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Does a ring homomorphism from $mathbb{Z}[frac{i}{2}]$ to a finite field with characteristic $pequiv 3 bmod 4$ such that the unity is mapped onto the unity exist?

Thank for your help.

Yes, there does exist such a homomorphism. Let $p$ be any prime congruent to $3$ mod $4$, and let $n$ be any even natural number. Then

• the finite field $mathbb{F}_{p^n}$ has characteristic $p$


• there is a root $alpha$ of the polynomial $x^2+1$ in $mathbb{F}_{p^n}$


• since $p$ is odd, there exists a $frac{1}{2}inmathbb{F}_{p^n}$

Then we can form the ring homomorphism $f:mathbb{Z}[frac{i}{2}]tomathbb{F}_{p^n}$ defined by $f(1)=1$ and $f(frac{i}{2})=frac{alpha}{2}$.

However, again with $pequiv 3bmod 4$, there is no homomorphism $f:mathbb{Z}[frac{i}{2}]tomathbb{F}_{p^n}$ when $n$ is odd.

Let $K$ be a field of characteristic $p$, where $p$ is a prime natural number. Write $R$ for the ring $mathbb{Z}left[frac{text{i}}{2}right]$. Suppose $varphi:Rto K$ is a ring homomorphism sending $1 in R$ to $1 in K$. Assume that $omega$ is the image of $u:=dfrac{text{i}}{2}$. Then, we have $$4omega^2=(2omega)^2=big(2varphi(u)big)^2=big(varphi(2u)big)^2=big(varphi(text{i})big)^2=varphileft(text{i}^2right)=varphi(-1)=-1,.$$ (For this to make sense, $p neq 2$.)

If $pequiv 1pmod{4}$, then clearly $omega$ is in the prime field $mathbb{F}_p subseteq K$. Elements of $R$ take the form $dfrac{a+btext{i}}{2^n}$, where $a,binmathbb{Z}$ and $n in mathbb{N}_0$. Hence, $varphi:Rto mathbb{F}_p$ sends $dfrac{a+btext{i}}{2^n} in R$ to $dfrac{a+2bomega}{2^n} in mathbb{F}_p$. (Of course, there are two such $varphi$'s.)

If $p equiv 3pmod{4}$, then $omega$ lies within a subfield $E$ of $K$ isomorphic to $$mathbb{F}_{p^2}congmathbb{F}_{p}[x]/leftlangle x^2+1rightranglecong mathbb{F}_p[text{i}],.$$ Consequently, for $varphi$ to exist, $K$ must have such a subfield $E$ (and therefore, in the case that $K$ is finite, we must have $|K|=p^{2t}$, or $Kcong mathbb{F}_{p^{2t}}$, for some $t in mathbb{N}$). As elements of $R$ take the form $dfrac{a+btext{i}}{2^n}$, where $a,binmathbb{Z}$ and $n in mathbb{N}_0$, $varphi:Rto E$ sends $dfrac{a+btext{i}}{2^n} in R$ to $dfrac{a+2bomega}{2^n} in E$. (Again, there are two such $varphi$'s for a given $E$ or $K$.)

To answer your question, if $p equiv 3pmod{4}$, such a homomorphism exists if and only if the finite field in question is of order $p^{2t}$ for some $t in mathbb{N}$. If we omit the unitary homomorphism condition (i.e., $1in R$ does not have to be mapped to $1in K$), then we have one extra map for any prime $p$---the zero map.

EDIT: I just want to add a little note on some discussion I had with a user regarding the comments below. To be clear, the term "ring homomorphisms" when rings involved may be nonunital will make no sense if we enforce the "unitary condition" (i.e., the multiplicative identity of one ring must be mapped to the multiplicative identity of another ring). Hence, a homomorphism between two unital rings doesn't have to have this unitary property, but it will be called "unitary" if it satisfies this unitary condition. It is not true that all homomorphisms between two unital rings must be unitary.

Let $f:Rto S$ be a ring homomorphism of rings $R$ and $S$. Assume that the two rings are unital. It doesn't matter (to him) wheter $f$ sends $1_R$ to $1_S$. In his argument, let $g$ be the restriction of $f$ to its image (i.e., $g:Rto text{im}(f)$ is identical to $f$). There is no denial (from me) that $text{im}(f)$ is a unital ring and $g$ would be a unitary ring homomorphism. My point is, even that is so, it doesn't mean $f$ is also unitary. For better clarification, I provided an example of a homomorphism from $mathbb{Z}$ to $mathbb{Z}/6mathbb{Z}$. While this homomorphism isn't unitary, its restriction to the image is.