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Algebraic Integer

'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

Reply 1

Indeterminate

Original post by 2710

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.

(edited 11 years ago)

Reply 2

2710

Original post by Indeterminate

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?

Reply 3

Indeterminate

Original post by 2710

This is because y is already a root of a polynomial that's not monic, hence it cannot be an algebraic integer by definition.

Reply 4

Mark13

Original post by 2710

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.

(edited 11 years ago)

Reply 5

Mark13

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...

Reply 6

2710

Original post by Mark13

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?

Reply 7

Mark13

Original post by 2710

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...

Reply 8

2710

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

Reply 9

Mark13

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.

(edited 11 years ago)

Reply 10

2710

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.

The definition im reading does not say anything about minimal

Reply 11

Mark13

Original post by 2710

The definition im reading does not say anything about minimal

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.

Reply 12

2710

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

Reply 13

Indeterminate

Original post by Mark13

Oops, yes

i meant to say that it doesn't have integer coefficients but that makes no difference I guess

Reply 14

2710

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.

Can you link me?

Reply 15

Mark13

Original post by 2710

http://en.wikipedia.org/wiki/Algebraic_integer

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]."