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)

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.

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?

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.

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 $\displaystyle K$.

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 x^pn - a^pn 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.