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Following Todd Trimble's comment and Stefan Kohl's remark, we can show that $R$ is isomorphic to $\mathbf{Z}^{\mathbf{Z}}$ as a $\mathbf{Z}$-module. Since the product of countably many copies of $\mathbf{Z}$ is not free as a $\mathbf{Z}$-module — see this MO question — we deduce that $R$ is not free over $\mathbf{Z}$, which implies in particular that $R$ is not free over $\mathbf{Z}[x]$.

Let us endow $\mathbf{Z}$ with the following well-ordering : $0, 1, -1, 2, -2, 3, -3\dots$ The map $\phi : \mathbf{Z}^{\mathbf{Z}} \to R$ will have the form $(a_n)_{n \in \mathbf{Z}} \mapsto f$ where $f(n)$ is defined as a finite linear combination of the $a_m$ where $m$ runs trough the indices which are at most $n$ with respect to this well-ordering.

In order to give the idea of the construction, assume we work with functions defined on $\mathbf{N}$ instead of $\mathbf{Z}$, and give $\mathbf{N}$ the usual ordering. Define a map $(a_n)_{n \in \mathbf{N}} \mapsto f$ by putting $f(0)=a_0$, $f(1)=a_1$, $f(2)=a_0+2a_2$, $f(3)=3a_1-2a_0+6a_3$… The coefficients are chosen using the Bézout identity in such a way that the needed congruences hold, e.g. $f(3) \equiv f(0) \pmod{3}$.

In more detail, assume $f(n-k),\ldots,f(n-1)$ have been already defined, and let us define $f(n)$. (We proceed in an analogous way when defining $f(n)$ assuming $f(n+1),\ldots,f(n+k)$ are already defined.) Put $M=\operatorname{lcm}(1,\ldots,k) = p_1^{\alpha_1} \cdots p_r^{\alpha_r}$. We must have $p_i^{\alpha_i} \leq k$ for each $i$. There exist integers $\lambda_1,\ldots,\lambda_r \in \mathbf{Z}$ such that $\lambda_i \equiv 1 \pmod{p_i^{\alpha_i}}$ and $\lambda_i \equiv 0 \pmod{p_j^{\alpha_j}}$ for $j \neq i$. Then we put $f(n) = \sum_{i=1}^r \lambda_i f(n-p_i^{\alpha_i})+M \cdot a_n$. We check that $f(n) \equiv f(n-j) \pmod{j}$ for each $j$, and $f(n)$ is clearly linear with respect to the sequence $(a_n)$.

This defines our linear map $\phi : \mathbf{Z}^{\mathbf{Z}} \to R$. Now, it is not hard to see that $\phi$ is bijective by working backwards, starting from $f \in R$ and defining $a_n$ inductively, starting from $a_0=f(0)$, $a_1=f(1)$ and so on. We have to check the following : if $f \in R$ then using the notations above, we have $f(n) \equiv \sum_{i=1}^r \lambda_i f(n-p_i^{\alpha_i}) \pmod{M}$. Indeed, this holds modulo each $p_i^{\alpha_i}$ using the property of the $\lambda_i$'s and the assumption on $f$.

To sum up, the general picture is that (after fixing the well-ordering of $\mathbf{Z}$ explained above), the $\mathbf{Z}$-module $R$ is the homomorphic image of $\mathbf{Z}^{\mathbf{N}}$ under some infinite lower-triangular matrix

$$ \begin{equation*} \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ 0 & 1 & 0 & 0 & \\ 0 & 1 & 2 & 0 & \\ 3 & -2 & 2 & 6 & \\ \vdots & & & & \ddots \end{pmatrix} \end{equation*}. $$

Following Todd Trimble's comment and Stefan Kohl's remark, we can show that $R$ is isomorphic to $\mathbf{Z}^{\mathbf{Z}}$ as a $\mathbf{Z}$-module. Since the product of countably many copies of $\mathbf{Z}$ is not free as a $\mathbf{Z}$-module — see this MO question — we deduce that $R$ is not free over $\mathbf{Z}$, which implies in particular that $R$ is not free over $\mathbf{Z}[x]$.

Let us endow $\mathbf{Z}$ with the following well-ordering : $0, 1, -1, 2, -2, 3, -3\dots$ The map $\phi : \mathbf{Z}^{\mathbf{Z}} \to R$ will have the form $(a_n)_{n \in \mathbf{Z}} \mapsto f$ where $f(n)$ is defined as a finite linear combination of the $a_m$ where $m$ runs trough the indices which are at most $n$ with respect to this well-ordering.

