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 2019-07-01, 20:00 #1 wildrabbitt   Jul 2014 3·149 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 2019-07-01 at 20:04
 2019-07-01, 20:10 #2 VBCurtis     "Curtis" Feb 2005 Riverside, CA 34·59 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 primary-school division can proceed.
 2019-07-01, 20:15 #3 CRGreathouse     Aug 2006 32·5·7·19 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).
2019-07-01, 21:05   #4
wildrabbitt

Jul 2014

3·149 Posts

Quote:
 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 primary-school division can proceed.
Thanks. You weren't misunderstanding my question. I believe that there's something hidden with maths like it's a mystery or I'm sereptticiously tricking myself into making simple things seem hard so I can feel better about how good I am at maths. Perhaps also that I think that the simple form should be smaller than it's numerator and multiplying the numerator by a number greater than 1 wouldn't have occurred to me.

So,

$$\frac{6+23\sqrt{2}}{2+\sqrt{2}}=\frac{(6+23\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})} =\frac{12-46+(46-6)\sqrt{2}}{2}=\frac{-34+40\sqrt{2}}{2}=-17+20\sqrt{2}$$

/* editted out the mistakes */

Quote:
 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).
I take your point but isn't it okay to say that in the ring $$Z[\sqrt{2}]$$,$$6+23\sqrt{2}$$ has a factor $$2+\sqrt{2}$$?

Thanks very much to both of you.

Last fiddled with by wildrabbitt on 2019-07-01 at 21:39

2019-07-01, 21:42   #5
wildrabbitt

Jul 2014

3×149 Posts

I guess I prompted CR's post

by writing
Quote:
 I've worked out that the algebraic integer 6+232√ is divisible by 2+2√. I find finding these factors by looking at norms quite tiring.

I should have said has a factor instead of divisble.

 2019-07-01, 22:32 #6 Dr Sardonicus     Feb 2017 Nowhere 106738 Posts Using the computational might of Pari-GP, ? f=x^2-2;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 2019-07-01 at 22:33 Reason: rewording
2019-07-02, 15:02   #7
CRGreathouse

Aug 2006

32×5×7×19 Posts

Quote:
 Originally Posted by wildrabbitt I take your point but isn't it okay to say that in the ring $$Z[\sqrt{2}]$$,$$6+23\sqrt{2}$$ has a factor $$2+\sqrt{2}$$?
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).

 2019-07-03, 16:17 #8 Dr Sardonicus     Feb 2017 Nowhere 10001101110112 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 Pari-GP 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.
2019-07-04, 13:18   #10
Dr Sardonicus

Feb 2017
Nowhere

3·17·89 Posts

Quote:
 Originally Posted by wildrabbitt 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.
In a field, every division (by any nonzero element) "comes out even" with a remainder of zero.

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.

 2019-07-04, 15:42 #11 CRGreathouse     Aug 2006 10111011000012 Posts In this context, fields are boring because all nonzero elements are units.

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