Let $R$ be a commutative ring and $I$ be an ideal of $R.$ Noting that $I^j I^k \subset I^{j+k},$ define a ring structure on the direct sum $$\operatorname{Rees}_{R}(I) :=\bigoplus_{j \geq 0} I^j = R \oplus I \oplus I^2 \oplus \cdots .$$ The homomorphism sending $R$ identically to the first term in this direct sum makes $\operatorname{Rees}_R(I)$ into and $R$-algebra, called the Rees algebra of $I$. Prove that if $a \in R$ is a nonzero divisor, then the Rees algebra of $(a)$ is isomorphic to the polynomial ring $R[x]$ (as an $R$-algebra).
I have no idea on how to solve this. Any help is much appreciated.
In fact, $A := \operatorname{Rees}_R((a))$ is isomorphic to $R[x]$ as a graded $R$-algebra, if you're familiar with that notion. If not, then don't worry about that.
Hint: Observe that $(0, a, 0, 0, 0, \ldots) \in A$ generates $A$ as an $R$-algebra. This gives you an algebra epimorphism $\phi: R[x] \to A$, given by $\phi(x) = (0, a, 0, 0, \ldots)$. Show that this is an isomorphism. If $f \in \ker(\phi)$, what can you say about $f$?