# Homomorphism of a set to its power set.

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?

• Do you care if $f(1)=1'$? If not you can just take $f(z) = \emptyset$ for all $z \in \mathbb{Z}$.
– kccu
Sep 23, 2019 at 14:01
• @kccu Considering $f(1)=1$ or, not is a different matter but, what you are considering is trivial case which will serve no purpose. I need a non trivial one
– gete
Sep 23, 2019 at 14:07
• Did you perhaps mean to write $\mathbb{N}$ instead of $\mathbb{Z}$ or is that some custom notation I am not aware of? Sep 23, 2019 at 14:12

I'll assume you meant $$\mathbb{N}$$ everywhere. Then the set of such homomorphisms is pretty limited. For every $$n \geq 1$$ we have $$f(n) = f(1)$$:

$$f(n) = f(1 + \ldots + 1) = f(1) \cup \ldots \cup f(1) = f(1)$$

Since by definition $$f(0) = \emptyset$$ we are only left with one degree of freedom by setting $$f(1)$$ -- depending on whether you see restricting $$f(1)$$ included in your homomorphism definition.

The above equality can also be seen in a different, more general light. Every semiring homomorphism $$f$$ gives rise to a monoid homomorphism $$f_\mathrm{Mon}: (S, +) \to (S', \oplus)$$ wrt. the additive operation. This is just because a semiring "contains" a monoid. The additive monoid here is $$(\mathbb{N}, +)$$, which is generated by $$1\in\mathbb{N}$$. This in turn means that $$f_\mathrm{Mon}$$ is already fully determined by its image of $$0$$ and $$1$$. Since we have $$f = f_\mathrm{Mon}$$, we can conclude the same for $$f$$.
In fact you could have made similar considerations for the multiplicative monoid to conclude that $$f$$ is determined by its image on $$0$$, $$1$$ and primes.

• would you please elaborate as what degree of freedom mean here?
– gete
Sep 23, 2019 at 14:22
• @gete Oh, nothing too formally. I meant that the set of all possible values for $f(1)$ (which is $\mathcal{P}(\mathbb{N})$) is in bijection to all possible homomorphisms. In symbols: $\mathcal{P}(\mathbb{N}) \cong \text{ set of homomorphisms } S \to S'$. Sep 23, 2019 at 14:26
• Sep 24, 2019 at 16:13

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.

• But it seems to be a trivial case. I need a non trivial sort
– gete
Sep 23, 2019 at 14:52
• My answer proves that there is no nontrivial examples. Sep 23, 2019 at 16:10
• Noted. Thanks....
– gete
Sep 23, 2019 at 16:21
• There is a minor typos error in your text. You may get it corrected for future readers.
– gete
Sep 24, 2019 at 0:44
• I didn't see any typo, did I miss something? Sep 24, 2019 at 5:37