Let R be a commutative ring Show that Rx has subring isomorp

Let R be a commutative ring. Show that R[x] has subring isomorphic to R.

Solution

First suppose that x, y L(a).
Then xa = 0 and ya = 0 so that
(x + y)a = xa + ya = 0 + 0 = 0
which shows that L(a) is closed under addition.
Notice that 0a = 0
so that 0 L(a) and finally note that if x L(a),
then 0 = 0 = (xa) = (x)a which proves
that x L(a) as well. We have shown that L(a) is a subgroup under addition.
Now we show it is closed under multiplication from elements of R.
Indeed, to show this, suppose
that r R and x L(a). Then (rx)a = r(xa) = r0 = 0
which shows that rx = xr L(a) and so
we have shown that L(a) is an ideal as desired.

Answer