Is this correct? Field extension of the rationals

In summary: It's called the minimal polynomial of the root over the field F. The minimal polynomial of a root in F[X] is the monic polynomial with that root as a root over F.
  • #1
nonequilibrium
1,439
2
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 [itex]\boxed{ \mathbb Q( \sqrt 2 ) \cong \frac{\mathbb Q[X]}{(X^2-2)}} [/itex]. (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?
 
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  • #2
Hi mr. vodka! :smile:

mr. vodka said:
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 [itex]\boxed{ \mathbb Q( \sqrt 2 ) \cong \frac{\mathbb Q[X]}{(X^2-2)}} [/itex]. (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 [itex]X^2-2[/itex] is an irreducible polynomial in [itex]\mathbb{Q}[X][/itex]. This means that
[itex]\mathbb{Q}/(X^2-2)[/itex] is a field. So the following function

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

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

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

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

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

If a is the root, then we can even define

[tex]f:F(a)\rightarrow K[/tex]

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
Thank you very much! :)
 
  • #4
It also might be useful to learn about minimal polynomials. If [itex] \alpha[/itex] is the root of some polynomial in [itex] F[X] [/itex], then there is a unique monic irreducible polynomial [itex]m_{\alpha, F}(X) \in F[X][/itex], which is known as the minimal polynomial of [itex] \alpha[/itex] over [itex]F[/itex]. It's said to be minimal because if f(X) is any other polynomial with [itex] \alpha[/itex] as a root, then [itex] m_{\alpha, F}(X) \, | \, f(X)[/itex]. (What's more, a monic polynomial over F with [itex] \alpha[/itex] as a root is the miminal polynomial of alpha over F iff it is irreducible over F.) It's then a theorem that [itex] F(\alpha) \cong F[X]/(m_{\alpha, F}(X))[/itex].
 
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  • #5
Hm, thanks, but I seem to be confused:

Take as a root [itex]\sqrt[3] 2[/itex], then its minimal polynomial is [itex]f = X^3-2[/itex] (?). This has one real solution and two complex solutions. For that reason it would seem that [itex]\mathbb Q[X]/(f)[/itex] would also have these two complex solutions as elements (as f is also the minimal polynomial for those complex solutions) while [itex]\mathbb Q(\sqrt[3] 2)[/itex] only adds one element, not three, but of course I'm wrong cause they're isomorphic... Where do I err?
 
  • #6
mr. vodka said:
Hm, thanks, but I seem to be confused:

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

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

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

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

But the polynomial [itex]Z^2+[X]Z+[X^2][/itex] doesn't necessarily have roots in [itex]\mathbb{Q}[X]/(X^2-2)[/itex].
 
  • #7
mr. vodka said:
while [itex]\mathbb Q(\sqrt[3] 2)[/itex] 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, [itex]1 + \sqrt[3] 2[/itex] and [itex]3 + 2 \sqrt[3]2 + \sqrt[3] 4[/itex].

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

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

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:
[tex]f(t) = (t - X) (t^2 + Xt + X^2)[/tex]​
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
[tex]\mathbb{Q}(\omega, \sqrt[3] 2)[/tex]
Incidentally, I'm pretty sure it is also isomorphic to:
[tex]\mathbb{Q}(\omega + \sqrt[3] 2)[/tex]


(aside: for some f, [itex]\mathbb{Q}[X] / f(X)[/itex] does have more than one root of f. For example, if f is a quadratic polynomial)
 
  • #8
(you made a typo in the (X² - 1) instead of (X³ - 1) )

So you say that I can choose to view [itex]\mathbb Q[X]/(X^3-2)[/itex] as adding either the real or the complex solutions? But by writing the factorization
[itex]Z^3-2= (Z-[X])(Z^2+[X]Z+[X^2])[/itex]
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
mr. vodka said:
(you made a typo in the (X² - 1) instead of (X³ - 1) )

So you say that I can choose to view [itex]\mathbb Q[X]/(X^3-2)[/itex] as adding either the real or the complex solutions? But by writing the factorization
[itex]Z^3-2= (Z-[X])(Z^2+[X]Z+[X^2])[/itex]
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?
 
  • #10
Indeed, I was too rash there.

Also thank you Hurkyl.

All very interesting and enlightening
 
  • #11
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.
 

What is a field extension?

A field extension is a mathematical concept that involves extending the elements of one field to create a larger field that includes both the original field and new elements.

What are the rationals?

The rationals, also known as the rational numbers, are a set of numbers that can be expressed as a ratio of two integers. They include all fractions and terminating decimals.

How is a field extension of the rationals created?

A field extension of the rationals is created by adding new elements, such as irrational numbers, to the set of rationals. This results in a larger field that includes both the rationals and the new elements.

What is the significance of a field extension of the rationals?

A field extension of the rationals allows for a more complete and precise understanding of numbers and their properties. It also allows for the study of more complex mathematical concepts and structures.

What are some examples of field extensions of the rationals?

Some examples of field extensions of the rationals include the real numbers, complex numbers, and algebraic numbers. These sets include both rational and irrational numbers, creating larger and more comprehensive fields.

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