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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

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    $\begingroup$ Do you care if $f(1)=1'$? If not you can just take $f(z) = \emptyset$ for all $z \in \mathbb{Z}$. $\endgroup$
    – kccu
    Sep 23, 2019 at 14:01
  • $\begingroup$ @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 $\endgroup$
    – gete
    Sep 23, 2019 at 14:07
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    $\begingroup$ Did you perhaps mean to write $\mathbb{N}$ instead of $\mathbb{Z}$ or is that some custom notation I am not aware of? $\endgroup$
    – ComFreek
    Sep 23, 2019 at 14:12

2 Answers 2

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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.

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  • $\begingroup$ would you please elaborate as what degree of freedom mean here? $\endgroup$
    – gete
    Sep 23, 2019 at 14:22
  • $\begingroup$ @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: $\mathcal{P}(\mathbb{N}) \cong \text{ set of homomorphisms } S \to S'$. $\endgroup$
    – ComFreek
    Sep 23, 2019 at 14:26
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – ComFreek
    Sep 24, 2019 at 16:13
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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.

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

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