In order to give the idea of the construction, assume we work with functions defined on $\mathbf{N}$ instead of $\mathbf{Z}$, and give $\mathbf{N}$ the usual ordering. Define a map $(a_n)_{n \in \mathbf{N}} \mapsto f$ by putting $f(0)=a_0$, $f(1)=a_1$, $f(2)=a_0+2a_2$, $f(3)=3a_1-2a_0+6a_3$… The coefficients are chosen using the Bézout identity in such a way that the needed congruences hold, e.g. $f(3) \equiv f(0) \pmod{3}$.

In more detail, assume $f(n-k),\ldots,f(n-1)$ have been already defined, and let us define $f(n)$. (We proceed in an analogous way when defining $f(n)$ assuming $f(n+1),\ldots,f(n+k)$ are already defined.) Put $M=\operatorname{lcm}(1,\ldots,k) = p_1^{\alpha_1} \cdots p_r^{\alpha_r}$. We must have $p_i^{\alpha_i} \leq k$ for each $i$. There exist integers $\lambda_1,\ldots,\lambda_r \in \mathbf{Z}$ such that $\lambda_i \equiv 1 \pmod{p_i^{\alpha_i}}$ and $\lambda_i \equiv 0 \pmod{p_j^{\alpha_j}}$ for $j \neq i$. Then we put $f(n) = \sum_{i=1}^r \lambda_i f(n-p_i^{\alpha_i})+M \cdot a_n$. We check that $f(n) \equiv f(n-j) \pmod{j}$ for each $j$, and $f(n)$ is clearly linear with respect to the sequence $(a_n)$.

This defines our linear map $\phi : \mathbf{Z}^{\mathbf{Z}} \to R$. Now, it is not hard to see that $\phi$ is bijective by working backwards, starting from $f \in R$ and defining $a_n$ inductively, starting from $a_0=f(0)$, $a_1=f(1)$ and so on. We have to check the following : if $f \in R$ then using the notations above, we have $f(n) \equiv \sum_{i=1}^r \lambda_i f(n-p_i^{\alpha_i}) \pmod{M}$. Indeed, this holds modulo each $p_i^{\alpha_i}$ using the property of the $\lambda_i$'s and the assumption on $f$.

To sum up, the general picture is that (after fixing the well-ordering of $\mathbf{Z}$ explained above), the $\mathbf{Z}$-module $R$ is the homomorphic image of $\mathbf{Z}^{\mathbf{N}}$ under some infinite lower-triangular matrix

$$ \begin{equation*} \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ 0 & 1 & 0 & 0 & \\ 0 & 1 & 2 & 0 & \\ 3 & -2 & 2 & 6 & \\ \vdots & & & & \ddots \end{pmatrix} \end{equation*}. $$

Following Todd Trimble's comment and Stefan Kohl's remark, we can show that $R$ is isomorphic to $\mathbf{Z}^{\mathbf{Z}}$ as a $\mathbf{Z}$-module. Since the product of countably many copies of $\mathbf{Z}$ is not free as a $\mathbf{Z}$-module — see this MO question — we deduce that $R$ is not free over $\mathbf{Z}$, which implies in particular that $R$ is not free over $\mathbf{Z}[x]$.

Let us endow $\mathbf{Z}$ with the following well-ordering : $0, 1, -1, 2, -2, 3, -3\dots$ The map $\phi : \mathbf{Z}^{\mathbf{Z}} \to R$ will have the form $(a_n)_{n \in \mathbf{Z}} \mapsto f$ where $f(n)$ is defined as a finite linear combination of the $a_m$ where $m$ runs trough the indices which are at most $n$ with respect to this well-ordering.

In order to give the idea of the construction, assume we work with functions defined on $\mathbf{N}$ instead of $\mathbf{Z}$, and give $\mathbf{N}$ the usual ordering. Define a map $(a_n)_{n \in \mathbf{N}} \mapsto f$ by putting $f(0)=a_0$, $f(1)=a_1$, $f(2)=a_0+2a_2$, $f(3)=3a_1-2a_0+6a_3$… The coefficients are chosen using the Bézout identity in such a way that the needed congruences hold, e.g. $f(3) \equiv f(0) \pmod{3}$.

