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We first establish the nilradical as the intersection of prime ideals:Every nilpotent element belongs to every prime ideal. In the other direction, if $a$ is not nilpotent, the multiplicative subset $T=\{1,a,a^2,\dots \}$ avoids zero. By the Zorn lemma, there is an ideal $J$ maximal among those disjoint from $T$. It is a prime ideal: if $uv\in J$ and neither $u$ nor $v$ belongs to $J$, then $J+(u)$ and $J+(v)$ each meet $T$. Multiplying elements in these two intersections gives an element of $T$ lying in $J$, a contradiction. Since $a\notin J$, this proves the identity. The identity also holds for the zero ring, with the intersection over an empty family interpreted as the whole ring.

A Zariski-closed set $V_R(I)$ contains the image of $f$ exactly whenThe last equality follows because $\phi (r)$ is a nilpotent element exactly when $r^n\in \ker \phi$ for some $n\ge 1$. Consequently the smallest Zariski-closed set containing the image, namely its closure, isThis is all of $\mathrm{Spec}R$ precisely when the kernel of a ring homomorphism is contained in every prime ideal of $R$. Hence

The equivalence relation defining the localization at a prime ideal givesThereforeEquivalently, all zero divisors of $R$ must lie in $P$. To prove necessity, if $s\notin P$ and $sr=0$ with $r\ne 0$, then $r/1=0$ in $R_P$, contradicting the hypothesis that the map is injective. Conversely, if every denominator annihilates no nonzero element, the displayed kernel of a ring homomorphism is zero, so the map is injective.

An integral domain automatically satisfies this criterion at every prime ideal, but the criterion itself does not require $R$ to be an integral domain.

Yes. Taking the constant coefficient defines a surjective ring homomorphismIts kernel of a ring homomorphism is $(x)$, so the first isomorphism theorem for rings gives $R[[x]]/(x)\cong R$.

We prove the needed preservation result. If $A$ is a Noetherian ring and $J\subseteq A$ is an ideal, any ascending chain of ideals in $A/J$ lifts under the quotient ring homomorphism to an ascending chain of ideals in $A$. The lifted chain stabilizes; surjectivity makes the original chain stabilize as well. Hence every quotient ring of a Noetherian ring is Noetherian.

Applying this with $A=R[[x]]$ and $J=(x)$ gives