Can multiplication of two primes be seen as a strong cipher? If we were define such a cipher:


A reversible function that would accept a message $M$ and an initialization vector $\text{IV}_1$ $\operatorname{map}(\text{IV}_1, M)$ which can map an input $M$ to a random prime $p_1$
A key derivation function $\operatorname{kdf}$ with a similar blueprint, with key $k$ and initialization vector $\text{IV}_2$ where $\operatorname{kdf}(\text{IV}_2, k) = p_2$
The encryption can then be defined simply as $p_1 . p_2 = C$
Decryption would then be $C / \operatorname{kdf}(\text{IV}_2, k) = C / p_2 = p_1$ and $M = \operatorname{map}^{'}(\text{IV}_1, p_1)$


I see this would be very similar to OTP; and can it be seen secure considering the idea is almost the same (and assuming factorization is hard)

Intuitively I can speculate, it might pave some way for some commutative and homomorphic properties as well...
This answer is supplementary to the one provided by Mikero. 

Assuming that the $\operatorname{map}$ and $\operatorname{kdf}$ functions are properly randomized and provide appropriate output (e.g. large enough to be non-repeating), then no two ciphertexts will share common factors. So being able to use $\operatorname{gcd}$ to break ciphertexts is not applicable in the basic form of the proposal.


  I see this would be very similar to OTP; and can it be seen secure considering the idea is almost the same (and assuming factorization is hard)


Emphasizing the point raised by Mikero: This is not a one-time pad, as it lacks information-theoretic security. 

An efficient algorithm for factoring can recover $p_1$ from $p_1 p_2$ and compute $M$ by inverting the $\operatorname{map}$ function. 

An adversary that knows or can guess $M$ can recover $p_2$ from the ciphertext by computing $p_1 = \operatorname{kdf}(\text{IV}_1, M)$ and then $p_2 = \frac{p_1 p_2}{p_1}$. Since each $p_2$ is randomized, this should not help them to decrypt other ciphertexts that they do not already know the plaintext of. However, it would allow them to mount an attack on $\operatorname{kdf}$. 


  Intuitively I can speculate, it might pave some way for some commutative and homomorphic properties as well...


It is conceivable that the ability to meaningfully multiply ciphertexts could be worked into the scheme. However, it would require a very clever $\operatorname{map}$ function - one which would also need to preserve the multiplication operator on it's own outputs.

The ability to add ciphertexts together appears to be barred by the requirement for each "key factor" ($p_2$) to be different. Something like $p_1 p_2 + p_3 p_4$ will not let you decrypt to $p_1 + p_3$ by dividing by $p_2$ and/or $p_4$

As for commutativity, decryption could be commutative, but encryption can't successfully be performed as such. 

For decryption: $$\frac{p_1 p_2 p_3}{p_2} = p_1 p_3\\\frac{p_1 p_3}{p_3} = p_1\\\frac{p_1 p_2 p_3}{p_3} = p_1 p_2\\\frac{p_1 p_2}{p_2} = p_1$$

Clearly, the sequence of factors that you divide by has no influence on the output.

However, for encryption, let's try the Three-pass Protocol (one of the things you might want a commutative cipher for): $$c_0 = p_1 p_2\\c_1 = c_0 p_3 = p_1 p_2 p_3\\c_2 = \frac{c_1}{p_2} = \frac{p_1 p_2 p_3}{p_2} = p_1 p_3\\p^{'}_1 = \frac{c_2}{p_3} = \frac{p_1 p_3}{p_3}$$

An attacker who has seen $c_0$ and $c_1$ can compute $\operatorname{gcd}(c_0, c_1)$ to recover $p_3$. When they obtain $c_2$, they can then recover $p_1$ by dividing by $p_3$. 

This demonstrates that the proposal does not provide a secure commutative cipher.