In more detail, assume $f(n-k),\ldots,f(n-1)$ have been already defined, and let us define $f(n)$. (We proceed in an analogous way when defining $f(n)$ assuming $f(n+1),\ldots,f(n+k)$ are already defined.) Put $M=\operatorname{lcm}(1,\ldots,k) = p_1^{\alpha_1} \cdots p_r^{\alpha_r}$. We must have $p_i^{\alpha_i} \leq k$ for each $i$. There exist integers $\lambda_1,\ldots,\lambda_r \in \mathbf{Z}$ such that $\lambda_i \equiv 1 \pmod{p_i^{\alpha_i}}$ and $\lambda_i \equiv 0 \pmod{p_j^{\alpha_j}}$ for $j \neq i$. Then we put $f(n) = \sum_{i=1}^r \lambda_i f(n-p_i^{\alpha_i})+M \cdot a_n$. We check that $f(n) \equiv f(n-j) \pmod{j}$ for each $j$, and $f(n)$ is clearly linear with respect to the sequence $(a_n)$.

This defines our linear map $\phi : \mathbf{Z}^{\mathbf{Z}} \to R$. Now, it is not hard to see that $\phi$ is bijective by working backwards, starting from $f \in R$ and defining $a_n$ inductively, starting from $a_0=f(0)$, $a_1=f(1)$ and so on. We have to check the following : if $f \in R$ then using the notations above, we have $f(n) \equiv \sum_{i=1}^r \lambda_i f(n-p_i^{\alpha_i}) \pmod{M}$. Indeed, this holds modulo each $p_i^{\alpha_i}$ using the property of the $\lambda_i$'s and the assumption on $f$.

To sum up, the general picture is that (after fixing the well-ordering of $\mathbf{Z}$ explained above), the $\mathbf{Z}$-module $R$ is the homomorphic image of $\mathbf{Z}^{\mathbf{N}}$ under some infinite lower-triangular matrix

\begin{equation*} \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ 0 & 1 & 0 & 0 & \\ 0 & 1 & 2 & 0 & \\ 3 & -2 & 2 & 6 & \\ \vdots & & & & \ddots \end{pmatrix} \end{equation*}

Following Todd Trimble's comment and Stefan Kohl's remark, we can show that $R$ is isomorphic to $\mathbf{Z}^{\mathbf{Z}}$ as a $\mathbf{Z}$-module. Since the product of countably many copies of $\mathbf{Z}$ is not free as a $\mathbf{Z}$-module — see this MO question — we deduce that $R$ is not free over $\mathbf{Z}$, which implies in particular that $R$ is not free over $\mathbf{Z}[x]$.

Let us endow $\mathbf{Z}$ with the following well-ordering : $0, 1, -1, 2, -2, 3, -3\dots$ The map $\phi : \mathbf{Z}^{\mathbf{Z}} \to R$ will have the form $(a_n)_{n \in \mathbf{Z}} \mapsto f$ where $f(n)$ is defined as a finite linear combination of the $a_m$ where $m$ runs trough the indices which are at most $n$ with respect to this well-ordering.

In order to give the idea of the construction, assume we work with functions defined on $\mathbf{N}$ instead of $\mathbf{Z}$, and give $\mathbf{N}$ the usual ordering. Define a map $(a_n)_{n \in \mathbf{N}} \mapsto f$ by putting $f(0)=a_0$, $f(1)=a_1$, $f(2)=a_0+2a_2$, $f(3)=3a_1-2a_0+6a_3$… The coefficients are chosen using the Bézout identity in such a way that the needed congruences hold, e.g. $f(3) \equiv f(0) \pmod{3}$.

In more detail, assume $f(n-k),\ldots,f(n-1)$ have been already defined, and let us define $f(n)$. (We proceed in an analogous way when defining $f(n)$ assuming $f(n+1),\ldots,f(n+k)$ are already defined.) Put $M=\operatorname{lcm}(1,\ldots,k) = p_1^{\alpha_1} \cdots p_r^{\alpha_r}$. We must have $p_i^{\alpha_i} \leq k$ for each $i$. There exist integers $\lambda_1,\ldots,\lambda_r \in \mathbf{Z}$ such that $\lambda_i \equiv 1 \pmod{p_i^{\alpha_i}}$ and $\lambda_i \equiv 0 \pmod{p_j^{\alpha_j}}$ for $j \neq i$. Then we put $f(n) = \sum_{i=1}^r \lambda_i f(n-p_i^{\alpha_i})+M \cdot a_n$. We check that $f(n) \equiv f(n-j) \pmod{j}$ for each $j$, and $f(n)$ is clearly linear with respect to the sequence $(a_n)$.

