Let R be a commutative ring with characteristic k. Then kr=0 for all \(\displaystyle{r}\in{R}\). Now, let f(x) in R[x]ZSK.

Let \(\displaystyle{f{{\left({x}\right)}}}={a}_{{n}}{x}^{{n}}+{a}_{{{n}-{1}}}{x}^{{{n}-{1}}}+\ldots+{a}_{{1}}{x}+{a}_{{0}}\) for some a_i in RZSK

Conider,

\(\displaystyle{k}{f{{\left({x}\right)}}}={\left({k}{a}_{{n}}\right)}{x}^{{n}}+{\left({k}{a}_{{{n}-{1}}}{x}^{{{n}-{1}}}+\ldots+{\left({k}{a}_{{1}}\right)}{x}+{k}{a}_{{0}}\right.}\)

=0+0+...+0

=0

Hence, charecteristic of R[x] is at most k. Since, for all \(\displaystyle{r}\in{R},{r}\in{R}{\left[{x}\right]}\), the characteristic of R[x] must be at least k. Hence, the characteristic of R[x] is exactly k.

Let \(\displaystyle{f{{\left({x}\right)}}}={a}_{{n}}{x}^{{n}}+{a}_{{{n}-{1}}}{x}^{{{n}-{1}}}+\ldots+{a}_{{1}}{x}+{a}_{{0}}\) for some a_i in RZSK

Conider,

\(\displaystyle{k}{f{{\left({x}\right)}}}={\left({k}{a}_{{n}}\right)}{x}^{{n}}+{\left({k}{a}_{{{n}-{1}}}{x}^{{{n}-{1}}}+\ldots+{\left({k}{a}_{{1}}\right)}{x}+{k}{a}_{{0}}\right.}\)

=0+0+...+0

=0

Hence, charecteristic of R[x] is at most k. Since, for all \(\displaystyle{r}\in{R},{r}\in{R}{\left[{x}\right]}\), the characteristic of R[x] must be at least k. Hence, the characteristic of R[x] is exactly k.