How many maps $\phi:\Bbb{N}\to\Bbb{N}$ are there with $\phi(ab) = \phi(a)+\phi(b)$? I am stuck on the following problem when I was trying to solve an entrance exam paper:

How many maps $\phi \colon \Bbb N \cup \{0\} \to \Bbb N \cup \{0\}$ are there with the property that $\phi(ab)=\phi(a)+\phi(b)$ for all $a,b \in \Bbb N \cup \{0\}$ ?

The options are as follows :


*

*none


*finitely many


*countably many


*uncountably many

Option 1 is not possible as if we take $a=1,b=0$ then we get $\phi(1)=0$ which is possible. But, I am not sure about the other options and could not decide which one holds good.
Can someone explain?
 A: Hint: Consider $\phi(0\cdot b)$.
A: Interestingly, there are no such maps if we remove that zero from the picture. That is, there are no maps $\phi:\mathbb{N}\rightarrow\mathbb{N}$ such that $\phi(ab)=\phi(a)+\phi(b)$ for all $a,b\in\mathbb{N}$. This is because, by taking $a=1=b$, we again see that $\phi(a)=0$ for all $a$, a contradiction as $0\not\in\mathbb{N}$.
This argument does not imply the other answers, because there are in fact infinitely many such maps $\phi:\mathbb{N}\rightarrow\mathbb{N}\cup\{0\}$ (the above just proves that the element $1$ us mapped to $0$). To see that there are infinitely many such maps, notice that you can map primes $p$ to $1$ or $0$ (and other places too!), and so long as every prime is mapped to $1$ or $0$ you have a homomorphism.
So, a brief summary:


*

*If $\phi:\mathbb{N}\cup\{0\}\rightarrow\mathbb{N}\cup\{0\}$, there is a unique such homomorphism.

*If $\phi:\mathbb{N}\rightarrow\mathbb{N}\cup\{0\}$, there is are infinitely many such homomorphism.

*If $\phi:\mathbb{N}\rightarrow\mathbb{N}$, there are no such homomorphisms.

*If $\phi:\mathbb{N}\cup\{0\}\rightarrow\mathbb{N}$, there are no such homomorphisms (this follows from 3).
A: The other answers have shown that such a map must be identically zero. If we only require the map to be a homomorphism on the positive integers, the question is slightly more interesting: the positive integers as a multiplicative monoid are generated by the primes, and FTArithmetic means that any set map from the primes to $\mathbb N_0$ extends uniquely to a homomorphism. 
A: I may be missing something- but it seems that the only map with this property is identically $0$. The proof is as follows:
$\phi(0.0)=\phi(0)+\phi(0)$ which implies that $\phi(0)=0$.
Now, for any $a \in \mathbb{N}$, $\phi(a.0)=\phi(a)+\phi(0)$ which means $\phi(0)=\phi(a)+\phi(0)$ which implies $\phi(a)=0$.
A: Certainly the map $\phi(n) = 0$ works.
Taking $a=0$ gives $\phi(0) = \phi(0) + \phi(b)$, whence $\phi(b) = 0$.  So in fact the identically zero map is the only solution.
A: Exactly one.
$a=0$ leads to: $\varphi\left(0\right)=\varphi\left(0\right)+\varphi\left(b\right)$
hence $\varphi\left(b\right)=0$. This for every $b$.
And function $x\mapsto0$ indeed suffices.
