TITLE: Homomorphism of a set to its power set.
QUESTION [5 upvotes]: Let $(S, +, \cdot, 0)$ and $(S', \oplus, \otimes, 0')$ be two  semirings. Then $f: S\rightarrow S'$ is said to be a homomorphism if for all $a, b\in S,$ $f(a+b)=f(a)\oplus f(b)$, $f(a.b)=f(a)\otimes f(b)$ and $f(0)=0'.$

Let $\Bbb Z$ be a set of non negative integers and $P(\Bbb Z)$ be its power set. Then $(\Bbb Z, +, \cdot, 0 )$ and $ (P(\Bbb Z), \cup, \cap, \emptyset))$ are semirings, where the operations on $\Bbb Z$ are usual addition and multiplication, while the operations on $P(Z)$ are usual set union and intersection.

Now, i wish to define a map $\phi: \Bbb Z\rightarrow P(\Bbb Z)$ such that $\phi $ is a homomorphism. Is no such homomorphism possible? If possible, how should $\phi$ be defined?

Edited:
Also, see a related question https://www.google.com/url?sa=t&source=web&rct=j&url=https://mathoverflow.net/questions/342038/define-a-homomorphism-of-a-set-of-graphs-to-its-power-set&ved=2ahUKEwi2tYaYn-fkAhWO63MBHa0DDNQQFjAAegQIBhAB&usg=AOvVaw1ojiTWIdWV636xCohN1NRU

REPLY [3 votes]: Let $f: (\Bbb Z, +, \cdot, 0 ) \to(P(\Bbb Z), \cup, \cap, \emptyset))$ be a semiring morphism. Then $f(0) = \emptyset$ and for all $n \in \Bbb N$, $f(n + (-n)) = f(0) = \emptyset = f(n) \cup f(-n)$. It follows that $f(n) = f(-n) = \emptyset$. Therefore, $f(x) = \emptyset$ for all $x \in \Bbb Z$.
If you want that $f$ preserves the identity of the multiplication, you get $f(1) = \Bbb Z$, which contradicts the previous result.