Algebraic Ints: Prove a+b & ab Algebraic

[bobn]

Homework Statement

if a is an algebraic number satisfying a^3+a+1 = 0 and b is an algebraic number satisfying b^2+b-3 = 0 prove that both a+b and ab are algebraic

Homework Equations

The Attempt at a Solution

a is root of equation x^3+x+1 = 0 and similarly b, so there exists a x = ab,and also x = a+b. so, ( x- ab)(x - a+b) is the required one.

Its from I N Herstein's Topics in Algebra 5.1.13

 

[Dick]

Homework Statement

if a is an algebraic number satisfying a^3+a+1 = 0 and b is an algebraic number satisfying b^2+b-3 = 0 prove that both a+b and ab are algebraic

Homework Equations

The Attempt at a Solution

a is root of equation x^3+x+1 = 0 and similarly b, so there exists a x = ab,and also x = a+b. so, ( x- ab)(x - a+b) is the required one.

Its from I N Herstein's Topics in Algebra 5.1.13

( x- ab)(x - a+b) is the required one what? It may be a polynomial that has x=ab as a root. But to show ab is algebraic you have to find a polynomial with integer coefficients that has ab as a root.

 

[bobn]

Dick, thanks for the reply.

Sorry I am confused between polynomial and algebraic.

 

[Dick]

Dick, thanks for the reply.

Sorry I am confused between polynomial and algebraic.

I just told you the difference. An algebraic number is the root of a polynomial with integer coefficients. sqrt(2) is a root of x^2-2=0. That shows it algebraic. The coefficients of the polynomial are integers. x=pi is a root of x^2-pi^2=0. That doesn't show it's algebraic. pi^2 isn't an integer.

 

[bobn]

a, b are integers, a+b and ab also. so.

(x- a-b)*(x-ab) = x^2 - x*(ab+a+b) - ab*(a+b) = f(x)

f(ab) = f (a+b) = 0, so, ab, a+b satisfies polynomial f(x) and hence, ab and a+b are algebraic.

 

[Dick]

a, b are integers, a+b and ab also. so.

(x- a-b)*(x-ab) = x^2 - x*(ab+a+b) - ab*(a+b) = f(x)

f(ab) = f (a+b) = 0, so, ab, a+b satisfies polynomial f(x) and hence, ab and a+b are algebraic.

If you were given that a and b are integers, that would be fine. You aren't given that a and b are integers. So it's not fine. I think you'd better go back and review the definition of 'algebraic' and any examples about them in your book.

 

[bobn]

hey, its given as algebraic integers, I typed that as number.

Thanks for your replies.

 

[bobn]

but I couldn't find any integer satisfying b^2+b-3 = 0, but Herstein had mentioned that b is an Algebraic integer.

 

[HallsofIvy]

Yes, your problem says that a and b are "algebraic integers". It does NOT follow that a and b are integers. Do you know the definition of "algebraic integer"?

 

[bobn]

yea, all ai s should be integers, but not necessarily roots right.

 

[bobn]

I think this should do.

f(a) = a^3+a+1 = 0; g(b) = b^2+b-3 = 0.

h(a+b) = ci*xi. This should be zero since, all xi 's are linearly dependent since they can be expressed as f(a) and g(b) i.e., a^i*b^i = a^j*(a^3+a+1)*b^k*( b^2+b-3) = 0. Similarly k(ab) = 0

 

[Office_Shredder]

You need to work on expressing yourself clearly. What are the x_is? What is h(x), and why is h(a+b)=c_i*x_i (the c's here aren't defined either). And then you have k(ab) without saying what k is!

 

[bobn]

Say f(a) = a^3+a+1 = 0; g(b) = b^2+b-3 = 0. [Given]

There exists , h(x) a polynomial with in the form of Ci*Xi where Cis are integer coefficients and h(a+b) = 0. since, all xi 's are linearly dependent since they can be expressed as f(a) and g(b) i.e., for each a^i*b^i = a^j*(a^3+a+1)*b^k*( b^2+b-3) = 0. Similarly there exists a polynomial k(x) with integer coefficients such that k(ab) = 0.

 

[bobn]

Office thanks for your reply.

More over, h(x) will be of degree 6 since, a and b are of degree 3,2 respectively with gcd 1. so, a+b will be of degree 6.