Non Commutative Polynomial Ring

Question

If $R$ is a commutative ring, so is $R[x].$

Is it necessary to take a commutative ring $R$ in order to construct something like $R[x]$ i.e. will $R[x]$ still be a ring if $R$ is not commutative?

Answer 1

You can define the polynomial ring $R[x]$: An element of $R[x]$ is a formal sum $\sum_{j=0}^n a_j x^j$, modulo the equivalence relation that, if $m < n$ and $a_{m+1} = a_{m+2} = \cdots = a_n$, then $\sum_{j=0}^m a_j x^j$ and $\sum_{j=0}^n a_j x^j$ are the same polynomial.

Addition is given by $\sum_{j=0}^n a_j x^j + \sum_{j=0}^n b_j x^j = \sum_{j=0}^n (a_j+b_j) x^j$, and multiplication is given by $\left( \sum_{i=0}^m a_i x^i \right) \left( \sum_{j=0}^n b_j x^j \right) = \sum_{i=0}^m \sum_{j=0}^n a_i b_j x^{i+j}$. It is straightforward to check that this defines a ring (with identity, not necessarily commutative).

The big difference with the commutative case is that evaluation of polynomials is not a map of rings. If $R$ is commutative and $\theta \in R$, then the map $f(x) \mapsto f(\theta)$ is a map of rings $R[x] \mapsto R$. (And, more generally, if $S$ is a commutative $R$-algebra and $\theta \in S$, then $f(x) \mapsto f(\theta)$ is a map of rings $R[x] \to S$.)

For non-commutative rings this isn't true! For example, let $R$ be the quaternions. We have $(x+i) (x-i) = x^2+1$, but $(j-i) (j+i) \neq j^2 +1 $. This makes polynomials a lot less useful when their coefficients are not commutative.