# Flat $\mathbb{Z}$-modules and algebraic independence

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!

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