Let R=F[x]/<x 2 -1>={a+bta,b in F; t 2 =1}. Show that a+bt is a unit in R if and only if a 2 b 2 . Solution If ab+bt is a unit there exists c,d in F such that (a+bt)(c+dt)=1 ac+bd=1 ad+bc=0 If we multiply the first equation by b we get abc+b^2d=b But bc=-ad from the second equation and we replace it in the term abc We get -a^2d+b^2d=b d(b^2-a^2)=b If a^2=b^2, then we obtain d*0=b, so b=0, so a^2=0, so a=0 because it is in a field Conversely, suppose a^2b^2 Then we choose d=b*(b^2-a^2) -1 and c=-ad*b -1 =-a*(b^2-a^2) -1 Then we have ac+bd=-a^2*(b^2-a^2) -1 +b^2*(b^2-a^2) -1 =1 and ad+bc=0 which shows (a+bt)(c+dt)=1 so a+bt is a unit .

1. Let R=F[x]/<x 2
-1>={a+bta,b in F; t 2
=1}.
Show that a+bt is a unit in R if and only if a 2
b 2
.
Solution
If ab+bt is a unit there exists c,d in F such that
(a+bt)(c+dt)=1
ac+bd=1
ad+bc=0
If we multiply the first equation by b we get
abc+b^2d=b
But bc=-ad from the second equation and we replace it in the term abc
We get
-a^2d+b^2d=b
d(b^2-a^2)=b
If a^2=b^2, then we obtain d*0=b, so b=0, so a^2=0, so a=0 because it is in a field
Conversely, suppose a^2b^2
Then we choose d=b*(b^2-a^2) -1
and c=-ad*b -1
=-a*(b^2-a^2) -1
Then we have ac+bd=-a^2*(b^2-a^2) -1
+b^2*(b^2-a^2) -1
=1
and ad+bc=0
which shows (a+bt)(c+dt)=1
so a+bt is a unit