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Let me give the description first

If F is a field, Let $H(F)$ be the ring of quaternions over F, that is, the set of all $a_0+a_1i+a_2j+a_3k$ , where $a_0, a_1, a_2, a_3 \in F$ and where equality, addition, and multiplication are defined as for the real quaternions.

The exercise is

If $F=\Bbb C$, the complex numbers, show that $H(\Bbb C)$ is not a division ring.

however, I have proved that the quaternion is a noncommutative division ring. Since the complex number is also a field, it should be also true for complex number. Do I misunderstand anything?

added: ($a_0+a_1i+a_2j+a_3k)(\frac{a_0}{b}-\frac{a_1}{b}i-\frac{a_2}{b}j-\frac{a_3}{b}k)=1, b=a_0^2+a_1^2+a_2^2+a_3^2$

Reinherd
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    Dividing by $b$ becomes difficult if $a_0 = 1, a_1 = i, a_2 = a_3 = 0$ (remember to keep the $i$ from the coefficient field $\Bbb C$ and from the quaternion field apart, preferably by using different letters, which I have not done here). – Arthur Jan 09 '17 at 14:30
  • Related – Viktor Vaughn Jan 09 '17 at 15:59

1 Answers1

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To avoid confusion, let's use the notation $$H(\mathbb C)=\text{span}_{\mathbb C}\{1,e_1,e_2,e_3\}, $$ where $1,e_1,e_2,e_3$ are the generators for the quaternions (i.e. to translate from your language, $e_1=i,\ e_2=j,\ e_3=k$).

To show that $H(\mathbb C)$ is not a division ring, we just need to find a zero divisor. To that end, we have $$(1+ie_1)(1-ie_1)=1+e_1^2=0.$$

Aweygan
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