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

Last problem I had from my exam:

Let be an extension field of a field whose char is . Let be an element of which is separable over and for which there exists a positive integer such that . Show that .

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

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

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

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

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

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

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 .

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 is indeed inside .

The first case (trivial) is if , then .

So, if (i.e. is prime), then ... I'm not sure (Speechless).

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

. Clearly, , so one of the irreducible factors of is the monic irreducible polynomial of (let's call it ). Thus, should divide , so for some . But is separable over , so , thus , which implies that .

The part I'm not sure about is if I can say that should divide .

Is what I've got okay?

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.

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 divides , but not too sure about dividing because I wasn't sure how you would interpret that product in (i.e. if you expand the , in the inner terms would go to zero, but how would you interpret the inner terms in when we don't know if for is in ?) Like, how would you interpret the term in when we don't know that is in , so we can't take the product (since , we can let but still, what does that mean in when we don't know that is in )

That was basically my "issue" because the expansion of seems to imply that is in . 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 , which means that the product is "doable" in , which means that the powers of should be in .

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 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, Z_{3} 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.