Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Calculus and Beyond Homework Help
Show that ##\mathbb{Z}[\sqrt 3]## is a Unique factorisation domain
Reply to thread
Message
[QUOTE="chwala, post: 6867747, member: 287397"] Yes, i was able to follow your proof; i understand the integral domain part; the part i am interested on is the unit part where you have indicated that the product has to satisfy ##(a+b\sqrt3) ⋅ (c+d\sqrt3)=1## this is the part i want to understand. I seem to get it now, what they probably mean is that if ##a,b ε \mathbb{z}## then we have to check and see that ##a## and ##b## satisfy ##a±b\sqrt3=1## and not necessarily the way i was doing it (considered conjugates as products). For ##\mathbb{z}[\sqrt -5]##, we cannot determine the ##a## and ##b## values satisfying ##(a+b\sqrt-5) ⋅ (c+d\sqrt-5)=1## [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Calculus and Beyond Homework Help
Show that ##\mathbb{Z}[\sqrt 3]## is a Unique factorisation domain
Back
Top