This defines our linear map $\phi : \mathbf{Z}^{\mathbf{Z}} \to R$. Now, it is not hard to see that $\phi$ is bijective by working backwards, starting from $f \in R$ and defining $a_n$ inductively, starting from $a_0=f(0)$, $a_1=f(1)$ and so on. We have to check the following : if $f \in R$ then using the notations above, we have $f(n) \equiv \sum_{i=1}^r \lambda_i f(n-p_i^{\alpha_i}) \pmod{M}$. Indeed, this holds modulo each $p_i^{\alpha_i}$ using the property of the $\lambda_i$'s and the assumption on $f$.

To sum up, the general picture is that (after fixing the well-ordering of $\mathbf{Z}$ explained above), the $\mathbf{Z}$-module $R$ is the homomorphic image of $\mathbf{Z}^{\mathbf{N}}$ under some infinite lower-triangular matrix

$$ \begin{equation*} \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ 0 & 1 & 0 & 0 & \\ 0 & 1 & 2 & 0 & \\ 3 & -2 & 2 & 6 & \\ \vdots & & & & \ddots \end{pmatrix} \end{equation*}. $$

Following Todd Trimble's comment and Stefan Kohl's remark, we can show that $R$ is isomorphic to $\mathbf{Z}^{\mathbf{Z}}$ as a $\mathbf{Z}$-module. Since the product of countably many copies of $\mathbf{Z}$ is not free as a $\mathbf{Z}$-module — see this MO question — we deduce that $R$ is not free over $\mathbf{Z}$, which implies in particular that $R$ is not free over $\mathbf{Z}[x]$.

Let us endow $\mathbf{Z}$ with the following well-ordering : $0, 1, -1, 2, -2, 3, -3\dots$ The map $\phi : \mathbf{Z}^{\mathbf{Z}} \to R$ will have the form $(a_n)_{n \in \mathbf{Z}} \mapsto f$ where $f(n)$ is defined as a finite linear combination of the $a_m$ where $m$ runs trough the indices which are at most $n$ with respect to this well-ordering.

In order to give the idea of the construction, assume we work with functions defined on $\mathbf{N}$ instead of $\mathbf{Z}$, and give $\mathbf{N}$ the usual ordering. Define a map $(a_n)_{n \in \mathbf{N}} \mapsto f$ by putting $f(0)=a_0$, $f(1)=a_1$, $f(2)=a_0+2a_2$, $f(3)=3a_1-2a_0+6a_3$… The coefficients are chosen using the Bézout identity in such a way that the needed congruences hold, e.g. $f(3) \equiv f(0) \pmod{3}$.

In more detail, assume $f(n-k),\ldots,f(n-1)$ have been already defined, and let us define $f(n)$. (We proceed in an analogous way when defining $f(n)$ assuming $f(n+1),\ldots,f(n+k)$ are already defined.) Put $M=\operatorname{lcm}(1,\ldots,k) = p_1^{\alpha_1} \cdots p_r^{\alpha_r}$. We must have $p_i^{\alpha_i} \leq k$ for each $i$. There exist integers $\lambda_1,\ldots,\lambda_r \in \mathbf{Z}$ such that $\lambda_i \equiv 1 \pmod{p_i^{\alpha_i}}$ and $\lambda_i \equiv 0 \pmod{p_j^{\alpha_j}}$ for $j \neq i$. Then we put $f(n) = \sum_{i=1}^r \lambda_i f(n-p_i^{\alpha_i})+M \cdot a_n$. We check that $f(n) \equiv f(n-j) \pmod{j}$ for each $j$, and $f(n)$ is clearly linear with respect to the sequence $(a_n)$.

This defines our linear map $\phi : \mathbf{Z}^{\mathbf{Z}} \to R$. Now, it is not hard to see that $\phi$ is bijective by working backwards, starting from $f \in R$ and defining $a_n$ inductively, starting from $a_0=f(0)$, $a_1=f(1)$ and so on. We have to check the following : if $f \in R$ then using the notations above, we have $f(n) \equiv \sum_{i=1}^r \lambda_i f(n-p_i^{\alpha_i}) \pmod{M}$. Indeed, this holds modulo each $p_i^{\alpha_i}$ using the property of the $\lambda_i$'s and the assumption on $f$.

