# How to make $R:=\{ \sum_{i=0}^{\infty} c_i A^i$ with only finitely many $c_i \neq 0 \}$ a commutative ring with one?

Let $A$ be an $n$ by $n$ matrix over a field $K$, define $R:=\{ \sum_{i=0}^{\infty} c_i A^i$ with only finitely many $c_i \neq 0 \}$

Could anyone help me show you can turn $R$ into a commutative ring with $1$?

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@Benjamin Lim: I assume $A$ is a set matrix? – Freeman May 8 '12 at 14:34
@Benjamin Lim: Yes, unless you can see any ambiguity in the wording I used? – Freeman May 8 '12 at 14:36
@Benjamin Lim: Indeed, I've been trying to use that fact to find an identity/zero (you can take the down vote off I'd you like!) – Freeman May 8 '12 at 14:39
@Benjamin Lim: are you sure $A \in R$ there are only finitely many non-zero coefficients – Freeman May 8 '12 at 14:45
Oh dammit.. Was reading it wrong.. I'm dsylexic.. – Freeman May 8 '12 at 14:56

Let $A$ be some $n\times n$ matrix over a field $K$ and consider $R := \{ \sum_{i=0}^{\infty} c_i A_i\}$ where all but finitely many $c_i$ non-zero. Now the identity matrix is in here because it can be written as $1 \times A^0$. Therefore it follows that the identity matrix is the multiplicative identity of the ring $R$.
It is easy to see from here that $R$ can be made into a unital commmutative ring.