# Prime Degree Polynomial

• Aug 2nd 2007, 07:56 AM
ThePerfectHacker
Prime Degree Polynomial
I discovered an elegant theorem. But it is easy to prove using some heavy altirary.

Theorem: Let \$\displaystyle f(x)\$ be a polynomial with rational coefficients having a prime degree \$\displaystyle p\$. If \$\displaystyle f(x)\$ is irreducible (cannot be factored in terms of other non-constant polynomials having rational coefficients) then \$\displaystyle f(x)\$ has exactly \$\displaystyle p\$ complex zeros.
• Aug 2nd 2007, 08:01 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
I discovered an elegant theorem. But it is easy to prove using some heavy altirary.

Theorem: Let \$\displaystyle f(x)\$ be a polynomial with rational coefficients having a prime degree \$\displaystyle p\$. If \$\displaystyle f(x)\$ is irreducible (cannot be factored in terms of other non-constant polynomials having rational coefficients) then \$\displaystyle f(x)\$ has exactly \$\displaystyle p\$ complex zeros.

Correct me if I'm wrong, but
\$\displaystyle x^3 + x^2 - x + 1 = 0\$
does not have 3 complex zeros. :confused:

-Dan
• Aug 2nd 2007, 08:04 AM
ThePerfectHacker
Quote:

Originally Posted by topsquark
Correct me if I'm wrong, but
\$\displaystyle x^3 + x^2 - x + 1 = 0\$
does not have 3 complex zeros. :confused:

-Dan

When I say "complex" I allow real also. Since \$\displaystyle \mathbb{R} < \mathbb{C}\$.
• Aug 2nd 2007, 08:07 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
When I say "complex" I allow real also. Since \$\displaystyle \mathbb{R} < \mathbb{C}\$.

Ah, so if I'm not mistaken (again) do you mean to say that these p zeros are distinct?

-Dan
• Aug 2nd 2007, 09:19 AM
Rebesques
Quote:

When I say "complex" I allow real also.

Then, isn't asking for it to be irreducible over Q just too demanding? :confused: :eek:
• Aug 2nd 2007, 10:18 AM
ThePerfectHacker
Quote:

Originally Posted by topsquark
Ah, so if I'm not mistaken (again) do you mean to say that these p zeros are distinct?

Yes, exactly.

We can state this theorem in a different way (equivalent):
Let \$\displaystyle f(x)\$ be a polynomial with integer coefficients and prime degree \$\displaystyle p\$ so that \$\displaystyle f(x)\$ is irreducible (cannot be factord into non-constant polynomials with integer coefficients) then \$\displaystyle f(x)\$ has exactly \$\displaystyle p\$ zeros.

Quote:

Originally Posted by Rebesques
Then, isn't asking for it to be irreducible over Q just too demanding? :confused: :eek:

Why?
• Aug 2nd 2007, 11:34 AM
Rebesques
Because the thing I found most intriguing in the original statement, was the promise that the roots are not real :) Can we have that?
• Aug 2nd 2007, 12:22 PM
ThePerfectHacker
Quote:

Originally Posted by Rebesques
Because the thing I found most intriguing in the original statement, was the promise that the roots are not real :) Can we have that?

In topsquark's example, that polynomial satisfies my conditions. Now, as promised, it has exactly 3 zeros (1 real 2 not). But, yes, we need that condition that they can be complex too.

Note, it should not be supprising because in field theory we usually deal in one big extension field, usually the algebraic closure. Which in this case is the closure of \$\displaystyle \mathbb{Q}\$ which are the algebraic numbers.