To sum up, the general picture is that (after fixing the well-ordering of $\mathbf{Z}$ explained above), the $\mathbf{Z}$-module $R$ is the homomorphic image of $\mathbf{Z}^{\mathbf{N}}$ under some infinite lower-triangular matrix

\begin{equation*} \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ 0 & 1 & 0 & 0 & \\ 0 & 1 & 2 & 0 & \\ 3 & 0 & -2 & 6 & \\ \vdots & & & & \ddots \end{pmatrix} \end{equation*}

Following Todd Trimble's comment and Stefan Kohl's remark, we can show that $R$ is isomorphic to $\mathbf{Z}^{\mathbf{Z}}$ as a $\mathbf{Z}$-module. Since the product of countably many copies of $\mathbf{Z}$ is not free as a $\mathbf{Z}$-module — see this MO question — we deduce that $R$ is not free over $\mathbf{Z}$, which implies in particular that $R$ is not free over $\mathbf{Z}[x]$.

Let us endow $\mathbf{Z}$ with the following well-ordering : $0, 1, -1, 2, -2, 3, -3\dots$ The map $\phi : \mathbf{Z}^{\mathbf{Z}} \to R$ will have the form $(a_n)_{n \in \mathbf{Z}} \mapsto f$ where $f(n)$ is defined as a finite linear combination of the $a_m$ where $m$ runs trough the indices which are at most $n$ with respect to this well-ordering.

In order to give the idea of the construction, assume we work with functions defined on $\mathbf{N}$ instead of $\mathbf{Z}$, and give $\mathbf{N}$ the usual ordering. Define a map $(a_n)_{n \in \mathbf{N}} \mapsto f$ by putting $f(0)=a_0$, $f(1)=a_1$, $f(2)=a_0+2a_2$, $f(3)=3a_1-2a_0+6a_3$… The coefficients are chosen using the Bézout identity in such a way that the needed congruences hold, e.g. $f(3) \equiv f(0) \pmod{3}$.

In more detail, assume $f(n-k),\ldots,f(n-1)$ have been already defined, and let us define $f(n)$. (We proceed in an analogous way when defining $f(n)$ assuming $f(n+1),\ldots,f(n+k)$ are already defined.) Put $M=\operatorname{lcm}(1,\ldots,k) = p_1^{\alpha_1} \cdots p_r^{\alpha_r}$. We must have $p_i^{\alpha_i} \leq k$ for each $i$. There exist integers $\lambda_1,\ldots,\lambda_r \in \mathbf{Z}$ such that $\lambda_i \equiv 1 \pmod{p_i^{\alpha_i}}$ and $\lambda_i \equiv 0 \pmod{p_j^{\alpha_j}}$ for $j \neq i$. Then we put $f(n) = \sum_{i=1}^r \lambda_i f(n-p_i^{\alpha_i})+M \cdot a_n$. We check that $f(n) \equiv f(n-j) \pmod{j}$ for each $j$, and $f(n)$ is clearly linear with respect to the sequence $(a_n)$.

This defines our linear map $\phi : \mathbf{Z}^{\mathbf{Z}} \to R$. Now, it is not hard to see that $\phi$ is bijective by working backwards, starting from $f \in R$ and defining $a_n$ inductively, starting from $a_0=f(0)$, $a_1=f(1)$ and so on. We have to check the following : if $f \in R$ then using the notations above, we have $f(n) \equiv \sum_{i=1}^r \lambda_i f(n-p_i^{\alpha_i}) \pmod{M}$. Indeed, this holds modulo each $p_i^{\alpha_i}$ using the property of the $\lambda_i$'s and the assumption on $f$.

To sum up, the general picture is that (after fixing the well-ordering of $\mathbf{Z}$ explained above), the $\mathbf{Z}$-module $R$ is the homomorphic image of $\mathbf{Z}^{\mathbf{N}}$ under some infinite lower-triangular matrix

\begin{equation*} \begin{pmatrix} 1 & 0 & 0 & 0 & \cdots \\ 0 & 1 & 0 & 0 & \\ 0 & 1 & 2 & 0 & \\ 3 & -2 & 2 & 6 & \\ \vdots & & & & \ddots \end{pmatrix} \end{equation*}