Let $R$ be a non-trivial commutative ring. If $R$ has only one prime ideal $\mathfrak{p}$, is $R$ Noetherian?
As pointed out in the comments, this affirmation is false. E.g., take $k[x^{1/2},x^{1/4},x^{1/8},...]/(x)$, where $k$ is a field, as a counterexample. However, it is still not clear to me why.
Can anyone please show me the (single) prime ideal in $k[x^{1/2},x^{1/4},x^{1/8},...]/(x)$ and proof that it is not noetherian?
$k[x_1, x_2, ... ] / (x_1^2, x_2^2, ... )$ where $k$ is a field is a counterexample. Any prime ideal must contain $(x_1, x_2, ...)$ so there is only one prime ideal. However, that ring is not Noetherian.