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Let F be a finite field of characteristic $p \in \{2, 3, 5\}$. Consider the quaternionic ring, $Q_F = \{a_1 + a_ii + a_j j + a_kk|a_1, a_i, a_j, a_k \in F\}$. Prove that $Q_F$ is not a division ring.

I am not sure what I need to show that $Q_F$ is not a division ring. All I know so far is: division ring is a multiplicative group and $Q_F$ has a multiplicative properties. So I think that I need to show $Q_F$ is not a multiplicative group.

Attempt: Let $\alpha=1+i,\beta=1+i+j\in Q_F$. Then $$\begin{align*} \alpha\beta&=(1+i)(1+i+j)\\ &=(1-1)+(1+1)i+(1+1)j+(1-1)k\\ &=2i+2j \end{align*}$$

With characteristic $p=2$, $\alpha\beta=0$.

With characteristic $p=3$, $\alpha\beta=2(i+j)$.

With characteristic $p=5$, $\alpha\beta=3(i+j)$.

From my argument, I don't see anything that can tell me $Q_F$ is not a multiplicative group.

Can anyone give me a hit to do this question? Thanks!

Update: As I keep working with the method I have the following:

$$\begin{align*} \alpha\gamma&=(1+i)(i+2j)\\ &=(-1)+(1)i+(2)j+(2)k\\ &=-1+i+2j+2k \end{align*}$$

$p=2$, $\alpha\gamma=1+i$. $p=3$, $\alpha\gamma=2+i+2j+2k$. $p=5$, $\alpha\gamma=4+i+2j+2k$.

$$\begin{align*} \beta\gamma&=(1+i+j)(i+2j)\\ &=(-1-2)+(1)i+(2)j+(2-1)k\\ &=-3+i+2j+k \end{align*}$$

$p=2$, $\beta\gamma=1+i+k$. $p=3$, $\beta\gamma=i+2j+k$. $p=5$, $\beta\gamma=2+i+2j+k$.

I don't get any zero divisors, I may make some error somewhere because I should get zero divisors when $p=3,5$ also.

Simple
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  • In characteristic $2$ you have shown that the ring has zero divisors (elements that multiply to 0). A zero divisor has no multiplicative inverse. Thus is can't be a division ring, at least in characteristic $2$. – Matt Samuel Jun 04 '15 at 22:12
  • Well, Wedderburn's theorem says that a finite division ring is a field, that is, multiplication is commutative. But it's like using a gun for killing a mosquito. – egreg Jun 04 '15 at 22:14

1 Answers1

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For characteristic $2$ your argument is good.

Consider $a_0+a_1i+a_2j+a_3k$; then $$ (a_0+a_1i+a_2j+a_3k)(a_0-a_1i-a_2j-a_3k)=a_0^2+a_1^2+a_2^2+a_3^2 $$ In characteristic $3$: $(1+i+j)(1-i-j)=\dots$

In characteristic $5$: $(1+i+2j+2k)(1-i-2j-2k)=\dots$

More generally (but it's a sledgehammer), Wedderburn's theorem says that a finite division ring is a field.

egreg
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  • I guss that $Q_F $ won't form a division ring when we have prime characteristics because the problem gives $p=2,3,5$ . Right? – Simple Jun 04 '15 at 22:26
  • @Simple The characteristic of a field is always prime. Actually, any prime can be written as the sum of (up to) four squares, so the same method applies for any prime. However, one proof of the theorem that every prime can be written as the sum of four squares relies on the fact that $Q_F$ is not a division ring, so this might be circular. You can find this proof in Herstein's “Algebra”. – egreg Jun 04 '15 at 22:31
  • Ah, but such a lovely sledgehammer! – Robert Lewis Jun 04 '15 at 22:58
  • @egreg I don't understand why when you multiply $(a_1+a_2 i)(a_1-a_2i)=a_1^2+a_2^2$ , $i $ become 1 – Simple Jun 05 '15 at 01:05
  • @Simple It's the same as for complex numbers: $-i\cdot i=-(i^2)=-(-1)=1$. – egreg Jun 05 '15 at 08:11