Finding the Units in the Ring \mathbb{Z}[t][\sqrt{t^{2}-1}]

lunnatican4

lunnatican4

Answered question

2022-01-14

Finding the Units in the Ring Z[t][t21]

Answer & Explanation

Shawn Kim

Shawn Kim

Beginner2022-01-15Added 25 answers

Step 1
Given: Z[t][t21]
Write:
x=t+t21
and
y=tt21
Then xy=1, i.e. y=x1, and t=(x+y)2
t21=(xy)2
Thus we see that
Z[t][t21]Z[12][x, x1]
Now it is easy to check that if A is an integral domain, then the only units in
A[x, x1]
are of the form axn, where
aA
is a unit and n is an integer. Since the units in Z[12] are precisely the elements ±2m (for some m) we see that the units in
Z[12][x, x1]
are of the form
±2mxn=±2m(t+t21)n
It is easy to check that if such an element and its inverse both actually lie in
Z[t][t21}],
then necessarily m=0 (e.g. for norm reasons, or just looking explicitly at their denominators), and so we get the answer that the units are precisely the elements of the form
±(t+t21)n
(Here n runs over Z; this is the same set as ±(t±t21)n, where n is now non-negative.)
Neil Dismukes

Neil Dismukes

Beginner2022-01-16Added 37 answers

Step 1
For any non-negative integer n the element
±(t±t21)n
has norm 1, and there are no elements
a(t)+b(t)t21
of norm -1 because
a(1)2b(1)21210
Step 2
Suppose
a(t)+b(t)t21
is a unit which is not one of the above units such that degb is minimal. Then it has norm 1, so
a(t)2b(t)2(t21)=1
It is not hard to see that the units ±1
are the only units for which b=0, so WLOG b is nonzero. This implies that
dega=d+1
degb=d
for some non-negative integer d, and moreover the leading terms of a and b must agree up to sign. If the leading terms agree, then
(a(t)+b(t)t21)(tt21)
=(ta(t)b(t)(t21))+(tb(t)a(t))t21
is a unit with the property that the coefficient of t21 has degree strictly less than that of b which is not on the above list, which contradicts the assumption of minimality. Similarly, if the leading terms of a and b are opposite in sign, then
(a(t)+b(t)t21)(t+t21)
(ta(t)+b(t)(t21))+(tb(t)+a(t))t21
is a unit with the property that the coefficient of t21 has degree strictly less than that of b which is not on the above list, again contradicting the assumption of minimality. So no such units exist.
alenahelenash

alenahelenash

Expert2022-01-24Added 556 answers

Just a geometric translation of Matts

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