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Initially, I was trying to look at the subrings of $\mathbb{Z}[X]$. Since I have failed hard, I have tried to at least count them.

So I have tried to build an injection from $\{0,1\}^\mathbb{N}$ to the set of subrings of $\mathbb{Z}[X]$. Since $\{0,1\}^\mathbb{N}$ is uncountable, we would have the set of subrings of $\mathbb{Z}[X]$ uncountable too.

By associating the sequence $(e_n)_{n\in\mathbb{N}}$ to the ring $\mathbb{Z}[1,p_1 e_1 X²,p_2 e_2 X^3,...]$ where $p_k$ is the $k$-th prime number and $e_k\in\{0,1\}$, it fails; for instance $\mathbb{Z}[1,2X²,3X^3,7X^5]$ is equal to $\mathbb{Z}[1,2X²,3X^3,7X^5,23X^{10}]$ because $X^{10}\in \mathbb{Z}[1,2X²,3X^3,7X^5]$ since $X^{10}=(7X^5 - (2X² * 3X^3))^2$.

I am unable to repair it, maybe it is flawed from the start.

Actually, I do not even know if the set of subrings of $\mathbb{Z}[X]$ is uncountable. If possible I would like a proof or a reference to a proof see since I am curious about its countability.

Thomas Andrews
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Fnifni
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  • It is actually countable. Try to write it as the union of the polynomials of degree $k$ for varying $k$ – Exodd Jul 14 '22 at 19:32
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    @Exodd: It seems like your comment wants to show the ring itself is countable. That's not the question. – Torsten Schoeneberg Jul 14 '22 at 19:36
  • I think it is uncountable if and only if there is one subring which has no finite set of generators. Not quite sure about that, though. I assume you want subrings to contain $1?$ – Thomas Andrews Jul 14 '22 at 19:55
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    @ThomasAndrews there‘s such a subring, e.g. $\Bbb Z[2X,2X^2,2X^3, \dots]$ – Lukas Heger Jul 14 '22 at 19:59
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    Yes, subrings must contain 1, i.e. a ring is defined to have a multiplicative identity. – Fnifni Jul 14 '22 at 20:18
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    Then given any $S\subseteq \mathbb N^+,$ the subring $\mathbb Z[2x^k]_{k\in S}$ is a different ring, since the polynomials will have some elements whose coefficients are $2$ on $x^k$ for $k\in S,$ and coefficients all divisible by $4$ for $k\notin S.$ @LukasHeger – Thomas Andrews Jul 14 '22 at 20:22

1 Answers1

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Given $S\subseteq \mathbb N^+,$ let $R_S$ be the subring of polynomials $\sum_{k=0}^n a_kx^k$ with the condition that $a_k$ is even for $k>0,$ and $a_k$ is divisible by $4$ for $k\in S.$ $a_0$ can be any integer.

This is obviously closed under addition and contains the identity. A little more work shows it is closed under multiplication.

Given two sets $S_1\neq S_2,$ we can find, for $k\in S_1\Delta S_2,$ that $2x^k$ is in exactly one of the $R_{S_i},$ where $\Delta$ is the symmetric difference.

So these subrings are all distinct.

Instead of $2,4$ we could use $d,ed$ for any pair $e,d>1$ and $e\mid d.$

This can be used to prove that $R[x]$ has uncountably many subrings for any commutative ring $R$ with an ideal $I$ where $I^2\neq I.$

Not sure about other cases, like $R=\mathbb Z/p\mathbb Z.$ Seems like you might get only countably many in that case?

Indeed, another way to write the proof is to consider subrings of $\mathbb Z/4\mathbb Z[x],$ where the coefficients for $k>0$ are $0$ for $k\in S$ and and $2$ or $0$ for $k\notin S.$ Then we take the inverse images of the uncountable subrings of $\mathbb Z/4\mathbb Z [x]$ through the map $\mathbb Z[x]\to\mathbb Z/4\mathbb Z[x].$

MathematicianByMistake
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Thomas Andrews
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