoning (that's already bad isn't it ?).

First of all I showed that a set $M$ is algebraically independent if and only if every finite subset of $M$ is algebraically independent (using the first definition).

We can thus assume that $M$ is finite.

Let $M = \{ \alpha_1, \dots, \alpha_n\}$. Suppose that there exists a non-zero polynomial $f \in K[X_1, \dots, X_n]$ such that $f(\alpha_1, \dots, \alpha_n) =0$.

I want to show that there is some $\alpha_i \in M$ that is algebraic over $K(M-\{\alpha_i\})$.

To do so, I write $f(X_1, \dots,X_n) = \sum_i g_i (X_1,\dots,X_{n-1})X_n^i$. Since $f(\alpha_i, \dots, \alpha_n) = 0$, we get an algebraic dependence of $\alpha_n$ over $K[X_1,\dots,X_n]$. Thus $\alpha_n$ is algebraic over $K(M-\{\alpha_n\})$ and we are done.

Could someone proof check my reasoning ? I am thankful for any comments.

Richard Kruel
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1 Answers1

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This is essentially correct, there is just a minor problem: The coefficients $g_i(X_1,\dots,X_{n-1})$ may become zero once we substitute in $\alpha_1,\dots,\alpha_{n-1}$ and so we might only get the trivial relation of $\alpha_n$ over $K(M\setminus\{\alpha_n\})$, therefore we can't pick any $\alpha_i$. Consider for example $K=\Bbb Q,M=\{\alpha_1=\sqrt{2},\alpha_2=\pi\}$ and $f = (X^2-2)Y$. Then $f\ne0$ and $f(\alpha_1,\alpha_2)=0$, yet $\alpha_2$ is not algebraic over $K(M\setminus\{\alpha_2\})=\Bbb Q(\sqrt{2})$. We could solve this as follows:
Write $$f(X_1,\dots,X_n)=\sum_{j=0}^mg_j(X_1,\dots,X_{n-1})X_n^j$$ with $g_m\ne0$. If $g_m(\alpha_1,\dots,\alpha_{n-1})\ne0$ we get that $f(\alpha_1,\dots,\alpha_{n-1},X_n)$ is not the zero-polynomial, hence $\alpha_n$ is algebraic over $K(M\setminus\{\alpha_n\})$. If on the other hand $g_m(\alpha_1,\dots,\alpha_{n-1})=0$, we may proceed inductively considering the set $M'=\{\alpha_1,\dots,\alpha_{n-1}\}$. If we are left with the case $n=1$, the polynomials $g_j$ will be constants, so the second case here cannot occur.

leoli1
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