# Injective homomorphism

Are there any injective multiplicative homomophisms from $\mathbb{Z}$ to $\mathbb{F}[x]$ where $\mathbb{F}$ is a field of Characteristic $p$?

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There is always exactly one homomorphism $\mathbb{Z}\to R$ for any ring $R$. The characteristic of a ring can be defined via this unique homomorphism. Using these remarks it should be clear what the correct answers are. – Curufin Jan 22 '13 at 22:29
@Curufin: Why not post that as an answer? – Clive Newstead Jan 22 '13 at 22:33
@Curufin: I think you have assumed that $R$ is unital and the homomorphism is onto. With these assumptions you are right. But without them, one can construct several homomorphisms. For instance from $\mathbb{Z}$ into $\mathbb{Z}\oplus \mathbb{Z}$ there are at least three different homomorphisms. – Vahid Shirbisheh Jan 22 '13 at 22:40
@Vahid: Why do we need surjectivity as well as $1$? If somewhere in your definition of ring homomorphisms you require that they preserve multiplicative identities, then for any given (unital) ring $R$ there can be only one ring homomorphism $f: \mathbb{Z} \to R$ (since $f(n)$ is determined by $f(1)$ for each $n \in \mathbb{Z}$). – Clive Newstead Jan 22 '13 at 22:52
@CliveNewstead: You misinterpreted my statement a little. First, assuming surjectivity and $R$ being unital implies that the image of $1$ is the unit element of $R$ (in other words, the homomorphism preserves units). Secondly, you can assume the homomorphism is unital (preserves units) and drop the surjectivity assumption. – Vahid Shirbisheh Jan 22 '13 at 23:11

If you intended ring homomorphisms, then I second Curufin's comment. But I assumed that, since you wrote "multiplicative", you intend your homomorphisms to just preserve multiplication, not necessarily addition. In that case, such homomorphisms would exist as long as the characteristic of $\mathbb F$ is not 2. You can define a homomorphism by sending each prime number $q$ to an arbitrarily chosen polynomial $f_q(x)$ in $\mathbb F[x]$ and then extending the map to all integers using the prime factorization: send $\pm q_1^{k_1}\dots q_r^{k_r}$ to $\pm f_{q_1}(x)^{k_1}\dots f_{q_r}(x)^{k_r}$. To make this homomorphism injective, choose the $f_q$'s a little carefully; for example, choose distinct, irreducible polynomials for distinct primes $q$.
Ah, I see what he meant. In fancy language: We're looking for injective semi-group homomorphisms from $(\mathbb{Z},\cdot)$ to $(\mathbb{F}[x], \cdot)$. In fact it seems that every injective semi-group-homomorphism maps prime numbers to polynomials that are relativly prime to every other image of a prime number. – Curufin Jan 22 '13 at 23:00
Any multiplicative homomorphism is determined by where it sends the primes and where it sends $-1$, because every integer is a product of these. The examples in my answer (before I specialized to the injective case) are the ones that send $-1$ to $-1$. There are similar ones sending $-1$ to $1$; just delete the $\pm$ from the output (but not the input) in my examples. The only other option is to send $-1$ to $0$ but then, by multiplicativity, everything has to map to $0$ – Andreas Blass Jan 22 '13 at 23:23