Showing that an element is a unit in a ring

  • #1
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Homework Statement


Let ##S=\{a+bi+cj+dk \mid a,b,c,d \in \mathbb{Z}\}## be the ring of integral Hamiltonian quaternions, where multiplication is defined using the same rules as in ##\mathbb{H}##, the ring of real Hamiltonian quaternions. Define a function $$N:S\to\mathbb{Z}, N(a+bi+cj+dk)=a^2+b^2+c^2+d^2.$$

Show that if ##N(\alpha)=1##, then ##\alpha## is a unit

Homework Equations




The Attempt at a Solution


So suppose we know that ##N(\alpha) = \alpha\bar{\alpha}##, where the bar is the conjugate of the quaternion. If ##N(\alpha)=1##, then ##\alpha\bar{\alpha} = 1##. So to show that ##\alpha## is a unit, all I need to show is the other direction, that ##\bar{\alpha}\alpha=1##. How can I do this? In groups one implies the other since we have inverses, but this is not the case with rings.
 

Answers and Replies

  • #2
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If you already know the inverse, show that it is in ##S## and that it actually is the inverse.
 
  • #3
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If you already know the inverse, show that it is in ##S## and that it actually is the inverse.
Well all I know is that ##\alpha\bar{\alpha} = 1##. Clearly ##\bar{\alpha}\in S##, so all I need to show that ##\bar{\alpha}\alpha = 1##. But I don't see how to get that
 
  • #4
Dick
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Well all I know is that ##\alpha\bar{\alpha} = 1##. Clearly ##\bar{\alpha}\in S##, so all I need to show that ##\bar{\alpha}\alpha = 1##. But I don't see how to get that
If ##\alpha=a+bi+cj+dk## and you define ##\bar{\alpha}=a-bi-cj-dk## so that ##\alpha \bar \alpha=a^2+b^2+c^2+d^2##, doesn't that make ##\bar \alpha \alpha=a^2+(-b)^2+(-c)^2+(-d)^2##?
 
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