Ring of Laurent Polynomials is Noetherian Implies Polynomial Ring is Noetherian?

Question

Suppose $R[t, t^{-1}]$ is Noetherian. Is $R[t]$ necessarily Noetherian?

(Here, $R$ is a commutative ring with an identity.)

I am able to show that the converse is true, but am a bit stuck with this direction.

Answer 1

Rather than showing this directly, I think the easiest approach is to show that both are equivalent to $R$ being Noetherian. Indeed, there is a surjective homomorphism $R[t,t^{-1}]\to R$ which is the identity on $R$ and maps both $t$ and $t^{-1}$ to $1$. So, since a quotient of a Noetherian ring is Noetherian, if $R[t,t^{-1}]$ is Noetherian, $R$ must be Noetherian as well. Then $R[t]$ is Noetherian by the Hilbert basis theorem.