I know that this question is naive, but I really want to find out if this observation is correct. First, let $\alpha_1, \alpha_2, \dotsc, \alpha_k$ be real numbers. Consider the $\mathbb{Z}$-module homomorphism \begin{align*} f \colon \mathbb{Z}[x_1, x_2, \dotsc, x_k] &\rightarrow \mathbb{R} \,, \\ p(x_1, x_2, \dotsc, x_k) &\mapsto p(\alpha_1, \alpha_2, \dotsc,\alpha_k) \,. \end{align*} Now, we will say that a polynomial $g \in \mathbb{Z}[x_1, x_2, \dotsc, x_k]$ is unfactored if there is no polynomial $h \in \mathbb{Z}[x_1, x_2, \dotsc, x_k]$ and integer $a$ such that $g=ah$. Let $I$ be the set of unfactored polynomials in $\mathbb{Z}[x_1, x_2, \dotsc, x_k]$. Given the following exact sequence of $\mathbb{Z}$-modules: $$ \operatorname{Im} f \overset{g'}{\longrightarrow} \operatorname{Im} f \rightarrow \operatorname{Im} f / \operatorname{Im} g', $$ where $$ g'(p(\alpha_1, \alpha_2, \dotsc, \alpha_k)) = g(\alpha_1, \alpha_2, \dotsc, \alpha_k) p(\alpha_1, \alpha_2, \dotsc, \alpha_k) $$ one can notice several things:

  1. $\operatorname{Im} f$ is a torsion free $\mathbb{Z}$-module (since for every real number $\gamma$ there is no integer $b$ with $\gamma b = 0$) and, therefore, a flat $\mathbb{Z}$-module.

  2. Since $\operatorname{Im} f$ is flat, this sequence is pure exact iff $\operatorname{Im} f / \operatorname{Im} g'$ is flat.

However, a theorem of Kaplansky says that $A$ is a pure submodule of $B$ iff for every ideal $L$ of the ring $R$ (in this case $\mathbb{Z}$): $A \cap L \cdot B = L \cdot A$. Using this theorem, one can deduce that a short exact sequence as above is pure, or, equivalently, $\operatorname{Im} f / \operatorname{Im} g'$ is flat iff $g(\alpha_1, \alpha_2, \dotsc, \alpha_k) \notin \mathbb{Z}$ and there is no polynomial $h$ with $g(\alpha_1, \alpha_2, \dotsc, \alpha_k) = l \cdot h(\alpha_1, \alpha_2, \dotsc, \alpha_k)$, for some integer $l$. However, it is clear that these conditions hold iff $\alpha_1, \alpha_2, \dotsc, \alpha_k$ are algebraically independent. Knowing that a direct sum of modules is flat iff every direct summand is so, we can see that the algebraic independence of $\alpha_1, \alpha_2, \dotsc, \alpha_k$ is equivalent to the purity of the sequence: $$ \bigoplus_{g \in I} \operatorname{Im} f \overset{g'}{\longrightarrow} \operatorname{Im} f \rightarrow \operatorname{Im} f / \operatorname{Im} g' $$ and to the flatness of $\bigoplus_{g \in I} \operatorname{Im} f / \operatorname{Im} g'$. It is known that rings with countably generated ideals (definitely the case of $\mathbb{Z}$) have weak homological dimension less than or equal to $2$. Moreover, it is known that $\mathbb{Z}$ has weak homological dimension $1$. This means that $\bigoplus_{g \in I} \operatorname{Im} f / \operatorname{Im} g'$ is flat iff the sequence above is not a minimal flat resolution of it. Due to the characterization of minimal flat resolutions which uses flat covers, we get that if $$ \bigoplus_{g \in I} \operatorname{Im} f_{g} \rightarrow \bigoplus_{g \in I} \operatorname{Im} f / \operatorname{Im} g' $$ is not a flat cover, then $\bigoplus_{g \in I} \operatorname{Im} f / \operatorname{Im} g'$ is flat and, therefore, $\alpha_1, \alpha_2, \dotsc, \alpha_k$ are algebraically independent. This implies that if there exists an endomorphism $s \colon \bigoplus_{g \in I} \operatorname{Im} f_{g} \rightarrow \bigoplus_{g \in I} \operatorname{Im} f_{g}$ such that the map above is equal to its composition with $s$ and there exist two elements $x_1, x_2 \in \bigoplus_{g \in I} \operatorname{Im} f_{g}$ such that $s(x_1) = s(x_2)$, then $\alpha_1, \alpha_2, \dotsc, \alpha_k$ are algebraically independent (because $s$ is not an automorphism).

The reason for which I consider this result interesting is that it transforms a statement of the form $\forall$ implies algebraic independence into $\exists$ implies algebraic independence. In other words, it sufficises to show that there exists a function $s$ and there exists some elements $x_1$, $x_2$ satisfying certain conditions, instead of proving that a statement holds for all the elements of the polynomial ring $\mathbb{Z}[x_1, x_2, \dotsc, x_k]$. Thanks in advance!

  • $\begingroup$ And the question is...? $\endgroup$ – user26857 May 16 at 21:28
  • $\begingroup$ Whether or not it is correct. I mean, is there anything wrong or are there any mistakes? $\endgroup$ – Samboan M May 17 at 4:13

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