What Conditions Ensure the Minimal Polynomial of α Divides P in Galois Theory?

[giulio_hep]

TL;DR Summary

irreducible P with roots in extension L of degree <= 2

I'm not sure the following passage is so trivial as it was supposed to be: I mean, what does exactly prove it? That's my question.
The step is the following:

if $P$ has a root $\alpha$ in $\mathbf L$ - an extension of $\mathbf K$ of degree <= $\frac n 2$ where n is the degree of $P$ over $\mathbf K$ - the minimal polynomial of $\alpha$ over $\mathbf K$ divides $P$.

While it looks sort of acceptable/obviously true, I'm not sure of what would really be its proof.
Thank you.

 

[giulio_hep]

I believe that the above holds iff the extension $\mathbf L$ is algebraic over $\mathbf K$.

That would make sense to me and hopefully answers my question.

I was following the week 2 of Ekaterina Amerik introducing Galois theory but this (IMO important) detail is missing in her explanation (specifically the Corollary one at 7:38) of Stem fields (https://www.coursera.org/learn/galois/lecture/O7lYD/2-1-stem-field-some-irreducibility-criteria).

In conclusion, I think we have to specify that L must be a K algebra or at least an algebraic extension over K in the above statement...

 

[Office_Shredder]

Am I missing something stupid. Let $m(x)$ be the minimal polynomial in $K[x]$. Then by euclidian division algorithm, $p(x)=m(x)q(x)+r(x)$ where $r$ has degree smaller than $m$, or $r=0$. Then $p(\alpha)=0$ implies that $r(\alpha)=0$ which is a contradiction, unless $r(x)=0$. In which case $m$ divides $p$.

That is to say, if $p$ has a root, the minimal polynomial of that root always divides $p$. It doesn't matter what the degree of the field extension is, or whether said extension is algebraic.

 

[giulio_hep]

Well, as far as I can understand it, let say that K is Q of rational and L is R of real numbers, theoretically, without other assumptions more restrictive (and if I am not wrong). Notice that in this case the extension would be transcendental instead of algebraic.
The point is that the m(x) in your example would be in L which is R, so the solution $\alpha$ but more importantly the coefficients (a fact that would be problematic in my opinion) of m(x) and q(x) could be real (I mean really radicals like $\sqrt 2$) and not rational, hence they would not be valid factors for the polynomial over K = Q. And I believe, in the end, that this last condition in particular is excluded for an initial assumption, I guess (or otherwise for some weird deduction, but I can't see how).

Also the point regarding the degree of the extension is not well explained. I think you are right, the degree is only important in the other direction of the theorem (she proved before this passage), but in this corollary (and in this direction of the iff in the theorem) it appears irrelevant, indeed.

 

[Office_Shredder]

The minimal polynomial of $\alpha$ over $K$ is a polynomial with coefficients in $K$, not $L$, or at least that's how I interpret that sentence. Can you be a bit more specific about which field the coefficients of each polynomial in your original post are supposed to be in?

 

[giulio_hep]

The point is that the corallary says

if P has a solution $\alpha$ in L, where L is an extension of K...

which means that the solution and the minimal polynomial are in L, even if P is over K.

My example is the obvious $x^2+1$ in Q which is divided by $x+i$ and $x-i$ in the extension R

 

[giulio_hep]

To be more clear, the full corollary one is:

P is irreducible over K if and only if it does not have roots in extensions L of K of degree less or equal than n/2 where n is the degree of P.

So, if I take the direction starting from "P has roots in an extension L of K...", it follows what I have written above.

 

[giulio_hep]

Ops... but now I see that the degree of a transcendental extension is infinite so the fact that the extension is algebraic could be deduced (but only from the constraint on the degree of the extension) without assuming it...

 

[Office_Shredder]

The point is that the corallary says

if P has a solution $\alpha$ in L, where L is an extension of K...

which means that the solution and the minimal polynomial are in L, even if P is over K.

There's no such thing as *the* minimal polynomial. There can be many minimal polynomials for a given root, depending on which field the coefficients of the polynomial can live in.

For example $\alpha= (2)^{1/4}i$ has a minimal polynomial of $x- 2^{1/4}i$ in $\mathbb{C}[x]$, $x^2+\sqrt{2}$ in $\mathbb{R}[x], and$x^4-2$in$\mathbb{Q}[x]$. If$\alpha\in L$, then the minimal polynomial of$\alpha$in$L[x]$is just$x-\alpha##, so it's probably not the field you actually want to look at the minimal polynomial over.

My example is the obvious $x^2+1$ in Q which is divided by $x+i$ and $x-i$ in the extension R

I assume you mean in the extension C :)

For the actual statement, there's a direct correspondence between "p has a root in a field extension of degree $k$ and p is divisible by a polynomial of degree $k$ (which would be the minimal polynomial of the root).

I think there are two parts here
1) do you understand that correspondence?
2.) If p is of degree n, what is the highest degree polynomial that can divide p?

 

[giulio_hep]

The minimal polynomial of a root of a monic polynomial P over a field K must divide P and it is unique and the definition is also in wikipedia... In the mentioned course there was a lesson in the previous week specifically about the definition of Minimal polynomial (that is unitary, etc...).
I'm pretty sure that they mean a solution $\alpha$ and a well defined minimal polynomial in L.Btw, Sorry for the confusion between C and R in my example.
Of course if I choose R as extension of Q, then a correct example for P over Q is $x^2-2$ which is divided by two roots ($\sqrt 2$ and $-\sqrt 2$) in R each one with a minimal polynomial in R while P is irreducible in Q. (if hopefully I've not introduced other errors in my revised example)

 

[giulio_hep]

No, I was wrong, I'm reading again the previous lesson.
Yes, you are right about it, they mean a minimal polynomial in K for the root in the extension L.
In my example the minimal polynomial in Q for the root $\sqrt 2$ in R would be P itself $x^2-2$, which is irreducible in Q.

Even so, it implies that L must be an algebraic extension of K for such a minimal polynomial to exist.

 

[giulio_hep]

And well, regarding the degree, the fact is simply that if P is reducible, then it has a factor of degree less or equal than n over 2: therefore, as per the correspondence you have pointed out, $\exists$ a stem field of that degree and we can take it as the extension to prove the corollary and the converse is also true, assuming of course that an extension of finite (i.e. $\frac n 2$) degree must be algebraic. I think I'm getting it right now.

 

[Office_Shredder]

Even so, it implies that L must be an algebraic extension of K for such a minimal polynomial to exist.

Only algebraic extensions can have finite degree. If $\beta \in L$ is transcendental, then every set of the form $\{1,\beta,\beta^2,...,\beta^n\}$ is linearly independent over $K$ (otherwise you would get a polynomial which is zero at $\beta$, so $[L:K]$ must be infinite.