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Additional information: Knowing that finite fields must have $p^k$ elements, with $p$ prime and $k\geq1,k\in \mathbb{N}$. In the case of $k=1$, the field is $\mathbb{Z}_p$.

Problem: Determine all integral domains $D$ such that $d^5=d$, for all $d \in D$.

My answer: $D$ is an integral domain, so we can use the cancellation law for ring product, and then $d^5=d \Rightarrow d^4=1$, and therefore $d\neq 0$. Factoring $d^4-1=0$ we have: \begin{align*} d^4-1 &= 0 \\ (d^2+1)(d^2-1) &= 0 \\ (d^2+1)(d+1)(d-1) &= 0 \end{align*} Then $d=1$, $d=(-1)$, or $d^2=(-1)$. We have two options:

  • $d^2\neq (-1)$: in this case we have $d=1=(-1)$ and $D=\{0,1\}\cong \mathbb{Z}_2$, or $d_1=1$, $d_2=(-1)$ and $D=\{0,1,-1\}\cong \mathbb{Z_3}$ (in both case is easy to prove that every element different from $0$ has an inverse, so $D$ is a finite field, and the isomorphism follows from the given additional information).
  • $d^2=(-1)$: We have that the equation $d^2+1=0$ has at most two solutions in $D$. Let $a$ and $b$ be the solutions. Follows that $a^2=(-1)=b^2$ and therefore: \begin{align*} a^2-b^2 &= 0 \\ (a+b)(a-b) &= 0 \Rightarrow a=b\lor a=-b \end{align*}

(Here's where my doubt begins)

If $b=1$ then $a=1$ or $a=(-1)$, in any case we fall into the integral domains already mentioned (same for $b=(-1)$), so suppose that $b\neq \pm 1$. Them $D = \{0,1,-1,b,-b\}\cong \mathbb{Z_5}$, because, suppose that $b=-b$. Then: \begin{align*} 2b=0 & \Rightarrow b+b=0 \\ & \Rightarrow b(1+1)=0b \\ & \Rightarrow (1+1)=0 \\ & \Rightarrow 1=-1 \end{align*} which is absurd. Is that right?

Paulo Sigiani
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  • $1=-1$ isn't that absurd: It's true in $\mathbb{Z}_2$. So is there an integral domain where $1=-1$ and there is also one element $b$ with $b^2+1=0$? – aschepler Sep 28 '22 at 00:18
  • From How to ask a good question: " Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title." – jjagmath Sep 28 '22 at 00:37

1 Answers1

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Yeah you stumbled onto it: there's $\Bbb Z_2$ as well.

One way to see it is that $x^5\equiv x\pmod2$ since $0,1$ both work.

So that gives us three: $\Bbb Z_5,\Bbb Z_2$ and $\Bbb Z_3$.

One way for $\Bbb Z_3$, Fermat's little theorem guarantees $x^5\equiv x\pmod 3$.

I think that's it: $x^5-x=0$ can have at most $5$ solutions in an integral domain. So there's no more options (except $\Bbb F_{2^2}$, which doesn't work. See @rschwieb's comment).

suckling pig
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    The only other field with less than 5 elements to rule out is the field of four elements, but it's easy to see it doesn't satisfy $x^5-x$ (it satisfies $x^3=1$, hence $x^6-1$, which would not go with $x^5-x$ well.) – rschwieb Sep 28 '22 at 11:21
  • @rschwieb indeed, I forgot that field, I was thinking about $\Bbb Z_4$, which isn't an integral domain. But as you say, I'm safe. – suckling pig Sep 28 '22 at 13:11