Question

describe all units in the ring
1) Z
2)Q
3) Z x Q x Z
Please, help. I don' t get

Answer 1

@oldrin.bataku

Answer 2

units are elements with multiplicative inverses... what integers do you know \(a,a^{-1}\in\mathbb{Z}\) solve \(aa^{-1}=1\)?

Answer 3

well we know \(a\ne 0,a^{-1}\ne0\) off the bat. we also know if \(a>1\) then necessarily \(aa^{-1}>1\) (and vice versa for \(a^{-1}>1\)). a similar problem arises with \(a<-1\) and vice versa

Answer 4

the only two elements we have to look at are \(\pm1\) -- and, well, they both have integer multiplicative inverses! \(a=a^{-1}=\pm1\) yields \(aa^{-1}=1\) so \(\pm1\) are the only units in \(\mathbb{Z}\)

Answer 5

must be INTEGER multiplicative inverse?

Answer 6

alternatively, recall that the integers are extended to form rationals. we know that \(n\in\mathbb{Z}\) has multiplicative inverse \(1/n\in\mathbb{Q}\); the question is whether \(1/n\in\mathbb{Z}\) and we know this is clearly only true for \(n=\pm1\)

Answer 7

yes, when we ask about multiplicative inverses in \(\mathbb{Z}\) we mean that the inverse must also be an integer!

Answer 8

since a^(-1) = 1/a , is it not that with any of a , we still have a a^(-1) =1?

Answer 9

@Loser66 \(1/a\) is an integer for \(a=\pm1\); note \(1/1=1\) and \(1/(-1)=-1\)

Answer 10

oh, you mean a^-1 is integer, too. right?

Answer 11

yes of course

Answer 12

how about in Q?

Answer 13

now in \(\mathbb{Q}\) our situation is more obvious; we know every rational number \(a/b\in\mathbb{Q}\) where \(a,b\in\mathbb{Z}\) permits a multiplicative inverse \(b/a\in\mathbb{Q}\). this only has problems when \(a=0\) and incidentally if \(a=0\) then \(a/b=0\) so the only time \(a/b\in\mathbb{Q}\) ISN'T a unit is when \(a/b=0\) so every non-zero element of \(\mathbb{Q}\) is a unit

Answer 14

Got you,
the last one, please

Answer 15

question: I must follow the order of the expression, right? I mean Z x Q x Z , so the unit (if there is) must has the form of (aa^-, fraction*fraction^-, bb^-) right?

Answer 16

since the unit of Z x Z are:
(1,1)
(1,-1)
(-1,1)
and (-1,-1)
can we put the unit of Q in the middle of each of pairs?

Answer 17

yes I believe so, but note that \(\mathbb{Z},\mathbb{Q}\) are commutative rings so \(aa^{-1}=aa^{-1}\)

Answer 18

@ the first question

Answer 19

and yes, that second question sounds precisely correct! so the units of \(\mathbb{Z}\times\mathbb{Q}\times\mathbb{Z}\) are precisely \((\pm1,q,\pm1)\) where \(q\in\mathbb{Q}\setminus\{0\}\)

Answer 20

Thanks a ton, I got it now
One more question, please
I know if A = P^-D P
so, A^ (whatever) = P^- D^(whatever) P
is there any other way to find out \(\sqrt{A}\)?