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I saw the related posts, and I tried a different proof. Please have a look.

Let $D$ be any field with $|D|=6$.

$|D|=6<\infty \Longrightarrow Char(D)\neq 0\Longrightarrow Char(D)=prime\ number$

From the theorem of Lagrange in $(D,+)$ we have that: $$ord(1_R) \mid |D| \Longrightarrow Char(D)\mid |D|\Longrightarrow Char(D)\mid 6$$ So, $Char(D)=2$ or $3$.

  • $Char(D)=2$

$\forall x\in U(D):ord(x)=Char(D) \Longrightarrow ord(x)=2 \Longrightarrow 2x=0_D \Longrightarrow x=-x $ (*)

If $H= \{0_D, a,b,a+b\} $ from (*) $(H,+)\leq (D,+)$ and from Theorem Of Lagrange we have that $|H|\mid |D| \Longrightarrow4|6$, contradiction.

  • $Char(D)=3$

Can we find another additive subgroup of $(D,+)$ with 4 or 5 elements to work on the same say as above?

Chris
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1 Answers1

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Here is a proof using your idea of looking at the additive group of $D$.

Since the additive group of $D$ is abelian and has order $6$, it must be $C_6$.

Let $g$ be a generator for the additive group of $D$, that is, an element of additive order $6$.

Then $0 = 6 \cdot g = (2 \cdot 1_D)(3 \cdot g)$. Since $3 \cdot g \ne 0$, we must gave $2 \cdot 1_D=0$ because $D$ is a domain.

In the same manner, $0 = 6 \cdot g = (3 \cdot 1_D)(2 \cdot g)$ implies $3 \cdot 1_D=0$.

But then $1_D = 3 \cdot 1_D - 2 \cdot 1_D = 0$ and $D$ is the zero ring.

lhf
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  • Why $(D,+)$ = additive group with$ |D|=6$ implies that is $C_6$? And also we didn find an additive subgroup $H'$ of $D$ such that $|H'|\nmid |D|$ – Chris Apr 12 '16 at 17:49
  • @Chris, there are only two groups of order $6$: $S_3$ and $C_6$. – lhf Apr 12 '16 at 17:55