Is this correct? Field extension of the rationals

[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?

 

[micromass]

Hi mr. vodka!

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?

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).

 

[nonequilibrium]

Thank you very much! :)

 

[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)).

 

[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?

 

[micromass]

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?

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).

 

[Hurkyl]

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?

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)

 

[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

 

[micromass]

(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

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

 

[nonequilibrium]

Indeed, I was too rash there.

Also thank you Hurkyl.

All very interesting and enlightening

 

[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.