Kronecker's theorem - Find a field with roots for X^4 + 1

In summary: This is because in the complex plane, the inverse of a complex number is its complex conjugate divided by its magnitude. In summary, this conversation discusses using Kronecker's theorem to find a root for X^4 + 1 in a field extension, and then shows that the other two roots can also be found in the same extension. This is because the field is closed under multiplication and inversion, allowing for the complex conjugate of the root to also be a root.
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Silversonic
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Homework Statement



Consider [itex]X^4 + 1 [/itex] in the field [itex] \mathbb{Q} [X] [/itex]. I used Kronecker's theorem to find a root for X^4 + 1 in the field extension [itex] \mathbb{Q} [\frac{(1+i)}{\sqrt{2}}] [/itex]. I'm asked to show that this field extension allows X^4 + 1 to be factorised completely, thus adding a single root gives all roots.

Homework Equations





The Attempt at a Solution



I have the answer, but it uses an assumption and I'm not sure how it got. It says:

"Since X^4 + 1 has real coefficients and has [itex] \frac{(1+i)}{\sqrt{2}} [/itex] as a root, the complex conjugate [itex] \frac{(1-i)}{\sqrt{2}} [/itex] must also be a root."

It then carries on as if it's known [itex] \frac{(1-i)}{\sqrt{2}} [/itex] is in the field extension to show that the other two roots (the negative of both those roots) are in the field extension. But how exactly do we know that the complex conjugate of the root [itex] \frac{(1+i)}{\sqrt{2}} [/itex] is also in the field extension?

Is it because this is a field and, letting [itex] \frac{(1+i)}{\sqrt{2}} = z [/itex];

[itex]z \frac{\overline{z}}{|z|^2} = 1[/itex]

so

[itex]\frac{\overline{z}}{|z|^2} [/itex]

Is the inverse of [itex] z [/itex], since this is a field it's inverse must be contained in the set, and so

[itex] |z|^2 \times \frac{\overline{z}}{|z|^2} = \overline{z}[/itex]

Is in the set? This would need to assume [itex] |z|^2 [/itex] is contained in the field too.
 
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  • #2
Set [itex]z = \frac{1+i}{\sqrt{2}}[/itex] and notice that [itex]|z| = 1[/itex], which means that [itex]z^{-1} = \overline{z}[/itex].
 

FAQ: Kronecker's theorem - Find a field with roots for X^4 + 1

What is Kronecker's theorem?

Kronecker's theorem, also known as the existence theorem, states that any non-constant polynomial with coefficients in a field can be factored into linear factors in an extension field of that field.

What is the significance of Kronecker's theorem?

Kronecker's theorem is significant because it guarantees the existence of an extension field with roots for any polynomial, which allows for the study and solution of polynomials in a broader context.

What is the field with roots for X^4 + 1?

The field with roots for X^4 + 1 is the field of complex numbers, denoted by C. This is because the complex numbers contain all four roots of the polynomial, namely ±i and ±1.

How does one find the roots of X^4 + 1 in the field of complex numbers?

The roots of X^4 + 1 in the field of complex numbers can be found by factoring the polynomial into linear factors: (X + i)(X - i)(X + 1)(X - 1). This yields the four roots: i, -i, 1, and -1.

Can Kronecker's theorem be applied to polynomials with coefficients in other fields?

Yes, Kronecker's theorem can be applied to any non-constant polynomial with coefficients in a field. This means that the theorem can be extended to polynomials with coefficients in fields such as the rational numbers, integers, and finite fields.

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