# Is this correct? Field extension of the rationals

1. Jul 26, 2011

### nonequilibrium

By F[X] I mean the polynomials with coefficients in field F. By F(X) I mean the rational polynomials.

I have a feeling that $\boxed{ \mathbb Q( \sqrt 2 ) \cong \frac{\mathbb Q[X]}{(X^2-2)}}$. (if not readable: the RHS is with [X])
Is this true? If so, how can I prove it? I suppose it would suffice I could show that the RHS is the smallest field extension of the rationals that contains sqrt(2) (as the LHS is obviously just that).

Also, is there maybe even a more general result behind this?

2. Jul 26, 2011

### micromass

Staff Emeritus
Hi mr. vodka!

You are 100% correct that this is true. How do we prove it? Well, the crucial point is that $X^2-2$ is an irreducible polynomial in $\mathbb{Q}[X]$. This means that
$\mathbb{Q}/(X^2-2)$ is a field. So the following function

$$f:\mathbb{Q}\rightarrow \mathbb{Q}/(X^2-2):q\rightarrow [q]$$

is an injective field homomorphism. We extend this field homomorphism to

$$f:\mathbb{Q}(\sqrt{2})\rightarrow \mathbb{Q}/(X^2-2):a+b\sqrt{2}\rightarrow [a+bX]$$

Check that this is an isomorphism.

This result can be generalized. The correct generalization involves splitting fields. If P(X) is an irreducible polynomial over a field F, then there exists a field extension K of F such that P(X) has a root in K. Indeed, we define

$$K=F[X]/(P(X))$$

If a is the root, then we can even define

$$f:F(a)\rightarrow K$$

by defining f(a)=X. This is again surjective.

What does this have to do with splitting fields? Well, we can use this result to show that: if P(X) is a polynomial over a field F, then there exists a field extension K of F such that P(X) splits.

Why is this true? Well, repeatedly apply the above to the irreducible factors of P(X).

3. Jul 26, 2011

### nonequilibrium

Thank you very much! :)

4. Jul 26, 2011

### spamiam

It also might be useful to learn about minimal polynomials. If $\alpha$ is the root of some polynomial in $F[X]$, then there is a unique monic irreducible polynomial $m_{\alpha, F}(X) \in F[X]$, which is known as the minimal polynomial of $\alpha$ over $F$. It's said to be minimal because if f(X) is any other polynomial with $\alpha$ as a root, then $m_{\alpha, F}(X) \, | \, f(X)$. (What's more, a monic polynomial over F with $\alpha$ as a root is the miminal polynomial of alpha over F iff it is irreducible over F.) It's then a theorem that $F(\alpha) \cong F[X]/(m_{\alpha, F}(X))$.

Last edited: Jul 26, 2011
5. Jul 26, 2011

### nonequilibrium

Hm, thanks, but I seem to be confused:

Take as a root $\sqrt[3] 2$, then its minimal polynomial is $f = X^3-2$ (?). This has one real solution and two complex solutions. For that reason it would seem that $\mathbb Q[X]/(f)$ would also have these two complex solutions as elements (as f is also the minimal polynomial for those complex solutions) while $\mathbb Q(\sqrt[3] 2)$ only adds one element, not three, but of course I'm wrong cause they're isomorphic... Where do I err?

6. Jul 26, 2011

### micromass

Staff Emeritus
No, doing $\mathbb{Q}[X]/(f)$ only adjoins one root! You can choose which one though. So $\mathbb{Q}[X]/(f)$ is isomorphic to $\mathbb{Q}(\sqrt[3]{2})$, but also to $\mathbb{Q}(a)$ where a are the other roots.
In general, the other roots are not contained in $\mathbb{Q}[X]/(f)$.

Note, that the root of the polynomial $Z^3-2$ in $\mathbb{Q}[X]/(X^2-2)$ is [X]. So the polynomial splits as

$$Z^3-2=(Z-[X])(Z^2+[X]Z+[X^2])$$

But the polynomial $Z^2+[X]Z+[X^2]$ doesn't necessarily have roots in $\mathbb{Q}[X]/(X^2-2)$.

7. Jul 26, 2011

### Hurkyl

Staff Emeritus
That extension adds a lot more than just one element -- it includes, for example, $1 + \sqrt[3] 2$ and $3 + 2 \sqrt[3]2 + \sqrt[3] 4$.

But you're right, you won't find any of the other cube roots of 2 in that field.

The thing you're missing is that $\mathbb Q[X]/(f)$ doesn't contain any of the three complex roots of f: it contains neither $\sqrt[3] 2$, $\omega \sqrt[3] 2$, nor $\omega^2 \sqrt[3] 2$. (Where $\omega$ is the primitive cube root of unity -- i.e. $\omega = exp(2 \pi i / 3)$)

What is true is that there is a field homomorphism from this to the real numbers, sending X sending X to $\sqrt[2] 3$, and two field homomorphisms from this to the complex numbers: one sends X to $\omega \sqrt[2] 3$, and one sends X to $\omega^2 \sqrt[2] 3$.

We say that this field has one real embedding and one complex conjugate pair of complex embeddings.

Using these homomorphisms, we can think of X as either of the three complex roots at our leisure -- but we obviously cannot think of X as being all three at once.

Now, how does f(t) factor in this field? as:
$$f(t) = (t - X) (t^2 + Xt + X^2)$$​
It can be shown that the quadratic term is irreducible. So it does turn out that f has only one root in this field. But we can make a new field extension that adjoins yet another element (a square root of -3), and this field will have three roots for f.

The resulting field is called the "splitting field" of f. It turns out to be isomorphic to
$$\mathbb{Q}(\omega, \sqrt[3] 2)$$
Incidentally, I'm pretty sure it is also isomorphic to:
$$\mathbb{Q}(\omega + \sqrt[3] 2)$$

(aside: for some f, $\mathbb{Q}[X] / f(X)$ does have more than one root of f. For example, if f is a quadratic polynomial)

8. Jul 26, 2011

### nonequilibrium

(you made a typo in the (X² - 1) instead of (X³ - 1) )

So you say that I can choose to view $\mathbb Q[X]/(X^3-2)$ as adding either the real or the complex solutions? But by writing the factorization
$Z^3-2= (Z-[X])(Z^2+[X]Z+[X^2])$
it is clear that we have added the real solution, yet I don't see "where you made the choice"

EDIT: this is not a reply to Hurkyl, his post appeared between my replying and act of posting; I have yet to read Hurkyl's post

9. Jul 26, 2011

### micromass

Staff Emeritus
Why is it clear that we have added the real solution?? Why can't [X] represent one of the complex solutions?

10. Jul 26, 2011

### nonequilibrium

Indeed, I was too rash there.

Also thank you Hurkyl.

All very interesting and enlightening

11. Jul 27, 2011

### Bacle

I don't know if this is too informal, and too quick-and-dirty, but the ideal generated by an irreducible polynomial is maximal, so that the quotient is a field, and, in the quotient, the base element is considered/defined-to-be, the zero element, i.e., x^2-2=0 for x in the quotient field.