Polynomial Ring is a Ring

Theorem

Let $R$ be a ring.

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

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

Proof

We have by definition of polynomial ring 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:

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

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:

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

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:

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

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:

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

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.

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

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

Hence the result.

$\blacksquare$

Sources

• B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra (1970): $\S 3.1$: Theorem $3.5$