Real Polynomials and the Conjugate Roots Theorem

Let be an arbitrary polynomial with real coefficients, and assume that is a root of .
Then, , where is an integer, and are all real numbers, and we have , which gives:

and we obtain that , meaning it is a root of .

(see properties of conjugates, purely real)

Assume is a root of and .
Then, by factor theorem, for some .
Now, by CJRT, is also a root of .
Hence, we have .
Since , then .
That is, is a root of .

By using FT again, we get where .
Substituting this into our first equation for , we get:

where .

By properties of conjugate and modulus,

Where and , so is a real quadratic polynomial. All that remains is to show that is in .

From above, in , we have that

where is the zero polynomial.
Using DAP, we get:

where is in and the remainder is the zero polynomial or .
Now, every real polynomial is a complex polynomial, so we can also view this as a statement in .
As for any field, DAP over tells us that the quotient and remainder are unique.
Therefore, is the zero polynomial, and has real coefficients.

We now prove that is irreducible in .
Note that , so we also have . Now, since is not purely real, we have , so and are both roots of .
But from CPN, we know that cannot have any additional roots in , so has no purely real root.
Hence, by FT, has no linear factor in .
Assume for the sake of contradiction that is reducible in .
Then can be written as the product of two polynomials in of positive degree.
By DAP, both polynomials must have degree 1. This means that has a linear factor in , which is a contradiction, and we conclude that is irreducible in .