20190701, 20:00  #1 
Jul 2014
702_{8} Posts 
dividing an algebraic integer by another
Hi,
I've worked out that the algebraic integer \(6+23\sqrt{2}\) is divisible by \(2+\sqrt{2}\). I find finding these factors by looking at norms quite tiring. Is another way to work out \(\frac{6+23\sqrt{2}}{2+\sqrt{2}}\) in it's simplest form? A division algorithm for example. Please show me how it goes. Last fiddled with by wildrabbitt on 20190701 at 20:04 
20190701, 20:10  #2 
"Curtis"
Feb 2005
Riverside, CA
2^{5}×3^{2}×19 Posts 
It's really elementary, so I may be misunderstanding your question:
Multiply top and bottom by the conjugate of the bottom. That makes the denominator a real number, so regular primaryschool division can proceed. 
20190701, 20:15  #3 
Aug 2006
5,987 Posts 
Note that, in general, you don't get an algebraic integer (just like dividing an integer by another integer doesn't give an integer, in general, but a rational number).

20190701, 21:05  #4  
Jul 2014
111000010_{2} Posts 
Quote:
So, \(\frac{6+23\sqrt{2}}{2+\sqrt{2}}=\frac{(6+23\sqrt{2})(2\sqrt{2})}{(2+\sqrt{2})(2\sqrt{2})} =\frac{1246+(466)\sqrt{2}}{2}=\frac{34+40\sqrt{2}}{2}=17+20\sqrt{2}\) /* editted out the mistakes */ Quote:
Thanks very much to both of you. Last fiddled with by wildrabbitt on 20190701 at 21:39 

20190701, 21:42  #5  
Jul 2014
2·3^{2}·5^{2} Posts 
I guess I prompted CR's post
by writing Quote:
I should have said has a factor instead of divisble. 

20190701, 22:32  #6 
Feb 2017
Nowhere
2×7^{2}×61 Posts 
Using the computational might of PariGP,
? f=x^22;z=Mod((6+23*x)/(2+x),f) %1 = Mod(20*x  17, x^2  2) ? charpoly(%) %2 = x^2 + 34*x  511 Last fiddled with by Dr Sardonicus on 20190701 at 22:33 Reason: rewording 
20190702, 15:02  #7 
Aug 2006
5,987 Posts 
Yes, just like in the ring Z it's right to say that 15 has a factor 3. But if you want to divide integers in general (apart from denominator 0, of course), you get rational numbers. Similarly, if you want to divide algebraic integers in general (or for a particular ring), you get algebraic numbers (or those in your ring).

20190703, 16:17  #8 
Feb 2017
Nowhere
2·7^{2}·61 Posts 
If f is an irreducible polynomial in Q[x], and b is a nonzero polynomial in Q[x] of degree less than the degree of f, then the polmod Mod(b, f) is invertible. Thus, Mod(a/b,f) is defined for any polynomial a in Q[x]. In PariGP calculations, f is usually monic (leading coefficient is 1) with integer coefficients. It is often the defining polynomial of a number field.
In the above example, I took f = x^2  2, and I also found the characteristic polynomial of Mod(a/b, f). The point of doing that was that Mod(a/b, f) is an algebraic integer precisely when its characteristic polynomial is monic and has integer coefficients. In fields of degree greater than 2, there can be cases where algebraic integers have polynomial expressions (mod f) which have fractional coefficients. 
20190703, 19:04  #9 
Jul 2014
2×3^{2}×5^{2} Posts 
Right. Well thanks for those posts. I thought my thread had come to it's natural end but I'm glad it hasn't.
Please don't take this to mean I'm not interested in the latest replies, but since I wasn't expecting anymore I was busy looking into some more things. I hope therefore it can be considered not to be without due interest in matters raised in this thread that I ask the following ; As I understand it, A Euclidean Domain had a Euclidean Norm and a Euclidean Algorithm for division. I'm fine with that. What I'm confused about is that in the same way that Every Euclidean Domain is a UFD, every Field is a Euclidean Domain. It seems logical to me that every field therefore has a Euclidean Norm and a Euclidean Algorithm so I'm totally puzzled about the fact that in Thomas Hardy's book The Theory of Numbers, a distinction is made between Euclidean Fields and NonEuclidean fields. For example, k(sqrt(23)), the real quadratic field is said not to be Euclidean whereas k(sqrt(2)) the real quadratic field associated with root 2, is said to be Euclidean. Help, please. 
20190704, 13:18  #10  
Feb 2017
Nowhere
2×7^{2}×61 Posts 
Quote:
You're unclear on the definitions. As the term is used in Hardy and Wright, "Euclidean field" is a number field whose ring of algebraic integers has a Euclidean (division with quotient and remainder) algorithm. The remainder is either 0 or is "smaller" than the divisor. The usual function used to measure the "size" of integers is the absolute value of the norm. You might try reading The Euclidean Algorithm in Quadratic Number Fields. 

20190704, 15:42  #11 
Aug 2006
5,987 Posts 
In this context, fields are boring because all nonzero elements are units.

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
primes p(n1) dividing the sum 2+3+5+7+...+p(n)  enzocreti  enzocreti  1  20190326 09:07 
primitive roots when the base is a quadratic algebraic integer  devarajkandadai  Number Theory Discussion Group  0  20180208 05:15 
An algebraic quandry  Unregistered  Homework Help  21  20101220 02:00 
Algebraic factor issues base 24  michaf  Conjectures 'R Us  18  20080521 10:08 
Algebraic factors  henryzz  ElevenSmooth  13  20071218 09:12 