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

**Bingk** · September 3rd 2012, 05:12 AM

Last problem I had from my exam:

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

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

I'm not sure what to do. Any help would be much appreciated

**Vlasev** · September 3rd 2012, 07:58 AM

The characteristic of a ring $R$ is usually defined as the smallest positive integer $n$ such that adding $1_R$ $n$ times to itself gives $0_R$ . If no such number exists, we say the characteristic is 0. Another way to define it is as the smallest positive integer $n$ such that adding $r\in R$ to itself $n$ times produces $0_R$ , for any $r\in R$ . The two definitions are equivalent, can you prove it?

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

It's not hard at all to see that a field $F$ and any of its extensions $E$ 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.

**Bingk** · September 3rd 2012, 10:10 AM

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 $r = 1r$ .

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 $c$ is indeed inside $K$ .

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

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

.

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

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

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

Is what I've got okay?

**Deveno** · September 3rd 2012, 10:25 AM

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.

**Bingk** · September 3rd 2012, 12:55 PM

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 $g(x)$ divides $x^{p^n}-c^{p^n}$ , but not too sure about $g(x)$ dividing $(x-c)^{p^n}$ because I wasn't sure how you would interpret that product in $K[x]$ (i.e. if you expand the $(x-c)^{p^n}$ , in $F[x]$ the inner terms would go to zero, but how would you interpret the inner terms in $K[x]$ when we don't know if $c^m$ for $0 < m < p^n$ is in $K[x]$ ?) Like, how would you interpret the term $acx^{p^n-1}$ in $K[x]$ when we don't know that $c$ is in $K$ , so we can't take the product $ac$ (since $p|a$ , we can let $ac=0c$ but still, what does that mean in $K$ when we don't know that $c$ is in $K$ )

That was basically my "issue" because the expansion of $(x-c)^{p^n}$ seems to imply that $c$ is in $K$ . 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 $K$ , which means that the product is "doable" in $K$ , which means that the powers of $c$ should be in $K$ .

**Deveno** · September 3rd 2012, 11:37 PM

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.

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