Irreducible central elements generate maximal ideals?

  • #1
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http://img17.imageshack.us/img17/3937/vjj6.png [Broken]

I don't understand the very last bit, where it says a and b are central and so c is, hence a is not irreducible. Basically what replaces the "hence"? I can't seem to figure out the link between c being central meaning a isn't irreducible.

a being irreducible means it isn't a unit, hence D[x]a is proper but if it is not maximal it sits inside an ideal D[x]b where b is a central non-unit.

Hence a = cb for some c. Since a and b are non-units, to be irreducible c must be a unit. But then c being central contradicts it being a unit? Why? There are central units contained in H[x] for instance (the ring of quaternions), a real number is a central unit.
 
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Answers and Replies

  • #2
130
1
Hmm, left it for a few hours and came back to it. Noticed what I was doing wrong, not noticing that it was talking about irreducibility over F[x], instead of the whole ring D[x]. Therefore c cannot be a unit otherwise D[x]a does not sit properly in D[x]b, but since c is central and b is central, a = cb is a product of two non-unit elements of F[x]. Hence a is not irreducible over F[x].
 

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