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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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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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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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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 .