Does there exist an integral Noetherian domain $R$ such that there is an injective unital ring homomorphism $R[[x]]\rightarrow R$?
I thought of using Krull dimension to answer this but a ring can have lower Krull dimension than its subring (e.g. $\mathbb{Z}\subset \mathbb{Q}$).
I think the answer is yes, with $R= \mathbb C$ (which is surely noetherian and integral !)
Indeed consider $K$ the field of fractions of $\mathbb C[[x]]$ and $L$ its algebraic closure.
Note that $\mathbb C [[x]]$ has cardinality $|\mathbb C|^{|\mathbb N|} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0^2}=2^{\aleph_0} = |\mathbb C|$, therefore so does $K$ and therefore so does $L$.
It follows that the transcendence degree of $L$ over $\mathbb Q$ is precisely $2^{\aleph_0}$, just like $\mathbb C$; and $L$ is algebraically closed, therefore $L\cong \mathbb C$ (it's actually quicker to see that $L$ injects into $\mathbb C$ but the stronger result is true as well).
But clearly $ \mathbb C[[x]] \to K \to L \to \mathbb C$ is then an injection.
This is the same trick as to show that $\mathbb C(X)$ injects into $\mathbb C$.
Of course, as was pointed out in the comments, for cardinality reasons this doesn't happen if $|R|^{\aleph_0} > |R|$, which happens e.g. for $|R| = \aleph_0$, i.e. $R$ countable (but also, by König's theorem, for higher cardinalities such as $\aleph_\omega$, though of course you will rarely encounter rings that provably have this cardinality)
