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May I refer you to:

http://www.math.niu.edu/~beachy/abstract_algebra/study_guide/soln62.html#s6201

Can you please explain in detail, problem 1, the inequality:

$[K(u,v): K(u)] \leq n$ ?

EDIT:

Let's see. First the fact that $[K(v): K]=n$, tell us that we can find a minimum polynomial $p(x) \in K[x]$ such that $p(v)=0$, i.e $v$ is algebraic over the field $K$. Now clearly $K(u)$ is a field extension of $K$ therefore we have:

$p(x) \in (K(u))[x]$

and $p(v)=0$.

Now consider the field extension $K(u,v)$ of $K(u)$. Let $h(x)$ be the minimum polynomial over $(K(u))[x]$ such that $h(u)=h(v)=0$. In particular $h(v)=0$.

Question:

We can't claim that $h=p$ even though both are monic irreducible right? because we don't even know if $p$ satisfies $p(u)=0$, all we know is that $p(v)=0$. So all we can say is that $h$ divides $p$ right? (this is the part I'm kinda confused with).

And then $[K(u,v): K(u)] \leq n$. Sorry if this seems trivial to you, just started learning on my own elementary field theory.

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2 Answers 2

up vote 2 down vote accepted

The inequality just comes from the fact that the degree of $v$ over $K$ is $n$. This just means that the minimum polynomial of $v$ over $K$ has degree $n$ and since this polynomial also has coefficients in $K(u)$, the minimum polynomial of $v$ over $K(u)$ divides this one and thus has degree less than or equal to $n$. Thus the inequality $[K(u, v) : K(u)] \leq n$ follows.

Edit

So let me try to address some of the points you made. If $p(x) \in K[x]$ is the minimum polynomial of $v$ over $K$ then sure, $p(v) = 0$ and also $p(x) \in K(u)[x]$ because $p(x)$ has coefficients in $K$ so certainly its coefficients lie in the field extension $K(u)$.

But now you can also consider the minimum polynomial of $v$ over the field extension $K(u)$. You called it $h(x) \in K(u)[x]$. You know that $p(v) = 0 = h(v)$ but now the only thing you can conclude is that $h(x)$ divides $p(x)$ in $K(u)[x]$, not that $h(x) = p(x)$.

The problem with the irreducibility is that it is relative to the polynomial ring in which you are considering the polynomials to lie. To give you an easy example. Consider the field extension $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$. Then the minimum polynomial of $\sqrt{2}$ over $\mathbb{Q}$ is $f(x) = x^2 - 2$. But the minimum polynomial of $\sqrt{2}$ over $\mathbb{Q}(\sqrt{2})$ is $q(x) = x - \sqrt{2}$. In this case the polynomial

$$p(x) = x^2 - 2 = (x - \sqrt{2})(x + \sqrt{2}) = q(x)(x - \sqrt{2})$$

factors over the polynomial ring $\mathbb{Q}(\sqrt{2})[x]$ so it does not remain irreducible.

If you have more doubts feel free to ask and I'll try to clarify.

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@Adrian Barquero: I will post a complete answer to see if I'm understanding the stuff. Can you please check? –  user6495 Apr 2 '11 at 8:28
    
@user6495 Sure, no problem. –  Adrián Barquero Apr 2 '11 at 8:29
    
@Adrián Barquero: thanks, one last question: am I correct in saying that $h(u)=h(v)=0$, this is just definition right? i.e if we had instead $F(a_{1},a_{2},...,a_{n})$ then $h$ would have $a_{1},a_{2},...,a_{n}$ as roots, yes? –  user6495 Apr 2 '11 at 9:25
    
@user6495 No, you don't necessarily have $h(u) = 0$. You only know that $h(v) = 0$ because $h(x)$ is the minimum polynomial of $v$ over $K(u)$, but you don't know anything about $u$. –  Adrián Barquero Apr 2 '11 at 9:29
    
@Adrián Barquero: Now I'm confused. The definition of the degree $[F(u): F]$ is that is it the degree of the minimum polynomial in $F[x]$ such that $u$ is a root of $f$, yes? or I'm wrong? My question is what would be the definition if instead of $F(u)$ we have something like $F(u,v)$ ? isn't then the definition that $[F(u,v):F]$ is the degree of the minimum polynomial over $F[x]$ such that $u,v$ are roots of this polynomial? or what is the definition? thank you again! –  user6495 Apr 2 '11 at 9:34

Since the minimal polynomial of $v$ over $K(u)$ has the degree of at most $n$. Is it enough for you?

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