If R is a commutative ring, show that the characteristic of R[x] is the same as the characteristic of R.

If R is a commutative ring, show that the characteristic of R[x] is the same as the characteristic of R.
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smallq9

Let R be a commutative ring with characteristic k. Then kr=0 for all $r\in R$. Now, let $f\left(x\right)\in R\left[x\right]$.
Let $f\left(x\right)={a}_{n}{x}^{n}+{a}_{n-1}{x}^{n-1}+\dots +{a}_{1}x+{a}_{0}$ for some ${a}_{i}\in R$
Conider,
$kf\left(x\right)=\left(k{a}_{n}\right){x}^{n}+\left(k{a}_{n-1}{x}^{n-1}+\dots +\left(k{a}_{1}\right)x+k{a}_{0}$
=0+0+...+0
=0
Hence, charecteristic of R[x] is at most k. Since, for all $r\in R,r\in R\left[x\right]$, the characteristic of R[x] must be at least k. Hence, the characteristic of R[x] is exactly k.