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Let $F$ and $E$ be the fields of order $8$ and $32$ respectively. Construct a ring homomorphism $F\to E$ or prove that one cannot exist.

Any element $x$ of $F$ satisfies $x^8=x$ and any nonzero element $x\in F$ satisfies $x^7=1$. Let $f:F\to E$ be a homomorphism. Then $1=f(1)=f(x^7)=f(x)^7$ for all nonzero $x\in F$. Also $0=f(0)=f(2)=f(1+1)=f(1)+f(1)=0$ (here $2=1+1\in F$). Now on the one hand, $f(2^7)=f(2)^7=1$. On the other hand, $f(2^7)$ is $f$ applied to $2+\dots+2$ ($2^6$ terms), so $f(2^7)=f(2\cdot 2^6)=f(2)+\dots+f(2) \text{ ($2^6$ terms) }=0+\dots+0=0$. This is a contradiction.

Is this reasoning correct? Are there other ways to solve this? In general, for which finite fields there exists a homomorphism between them?

As I pointed out in the comments, the above argument is incorrect. How do I solve the problem then?

user557
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    As always, right after posting the question, I found a flaw in my argument: I use that $2^7=1$, but $x^7=1$ is valid only for $x\ne 0$ and $2=0$. – user557 Jul 16 '18 at 22:51
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    Good call on that flaw (and the exact same flaw in my answer, effectively)! – Cameron Buie Jul 17 '18 at 00:07
  • Does the definition of "ring homomorphism" you work with include that $f(1) =1$? All answers assume this and it is a natural thing to do. However, without that assumption, the zero map $f(x)=0$ for all $x$ is the only possibility. – Torsten Schoeneberg Jul 17 '18 at 17:00
  • @TorstenSchoeneberg Yes it does. – user557 Jul 17 '18 at 17:26

4 Answers4

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A ring homomorphism $F\to E$ induces a group homomorphism $F^\times \to E^\times$. These groups have orders $7$ and $31$ and so this homomorphism must be the trivial one. On the other hand, a ring homomorphism from a field is injective.

lhf
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Let $F$ be the field with only two elements. If $F_8$ and $F_{32}$ are fields of order $8$ and $32$ respectively, then they are both realizable as algebraic extensions of $F.$ Any homomorphism $\phi : F_8 \longrightarrow F_{32}$ must be injective, as all fields are simple. That is, if we assume that $F_8 \subset F_{32},$ by way of identifying $F_8$ with $\text{Img}(\phi)$ then this is to say that $F_{32}$ must be realizable as an algebraic extension of $F_8.$ So we get the following inclusions $F\subset F_8 \subset F_{32}.$ Since these are all finite extensions, and since every finite extension of a finite field is Galois, then the Galois correspondence gives us corresponding subgroups of orders $1,$ $3$ and $5$ respectively. Since any group of order $5$ has no subgroup of order $3$, then the existence of an embedding $F_8\subset F_{32}$ must be impossible.

Conclude that no such homomorphism exists.

Chickenmancer
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Here is a direct proof similar to your attempt. Let $x \notin\{0,1\}$ be an element of $\Bbb F_8$ and let $f: \Bbb F_8 \rightarrow \Bbb F_{32}$ be a (unital) ring homomorphism. If $f(x) \neq 0$, we have $f(x)^{31} = 1$; on the other hand, $f(x)^7 = f(x^7) =f(1)=1$. But since $\gcd(7,31)=1$, this means $f(x)=1$. Consequently $f(x-1)=f(x)-f(1)=0$, but since we chose $x-1 \neq 0$, this means that $f$ is the zero map (why?), contradiction. (If $f(x) =0$, it follows immediately that $f$ is the zero map.)

Torsten Schoeneberg
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  • What about the case where the suborder/subfield is the same but the characteristic different ? – user2284570 May 25 '25 at 21:52
  • @user2284570 I don't understand, can you give an example of that "case"? But whatever it is, it should be very easy to see that there are no nonzero ring homomorphisms between fields of different characteristics. – Torsten Schoeneberg May 25 '25 at 22:15
  • An element of a finite field can have an order that is smaller than the full larger field. This is how I see building homorphism between finite fields of different degree or even characteristic. – user2284570 May 26 '25 at 07:55
  • Well sure you can talk about homomorphisms between finite cyclic groups, but then the field structure is redundant, because as mentioned, none of those will ever define ring homomorphisms. So that's just a different kind of question. – Torsten Schoeneberg May 27 '25 at 03:37
  • My interest is mapping 2 points to a different finite field while preserving their unknown discrete logarithm. I m intrested in doing this even if this enlarge the field. I suppose the answer in such case is a quick no there s no cases where this is possible ? – user2284570 May 27 '25 at 11:31
  • I don't clearly understand what you are asking, and if you have a clear question, you should ask it on its own. – Torsten Schoeneberg May 27 '25 at 14:56
  • Sorry, I wanted to write finite field elements instead of points. But for asking the right question, am I right to understand that homorphisms allows to map between finite fields elements of different fields ? – user2284570 May 27 '25 at 20:46
  • Homomorphisms of what? Rings, fields, groups? What structure do you want your maps to preserve? – Torsten Schoeneberg May 27 '25 at 21:12
  • I want to preserve discrete logarithms between the 2 finite field elements without computing the discrete logarithm (something possible for mapping binary elliptic curves to hyperelliptic curve even if I know it’s barely related)… – user2284570 May 27 '25 at 21:23
  • I don't know much about discrete logarithms, but I would think they have bases, and it is not clear to me how to even make precise how a map between two finite fields would "preserve discrete logarithms" unless one of the fields is contained in the other. One could just preserve multiplicative orders of elements, but as said many comments ago, then this is just a question about relating finite cyclic groups. In any case, I will not discuss this here further, as it seems only superficially related to this question and answer. If you have a question to ask on this site, ask it as a question. – Torsten Schoeneberg May 27 '25 at 23:24
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Since such a homomorphism $\phi$ would be injective (the kernel is an ideal, and we are assuming $\phi$ nontrivial), it is in fact an embedding, and $\phi (\text{GF}(2^3))\cong \text{GF}(2^3)$ would be a subfield of $\text{GF}(2^5)$; but since $3\not\mid5$, this is impossible .