$I$ is an ideal in $R$ implies that $I[x]$ is an ideal in $R[x]$.

Question

Is the following statement right?

If $I$ is an ideal in the ring $R$, then $I[x]$ is an ideal in the polynomial ring $R[x]$.

If so, how can I prove it?

Answer 1

Suppose $I$ is an ideal in $R$. Clearly $I[x]$ is a subring of $R[x]$ since $I$ must be a subring of $R$. By definition, $\forall r\in R\;\forall x\in I\;rx,xr\in I$. Then since the coefficients of the product of two polynomials result from products of the coefficients, it is clear that $\forall p\in R[x]\;\forall q\in I[x]\;pq,qp\in I[x]$.