Let $R<S$ be an extension of commutative rings with identities (i.e. $R$ is a subring of $S$). We say $s_1,s_2,...,s_k$ are algebraically independent over $R$ iff there is no polynomial $f\in R[x_1,...,x_k]$ such that $f(s_1,...,s_k)=0$

let $n,M$ be positive integers (where $n>1$), and set $R=\mathbb{Z}_n$ and $S=\mathbb{Z}_n[x_1,x_2,...,x_M]$ (So $R$ is viewed as the ring of constant polynomials in $S$). Let $s_1,...,s_k\in S$ be algebraically independent over $R$.

Question 1: If $n$ is prime (so $R$ is a field now), must we have $k\leq M$ ? I think the answer is yes and it follows from steinitz theory of dependence relations. I can not trust myself though, because I read about this long time ago and I forgot the exact statements of the theorems of steinitz theory.

Now here is the question that I am more interested about

Question 2: If $n>1$ is just a positive integer (not necessarily prime), must we have $k\leq M$ ?

Thank you a lot.

Edit: $\mathbb{Z}_n$ means $\{0,1,...,n-1\}$ under addition and multiplication modulo $n$

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    $\begingroup$ By $\mathbb{Z}_n$, do you mean the projective limit (over $k$) of $\mathbb{Z}/n^k\mathbb{Z}$, or $\mathbb{Z}/n\mathbb{Z}$? $\endgroup$ – YCor Apr 3 '17 at 22:19
  • $\begingroup$ @YCor I mean $\mathbb{Z}/n\mathbb{Z}$ $\endgroup$ – Amr Apr 3 '17 at 22:31
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    $\begingroup$ Do you really want to consider $\sqrt{2}$ and $\sqrt{3}$ to be algebraically independent over $\Bbb{Q}$? I don't think that's the standard definition. $\endgroup$ – Axel Boldt Apr 3 '17 at 22:39
  • $\begingroup$ @AxelBoldt Thank you a lot. I formalized my intuition in an incorrect way. I will modify my question $\endgroup$ – Amr Apr 3 '17 at 22:41
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    $\begingroup$ This holds over any nontrivial commutative ring. It is a particular case of Theorem 2 in my post mathoverflow.net/a/144625 (indeed, it is the case when $C' = B $). $\endgroup$ – darij grinberg Apr 4 '17 at 4:22

If $p$ is prime and $s_1,\ldots,s_k \in \mathbb{F}_p[x_1,\ldots,x_M]$ are algebraically independent, then we must have $k\leq M$ since the field $\mathbb{F}_p(s_1,\ldots,s_k)$ is of transcendence degree $k$ over $\mathbb{F}_p$ and included in $\mathbb{F}_p(x_1,\ldots,x_M)$ which is of transcendence degree $M$.

Now in general, $s_1,\ldots,s_k$ being algebraically independent over $R$ means that the family of their monomials $s_1^{i_1}\cdots s_k^{i_k}$ (for $\underline{i}$ ranging over $\mathbb{N}^k$) is linearly independent over $R$.

Over $R = \mathbb{Z}/n\mathbb{Z}$ where $n$ is composite (please don't write $\mathbb{Z}_n$ for $\mathbb{Z}/n\mathbb{Z}$), when $n$ is squarefree, the problem is easily reduced to $n$ prime by the Chinese remainder theorem: if $n=p_1\cdots p_r$ then $\mathbb{Z}/n\mathbb{Z} = \mathbb{F}_{p_1} \times \cdots \times \mathbb{F}_{p_r}$ and the same holds rings of polynomials of $M$ variables thereof, and linear independence is checked componentwise.

Over $R = \mathbb{Z}/p^v\mathbb{Z}$, if I am not mistaken, elements $z_1,\ldots,z_N \in R^n$ are $R$-linearly independent iff their reduction mod $p$ (in $\mathbb{F}_p^n$) are $\mathbb{F}_p$-linearly independent (as seen by multiplying the coefficients by the appropriate power of $p$). So the same conclusion should hold for algebraic independence of polynomials.


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