Field extension, properties,proof

[ivos]

Let F be a field of characteristic 0. Let f,g be irreducible polynomials over F. Let u be root of f, v be root of g; u,v are elements of field extension K/F. Let F(u)=F(v).
Prove (with using basic polynomial theory only, without using linear algebra and vector spaces):
1) deg f = deg g (deg f is degree of the polynomial f)
2) any element of F(u) is a root of some irreducible polynomial h over F
3) deg h divides deg f (h is as in 2))

Are the propositions 1),2),3) valid for any field F, not only for fields with char F = 0?

 

[DonAntonio]

Let F be a field of characteristic 0. Let f,g be irreducible polynomials over F. Let u be root of f, v be root of g; u,v are elements of field extension K/F. Let F(u)=F(v).


*** Do you mean here " Assume that \mathbb F(u)=\mathbb F(v) "? ***


Prove (with using basic polynomial theory only, without using linear algebra and vector spaces):
1) deg f = deg g (deg f is degree of the polynomial f)


*** Assuming the above: since \mathbb F[x]/<f(x)>\cong\mathbb F(v)=\mathbb F(v)\cong \mathbb F[x]/<g(x)> , we get

at once equality of ideals <f(x)>\,=\,<g(x)> , since the above isomorphism shows that for

\,\,h(x)\in\mathbb F[x]\,,\,\,h(x)\in <f(x)>\Longleftrightarrow h(x)\in <g(x)> , and from here both

f(x)\in <g(x)> \,\,and\,\,g(x)\in <f(x)> \Longrightarrow f(x)=kg(x)\,,\,\,k\in\mathbb F ***



2) any element of F(u) is a root of some irreducible polynomial h over F


*** This is trivial since the extension \mathbb F(u) is algebraic over \mathbb F , by construction...***


3) deg h divides deg f (h is as in 2))



*** Right now I can't see how to show this without using linear algebra... ***

Are the propositions 1),2),3) valid for any field F, not only for fields with char F = 0?

With the given data they are true in any case.

DonAntonio

 

[ivos]

f(x)=k.g(x) is not valid (in generally)
F=Q, f(x)=x^2-2,g(x)=x^2-2*x-1, u=sqrt(2), v=sqrt(2)+1
Then Q(u)=Q(v), but f(x) is not equals k.g(x)...

 

[DonAntonio]

f(x)=k.g(x) is not valid (in generally)
F=Q, f(x)=x^2-2,g(x)=x^2-2*x-1, u=sqrt(2), v=sqrt(2)+1
Then Q(u)=Q(v), but f(x) is not equals k.g(x)...

You are right. Then I can't help you without linear algebra.

DonAntonio

 

[ivos]

I can prove 1), 2) without linear algebra:

Let f,g be irreducible polynomials over field F. Let u be root of f, v be root of g. Assume, that F(u)=F(v).
Then deg f = deg g.

Proof:
Its well known (it can be proven by basic polynomial algebra) that any element of F(u), in particular v, has polynomial expression degree at most n-1 in u over F , i.e.
v=t(u)=a0+a1*u+...+an-1*u^(n-1) (a0,...,an-1 are elements of F)
Let deg f=n and let u1=u,u2,...,un be roots of f.
Then polynomial h(x)=(x-t(u1))*((x-t(u2))*...*(x-t(un)) has degree n, has v as root and
its coefficients are symmetric functions of u1,u2,...,un, i.e. elements of F.
Since irreducible polynomial g divides h (they have common root v), it follows
deg g <= deg h = n = deg f. Exchange u and v gives deg f <= deg g.
So, deg g = deg f.

 

[DonAntonio]

I can prove 1), 2) without linear algebra:

Let f,g be irreducible polynomials over field F. Let u be root of f, v be root of g. Assume, that F(u)=F(v).
Then deg f = deg g.

Proof:
Its well known (it can be proven by basic polynomial algebra) that any element of F(u), in particular v, has polynomial expression degree at most n-1 in u over F , i.e.
v=t(u)=a0+a1*u+...+an-1*u^(n-1) (a0,...,an-1 are elements of F)


*** I'm not sure what you call "basic polynomial algebra" to: it's easy to prove this using the properties of Euclidean Domain of the polynomial

ring F[x] -- namely, the fact that we can divide with remainder two polynomials and etc. -- and then using that \mathbb F(u)\cong \mathbb F[x]/&lt;f(x)&gt; .

But for the above, I can't see how else you can prove that, and I am seriously and honestly interested in knowing this, since if you reach this result

the way I just described, I can't understand why would you want to avoid linear algebra while using ring theory...

DonAntonio ***


Let deg f=n and let u1=u,u2,...,un be roots of f.
Then polynomial h(x)=(x-t(u1))*((x-t(u2))*...*(x-t(un)) has degree n, has v as root and
its coefficients are symmetric functions of u1,u2,...,un, i.e. elements of F.
Since irreducible polynomial g divides h (they have common root v), it follows
deg g <= deg h = n = deg f. Exchange u and v gives deg f <= deg g.
So, deg g = deg f.

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