# A separable element in an extension field is also in the field

• Sep 3rd 2012, 05:12 AM
Bingk
A separable element in an extension field is also in the field
Last problem I had from my exam:

Let $\displaystyle F$ be an extension field of a field $\displaystyle K$ whose char is $\displaystyle p$. Let $\displaystyle c$ be an element of $\displaystyle F$ which is separable over $\displaystyle K$ and for which there exists a positive integer $\displaystyle n$ such that $\displaystyle c^{p^n} \in K$. Show that $\displaystyle c \in K$.

The first thing that bugged me was that it wasn't clear if $\displaystyle F$ or $\displaystyle K$ had char $\displaystyle p$, but then it seems that whichever has char $\displaystyle p$ means that the other would also have char $\displaystyle p$. I'm also tempted to assume that $\displaystyle p$ is prime, but that would be presumptuous of me (Nod)

I'm not sure what to do. Any help would be much appreciated (Nod)
• Sep 3rd 2012, 07:58 AM
Vlasev
Re: A separable element in an extension field is also in the field
The characteristic of a ring $\displaystyle R$ is usually defined as the smallest positive integer $\displaystyle n$ such that adding $\displaystyle 1_R$ $\displaystyle n$ times to itself gives $\displaystyle 0_R$. If no such number exists, we say the characteristic is 0. Another way to define it is as the smallest positive integer $\displaystyle n$ such that adding $\displaystyle r\in R$ to itself $\displaystyle n$ times produces $\displaystyle 0_R$, for any $\displaystyle r\in R$. The two definitions are equivalent, can you prove it?

Now, onto your question! IF $\displaystyle F$ is a field of characteristic $\displaystyle n$, then either $\displaystyle n = 0$ or $\displaystyle n$ is prime. (You can prove this)

It's not hard at all to see that a field $\displaystyle F$ and any of its extensions $\displaystyle E$ must have the same characteristic. They share a multiplicative identity.

As a next step, you should take the definitions of the things you are using and play around with them to see if you can make any progress.
• Sep 3rd 2012, 10:10 AM
Bingk
Re: A separable element in an extension field is also in the field
I'm familiar with the definition of the characteristic of a ring (although, just wanted to point out that your first definition assumes the ring has a multiplicative identity). The definitions are equivalent since $\displaystyle r = 1r$.

For the second part, I haven't formally proven it, but I see how it works. If the characteristic isn't prime, then you get zero divisors, which you can't have in fields.

Now, my real problem is trying to get that $\displaystyle c$ is indeed inside $\displaystyle K$.

The first case (trivial) is if $\displaystyle p=0$, then $\displaystyle c^{0^n}=c^1=c \in K$.

So, if $\displaystyle p \neq 0$ (i.e. $\displaystyle p$ is prime), then ... I'm not sure (Speechless).

I've tried several things, and I think this is the closest:

$\displaystyle f(x)=x^{p^n}-c^{p^n} \in K[x]$. Clearly, $\displaystyle f(c)=0$, so one of the irreducible factors of $\displaystyle f(x)$ is the monic irreducible polynomial of $\displaystyle c$ (let's call it $\displaystyle g(x)$). Thus, $\displaystyle g(x)$ should divide $\displaystyle x^{p^n}-c^{p^n} = (x-c)^{p^n}$, so $\displaystyle g(x)=(x-c)^{m}$ for some $\displaystyle 0 \leq m < p^n$. But $\displaystyle c$ is separable over $\displaystyle K$, so $\displaystyle m=0$, thus $\displaystyle g(x)=x-c \in K[x]$, which implies that $\displaystyle c \in K$.

The part I'm not sure about is if I can say that $\displaystyle g(x)$ should divide $\displaystyle (x-c)^{p^n}$.

Is what I've got okay?
• Sep 3rd 2012, 10:25 AM
Deveno
Re: A separable element in an extension field is also in the field
i think you mean m = 1 (because (x - c)0 = 1, which has no roots).

the minimal polynomial for c, m(x) always divides any polynomial f(x) for which f(c) = 0.

to see this, write:

f(x) = m(x)q(x) = r(x), where either r = 0, or deg(r) < deg(m).

then 0 = f(c) = m(c)q(c) + r(c) = 0q(c) + r(c) = r(c).

since deg(r) < deg(m) contradicts m being the minimal polynomial of c, we must have r = 0.
• Sep 3rd 2012, 12:55 PM
Bingk
Re: A separable element in an extension field is also in the field
oops! yeah! I meant 1, haha, it's 3 a.m. here :)
On that note, I think you meant m(x)q(x) + r(x) ;)

I definitely need to sleep soon.

I'm confused. I thought g(x) was the min. poly. of c in K[x], if it is, then you may read the next paragraph onwards. Otherwise, how do we justify that g(x) is in K[x]?

Also, I'm okay with the fact that the minimal polynomial of c divides any polynomial for which c is a root. My "issue" was interpreting that polynomial in another form. Basically, I was fine with $\displaystyle g(x)$ divides $\displaystyle x^{p^n}-c^{p^n}$, but not too sure about $\displaystyle g(x)$ dividing $\displaystyle (x-c)^{p^n}$ because I wasn't sure how you would interpret that product in $\displaystyle K[x]$ (i.e. if you expand the $\displaystyle (x-c)^{p^n}$, in $\displaystyle F[x]$ the inner terms would go to zero, but how would you interpret the inner terms in $\displaystyle K[x]$ when we don't know if $\displaystyle c^m$ for $\displaystyle 0 < m < p^n$ is in $\displaystyle K[x]$?) Like, how would you interpret the term $\displaystyle acx^{p^n-1}$ in $\displaystyle K[x]$ when we don't know that $\displaystyle c$ is in $\displaystyle K$, so we can't take the product $\displaystyle ac$ (since $\displaystyle p|a$, we can let $\displaystyle ac=0c$ but still, what does that mean in $\displaystyle K$ when we don't know that $\displaystyle c$ is in $\displaystyle K$)

That was basically my "issue" because the expansion of $\displaystyle (x-c)^{p^n}$ seems to imply that $\displaystyle c$ is in $\displaystyle K$. I think you could argue that since the inner terms goes to zero, and they are linearly independent, then each term must equal zero in $\displaystyle K$, which means that the product is "doable" in $\displaystyle K$, which means that the powers of $\displaystyle c$ should be in $\displaystyle K$.
• Sep 3rd 2012, 11:37 PM
Deveno
Re: A separable element in an extension field is also in the field
the fact that the "inner terms disappear" is a function of the characteristic, not the field itself.

we know that g(x) is in K[x] (by definition). since g(x) = (x - c)m in F[x] (since f(x) splits in F), and since c is separable, g(x) must be linear. that is all you need.

we don't "expand" f(x) in K[x], we expand it in F[x]. since g(x) is in K[x], and F is an extension of K, g(x) is also in F[x].

it's like this: separability of an element c MEANS it's minimal polynomial has no repeated roots. yes, these roots may lie in a larger field, but if they do they are DISTINCT.

however, f(x) has only ONE root in F[x]: c. the polynomials of the form xpn - apn in a field of characteristic p are SPECIAL.

now if c were NOT separable, then we'd be stuck. it could be that some lesser power of c lies in K, but not c itself.

you should "play around" with say, Z3 and the field of order 9. differences of two cubes (or ninth powers) in such a field are "nice", in that they are very easy to factor.