$Spec(R)$ is irreducible if and only if $Spec(R[T])$ is irreducible

Question

I want to show that for an arbitrary ring $R$ the following equivalence holds:

$Spec(R)$ is irreducible if and only if $Spec(R[T])$ is irreducible.

I have tried to show this by using the characterisation: $Spec(R)$ is irreducible if and only if its nilradical is prime. However, I couldn't show the statement above.

Answer 1

Suppose $P$ is prime in $R$. I claim that $P[T]$ is prime in $R[T]$. Let $p: R[T] \to R/P[T]$ be the quotient map. This has kernel $P[T]$, and since $P$ is prime, $R/P$ is an integral domain, and therefore so is $R/P[T]$.

Can you specialize this argument to the case $P=nil(R)$?