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# Field Extensions Watch

1. I ran into this problem while solving a larger one. Can anyone see where I've gone wrong?

Suppose that is a finite extension, and let be a basis for L over K.

and by induction on we get

which is certainly wrong.
2. (Original post by james22)
I ran into this problem while solving a larger one. Can anyone see where I've gone wrong?

Suppose that is a finite extension, and let be a basis for L over K.

and by induction on we get

which is certainly wrong.
There's no way of knowing if this is 2 or not, without some characterisation of the a_i. This will be equal to the degree of the minimal polynomial of over .

To expand, consider the splitting field of (and splitting fields are always finite extensions). If we take , which has of course has minimal polynomial over , then we get , with basis . This isn't yet the splitting field, but hereafter is irrelevant - whatever happens next, by the tower law the degree of the extension isn't going to be a power of 2!
3. (Original post by FireGarden)
There's no way of knowing if this is 2 or not, without some characterisation of the a_i. This will be equal to the degree of the minimal polynomial of over .

To expand, consider the splitting field of (and splitting fields are always finite extensions). If we take , which has of course has minimal polynomial over , then we get , with basis . This isn't yet the splitting field, but hereafter is irrelevant - whatever happens next, by the tower law the degree of the extension isn't going to be a power of 2!
OK, I see that it may not be 2, but surely it cannot be 1?

This would mean that [L:K(a_1)] is at least 2^(n-1)? This would mean that [L:K] is at least 2^n, but 2^n is bigger than n, which is a contradiction.
4. (Original post by james22)
OK, I see that it may not be 2, but surely it cannot be 1?

This would mean that [L:K(a_1)] is at least 2^(n-1)? This would mean that [L:K] is at least 2^n, but 2^n is bigger than n, which is a contradiction.
Well, one of the extensions is going to have degree 1. If is the basis over , then one of the is going to be and then that extension is of course of degree 1; the field won't change.

I think an issue is you're starting with the basis of the big field to construct it from the base field.

Take , where is a primitive 5th root of unity. This is clearly a simple extension, but the basis will be . Now if we tried to adjoin these basis elements individually to as it seems to appear in your inductive step, you'd get the degree of the extension being if you didn't notice something crucial: After the first element, you're done. and all its non-real powers are all primitive 5th roots; adjoining any one of them will give the same field. In your example, it could be like this case, where adjoining necessarily puts into your field to keep the operations closed. Adjoining the 'next element', as your inductive step is trying to, puts in an element that's now already there; the degree of that extension shall then be 1.

I hope that was clear..!
5. (Original post by FireGarden)
Well, one of the extensions is going to have degree 1. If is the basis over , then one of the is going to be and then that extension is of course of degree 1; the field won't change.

I think an issue is you're starting with the basis of the big field to construct it from the base field.

Take , where is a primitive 5th root of unity. This is clearly a simple extension, but the basis will be . Now if we tried to adjoin these basis elements individually to as it seems to appear in your inductive step, you'd get the degree of the extension being if you didn't notice something crucial: After the first element, you're done. and all its non-real powers are all primitive 5th roots; adjoining any one of them will give the same field. In your example, it could be like this case, where adjoining necessarily puts into your field to keep the operations closed. Adjoining the 'next element', as your inductive step is trying to, puts in an element that's now already there; the degree of that extension shall then be 1.

I hope that was clear..!
I see, I was too caught up in the linear algebra bit to notice that. I think that what I have done is still enough to solve my problem though, as I don't thinkt the errors matter too much (this thing I was trying to prove was much strong than I needed). I'll post the whole question and result when I've done it or failed again.
6. (Original post by FireGarden)
x
Thanks for your help, could you please tell me if my overall solution is correct?

Let be a finite extension of degree , and be an irreducible polynomial over . If , if and are coprime, then has no roots in .

Solution

Suppose f does have a solution in , .

I think this works. This was my first solution but I was put off because it only relies on not dividing , which is weaker than them being coprime.
7. (Original post by james22)
Thanks for your help, could you please tell me if my overall solution is correct?

Let be a finite extension of degree , and be an irreducible polynomial over . If , if and are coprime, then has no roots in .

Solution

Suppose f does have a solution in , .

I think this works. This was my first solution but I was put off because it only relies on not dividing , which is weaker than them being coprime.
This is correct, however, the hypothesis is giving you more; namely, your polynomial is actually irreducible over the extension. Of course, under an obvious condition on its degree.
This is correct, however, the hypothesis is giving you more; namely, your polynomial is actually irreducible over the extension. Of course, under an obvious condition on its degree.
I have used irreducibility to get that p is the minimum polynomial, not just any 0 polynomial.
9. (Original post by james22)
I have used irreducibility to get that p is the minimum polynomial, not just any 0 polynomial.
I am not sure I understand what you are saying.

You are using irreducibility to get the degree of the root.
I am not sure I understand what you are saying.

You are using irreducibility to get the degree of the root.
Same thing. If p was reducible then the degree of the extension would not be n.
11. (Original post by james22)
Same thing. If p was reducible then the degree of the extension would not be n.
So yeah, that's fine. I just said that the polynomial is also irreducible over the extension of your base field.

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