# Roots of different irreducible polynomials are algebraically independent

Let $F$ be a field, and let $f$ be a monic irreducible polynomial over $F$.

1. Let $\alpha$ be a root of some other monic irreducible $g\ne f$. Then is $f$ still irreducible in $F(\alpha)$?
2. Is it true that $f$ factors completely in $F(\alpha)$, where $\alpha$ is a root of $f$? Obviously $x-\alpha$ is a factor of $f$, but it is not clear to me that all the other roots are also in the field.
3. Is $F(\alpha)$ isomorphic to $F(\beta)$ if $\alpha,\beta$ are roots of the same irreducible $f$? Are they equal (as subsets of some larger field that splits $f$)?

These are questions I have made up in order to better understand finite field extensions. I don't have a reference, and I believe all of them to be true, although I have no proof, which is why I am asking the question.

Edit: (Moved to a new question)

• For (1), it obviously depends. I do not believe you cannot find examples... – Mariano Suárez-Álvarez Nov 26 '14 at 23:16
• @MarianoSuárez-Alvarez Call it faulty intuition, as the answers below show I was way off-base about everything here. I made a statement elsewhere that the union of two finite field extensions is a finite field extension, but I'm not confident that I know how to define such a thing, hence this question. If I had been right about some of this, the job would be easy, but it seems it's more complicated than I thought. – Mario Carneiro Nov 26 '14 at 23:40
• It's not bad form to ask a post-question question -- it's bad form to do so by editing this post rather than making a brand new post. Your new question can be answered, and the answers to it won't look anything like the answers to the questions you asked. – Hurkyl Nov 27 '14 at 0:05
• @Hurkyl Fair enough. Moved the edit to a new question. – Mario Carneiro Nov 27 '14 at 0:20

## 4 Answers

For your first problem, it need not be the case. For let $f(x)=x^4+16$. It is not hard to show that $f(x)$ is irreducible over the rationals. Note that $$x^4+16=(x^2+2\sqrt{2} x+4)(x^2-2\sqrt{2} x+4),$$ and let $g(x)=x^2-2$.

• I have made an edit to the question; please take a look at it if you can. – Mario Carneiro Nov 26 '14 at 23:54

Here are some hints.

1. Let $f$ and $g$ be the minimal polynomials for $\sqrt2$ and $1+\sqrt2$ over $\mathbb{Q}$. Is $f$ irreducible over $\mathbb{Q}(1 + \sqrt2)$?

2. Let $f(x) = x^3 - 2 \in \mathbb{Q}[x]$. This polynomial has a real root $\alpha = 2^{1/3}$ and two non-real roots $\alpha e^{2\pi i/3}$ and $\alpha e^{-2\pi i/3}$. Does $f$ factor completely over $\mathbb{Q}(\alpha)$?

3. Can you see that $F(\alpha)$ and $F(\beta)$ are both isomorphic to $F[x]/(f(x))$? Considering the example in 2. are $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\alpha e^{2\pi i/3})$ equal as subfields of $\mathbb{C}$?

1. Take $F = \mathbb{R}, g(x) = x^2 + 1, \alpha = i, f(x) = x^2 + 4.$

2. Take $F = \mathbb{Q}, f(x) = x^3 - 2, \alpha = \sqrt[3]{2}.$

3. Take $F = \mathbb{Q}, f(x) = x^p - 2, p$ is a prime, $\alpha = p$-th root of unity, $\beta = \sqrt[p]{2}.$ Then $F(\alpha)$ and $F(\beta)$ are not equal as a subfield of $\mathbb{C}.$

For 1 See Krish's answer.

For 2

$f(x)=x^3-2\in\mathbb{Q}[x]$, then

(a) $\,\,\,$$\sqrt[3]{2} is a root. Does f(x) split over \mathbb{Q}(\sqrt[3]{2})? (b) \,\,\,$$\omega\sqrt[3]{2}$ is also a root. Does $f(x)$ split over $\mathbb{Q}(\omega\sqrt[3]{2})$? NOTE: $\omega$ is a cube root of unity.

Recall: Definition of splitting field.

For 3

Yes, there's a theorem which states that if $\alpha$ and $\beta$ are the roots of the same irreducible polynomial, then $F(\alpha)\cong F(\beta)$

$f(x)=x^4-2$, the roots are $\pm\sqrt[4]{2},\pm i\sqrt[4]{2}$ and $f(x)$ is irreducible over $\mathbb{Q}$

So $\mathbb{Q}(\sqrt[4]{2})\cong\mathbb{Q}(i\sqrt[4]{2})$ but are they equal?