Polynomial Ring of Sequences is Ring

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Theorem

Let $R$ be a ring.

Let $P \left[{R}\right]$ be the polynomial ring over sequences in $R$.


Then $P \left[{R}\right]$ is itself a ring.


Proof

We have by definition of polynomial ring over sequences in $R$ that:

$P \left[{R}\right] = \left\{{\left \langle {r_0, r_1, r_2, \ldots}\right \rangle}\right\}$

where each $r_i \in R$, and all but a finite number of terms is zero.


Proof that Operations are Closed

We need to ensure that the operations as defined are closed.

Let $r = \left \langle {r_0, r_1, r_2, \ldots}\right \rangle, s = \left \langle {s_0, s_1, s_2, \ldots}\right \rangle \in P \left[{R}\right]$.

As all but a finite number of terms of $r$ and $s$ are zero, there exist $m, n \ge 0$ such that:

$\forall i > m: r_i = 0$
$\forall j > n: s_j = 0$

Let $l = \max \left\{{m, n}\right\}$.


We can express the operations on $P \left[{R}\right]$ as:

\((1):\quad\) \(\displaystyle \left \langle {r_0, r_1, \ldots, r_m}\right \rangle + \left \langle {s_0, s_1, \ldots, s_n}\right \rangle\) \(=\) \(\displaystyle \left \langle {r_0 + s_0, r_1 + s_1, \ldots, r_l + s_l}\right \rangle\)
\((2):\quad\) \(\displaystyle -\left \langle {r_0, r_1, \ldots, r_m}\right \rangle\) \(=\) \(\displaystyle \left \langle {-r_0, -r_1, \ldots, -r_m}\right \rangle\)
\((3):\quad\) \(\displaystyle \left \langle {r_0, r_1, \ldots, r_m}\right \rangle \left \langle {s_0, s_1, \ldots, s_n}\right \rangle\) \(=\) \(\displaystyle \left \langle {t_0, t_1, \ldots, t_{m+n} }\right \rangle\) where $\displaystyle t_k = \sum_{i \mathop + j \mathop = k} r_i s_j$


We have that:

$\forall i > l: r_i + s_i = 0$

and so

$r + s \in P \left[{R}\right]$

Equally clearly:

$-r \in P \left[{R}\right]$

Now consider:

$\displaystyle \left({r s}\right)_i = \sum_{j \mathop + k \mathop = i}r_j s_k$

Let $i > m + n$.

Then in any $r_j s_k$ such that $j + k = i$, either $j > m$ or k > n.

In the first case $r_j = 0$ and in the second $s_j = 0$.

In either case $r_j s_k = 0$.

So:

$\displaystyle \forall i > m + n: \left({r s}\right)_i = \sum_{j \mathop + k \mathop = i}r_j s_k = 0$

So:

$r s \in P \left[{R}\right]$

$\Box$


Proof of Additive Group

The addition operation $r + s$ is clearly commutative and associative, and:

$\left \langle {0, 0, \ldots}\right \rangle = 0_{P \left[{R}\right]}$

Equally clearly:

$\forall r \in P \left[{R}\right]: r + \left({-r}\right) = \left \langle {r_i + \left({-r_i}\right)}\right \rangle = \left \langle {0, 0, \ldots}\right \rangle = 0_{P \left[{R}\right]}$

and so $-r$ is the inverse of $r$ for addition.

So $\left({P \left[{R}\right], +}\right)$ is an abelian group, as it needs to be for $P \left[{R}\right]$ to be a ring.

$\Box$


Proof of Ring Product

We need to establish that the ring product is associative.

Let $r = \left \langle {r_0, r_1, \ldots}\right \rangle, s = \left \langle {s_0, s_1, \ldots}\right \rangle, t = \left \langle {t_0, t_1, \ldots}\right \rangle \in P \left[{R}\right]$.

Then:

\(\displaystyle \left({\left({r s}\right) t}\right)_n\) \(=\) \(\displaystyle \sum_{i \mathop + j \mathop = n} \left({r s}\right)_i t_j\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop + j \mathop = n} \left({\sum_{k \mathop + l \mathop = i} r_k s_l}\right) t_j\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{k \mathop + l \mathop + j \mathop = n} r_k s_l t_j\)

Similarly for $\left({r \left({s t}\right)}\right)_n$.

So the ring product is associative, and so forms a semigroup.

$\Box$


Proof of Distributivity

Finally we need to show that the ring product is distributive over ring addition.


\(\displaystyle \left({\left({r + s}\right) t}\right)_n\) \(=\) \(\displaystyle \sum_{i \mathop + j \mathop = n} \left({r_i + s_i} \right) t_j\)
\(\displaystyle \) \(=\) \(\displaystyle \sum_{i \mathop + j \mathop = n} r_i t_j + \sum_{i \mathop + j \mathop = n} s_i t_j\)
\(\displaystyle \) \(=\) \(\displaystyle \left({r t}\right)_n + \left({s t}\right)_n\)

Similarly for $\left({t \left({r + s}\right)}\right)_n$.

Hence the result.

$\blacksquare$


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