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# Algebraic Integer watch

1. 'Let y be a root of f(t) = t^3+0.5t+1 (irreducible/Q)

Then clearly y is not an algebraic integer.'

Can I ask why this is? I mean why is it 'clearly'? Is it not possible to have a polynomial g with integer coefficients st g(y) = 0?

Thanks
2. (Original post by 2710)
'Let y be a root of f(t) = t^3+0.5t+1 (irreducible/Q)

Then clearly y is not an algebraic integer.'

Can I ask why this is? I mean why is it 'clearly'? Is it not possible to have a polynomial g with integer coefficients st g(y) = 0?

Thanks
By definition, an algebraic integer is a root of a monic polynomial with integer coefficients. Monic means that the coefficient of the term with the highest power is 1. This is a monic polynomial, but 0.5 isn't an integer.
3. (Original post by Indeterminate)
By definition, an algebraic integer is a root of a monic polynomial with integer coefficients. Monic means that the coefficient of the term with the highest power is 1. This is a monic polynomial, but 0.5 isn't an integer.
An element y is considered an algebraic integer if there exists a monic polynomial f(t) e Z[t] st f(y) = 0

I know obviously that the above eqn does not have integer coefficients. But thats not what im asking. If you read my question, im asking, why can there not be a different eqn somewhere out there, with integer coefficients st f(y)=0?
4. (Original post by 2710)
An element y is considered an algebraic integer if there exists a monic polynomial f(t) e Z[t] st f(y) = 0

I know obviously that the above eqn does not have integer coefficients. But thats not what im asking. If you read my question, im asking, why can there not be a different eqn somewhere out there, with integer coefficients st f(y)=0?
This is because y is already a root of a polynomial that's not monic, hence it cannot be an algebraic integer by definition.
5. (Original post by 2710)
An element y is considered an algebraic integer if there exists a monic polynomial f(t) e Z[t] st f(y) = 0

I know obviously that the above eqn does not have integer coefficients. But thats not what im asking. If you read my question, im asking, why can there not be a different eqn somewhere out there, with integer coefficients st f(y)=0?
Suppose y is an algebraic integer. Then it has a minimal polynomial which has integer coefficients, and the minimal polynomial divides t^3+t/2+1. What goes wrong?

EDIT: Since we're given that it's irreducible over Q, we know what the minimal polynomial is, which makes things a lot more straightforward.
6. (Original post by Indeterminate)
This is because y is already a root of a polynomial that's not monic, hence it cannot be an algebraic integer by definition.
The definition of an algebraic integer is a complex number that satisfies a monic polynomial with integer coefficients - algebraic integers can (and do) satisfy non-monic polynomials. Anyway, t^3+t/2+1 is monic...
7. (Original post by Mark13)
The definition of an algebraic integer is a complex number that satisfies a monic polynomial with integer coefficients - algebraic integers can (and do) satisfy non-monic polynomials. Anyway, t^3+t/2+1 is monic...
Ok, so this is the minimal polynomial.

What is to say that there isnt a polynomial g st:

(t^3+t/2+1)g = monic with integer coefficients?

Because if there is one, then y would staisfy this eqn, it is monic, and also has integer coefficients. I know there isnt one, coz I know the answer, but I just want to know why there isnt one?
8. (Original post by 2710)
Ok, so this is the minimal polynomial.

What is to say that there isnt a polynomial g st:

(t^3+t/2+1)g = monic with integer coefficients?

Because if there is one, then y would staisfy this eqn, it is monic, and also has integer coefficients. I know there isnt one, coz I know the answer, but I just want to know why there isnt one?
See the edit to my previous post: you know the minimal polynomial of y must divide t^3+t/2+1, but you also know that t^3+t/2+1 is irreducible over Q, which tells you...
9. (Original post by Mark13)
See the edit to my previous post: you know the minimal polynomial of y must divide t^3+t/2+1, but you also know that t^3+t/2+1 is irreducible over Q, which tells you...
nono, you cannot use that reasoning. because t^3+t/2+1 IS the minimal polynomial. Right? And then the question in my post before stands
10. (Original post by 2710)
nono, you cannot use that reasoning. because t^3+t/2+1 IS the minimal polynomial. Right? And then the question in my post before stands
It's a standard fact that a complex number is an algebraic integer iff its minimal polynomial (over Q) has coefficients in Z. Go ahead and prove it if you want.
11. (Original post by Mark13)
It's a standard fact that a complex number is an algebraic integer iff its minimal polynomial has coefficients in Z. Go ahead and prove it if you want.
12. (Original post by 2710)
Whatever definition you're using, it's a standard result that a complex number is an algebraic integer if and only if its minimal polynomial over Q has coefficients in Z.
13. (Original post by Mark13)
Whatever definition you're using, it's a standard result that a complex number is an algebraic integer if and only if its minimal polynomial over Q has coefficients in Z.
http://en.wikipedia.org/wiki/Algebraic_integer
14. (Original post by Mark13)
The definition of an algebraic integer is a complex number that satisfies a monic polynomial with integer coefficients - algebraic integers can (and do) satisfy non-monic polynomials. Anyway, t^3+t/2+1 is monic...
Oops, yes

i meant to say that it doesn't have integer coefficients but that makes no difference I guess
15. (Original post by Mark13)
Whatever definition you're using, it's a standard result that a complex number is an algebraic integer if and only if its minimal polynomial over Q has coefficients in Z.
16. Pulled straight from that page:

"The following are equivalent definitions of an algebraic integer...

\alpha \in K is an algebraic integer if there exists a monic polynomial f(x) \in \mathbb{Z}[x] such that f(\alpha) = 0.
\alpha \in K is an algebraic integer if the minimal monic polynomial of \alpha over \mathbb Q is in \mathbb{Z}[x]."
17. (Original post by Mark13)
Pulled straight from that page:

"The following are equivalent definitions of an algebraic integer...

\alpha \in K is an algebraic integer if there exists a monic polynomial f(x) \in \mathbb{Z}[x] such that f(\alpha) = 0.
\alpha \in K is an algebraic integer if the minimal monic polynomial of \alpha over \mathbb Q is in \mathbb{Z}[x]."
Oh I see. I needed that 'equivalent definition'. Thanks!

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