6:54 But how do we make sure that EVERY irreducible polynomial in F[t] splits over E or is irreducible over E? There will be infinite of those. What if one such polynomial has only one root in E.
The polynomial p and F[t] are designated to be 'measured' against F and E. How this is all done was detailed in previous videos in the playlist. This is a definition, so there isn't a justified proof. Normal Extension is encapsulating this structure for invocation at a future time. If there is only one root in E, and the polyn p only has one root, I'd say that E is a normal Extension of F.
I'm not quite convinced by your "every" argument. You've shown that every polynomial with sqrt(3)-sqrt(2) as a root has all its roots in E. But what about a polynomial having some other element of E as a root? It's not obvious to me that it will also have all its roots in E.
Why can we conclude in the final example that every polynomial having p as a factor, splits completely??? We only know that p splits completely in this example...if f(x) = p(x) q(x) in E and p splits completely in E, can we say anything about splitting q in E?
Remember that we are talking about irreducible polynomials here. In the example p(x) = x^4 -10x^2 + 1, and it is irreducible over the rationals. Therefore, any polynomial, lets say f(x), that contains sqrt(3) -sqrt(2) would have to be in the form f(x) = p(x)q(x) for some polynomial q(x). Therefore, it is reducible. Since we are only concerning ourselves with irreducible polynomials we cannot say anything about q(x) in this case. Hope that helps.
Agreed, how do we know there isn't an irreducible polynomial that doesn't have sqrt(3)-sqrr(2) as a root, but that does have some other root in the field extension? Seems that something is missing in the proof.
Re: normality of ℚ/t⁴-10t²+1. Here's a sketch of a full proof. Let's for simplicity assume without proof that any element of this extension field can be presented as a+b√2+c√3+d√6, where a,b,c,d are rational. (This is true, but slightly annoying to demonstrate.) Consider a polynomial P over ℚ that's irreducible over ℚ, but has a+b√2+c√3+d√6 as a root. No polynomial over ℚ can distinguish √2 and -√2, nor √3 and -√3; therefore, P must also have as roots the following three elements of ℚ/t⁴-10t²+1: a-b√2+c√3-d√6, a+b√2-c√3-d√6, a-b√2-c√3+d√6. Now, we want to show that P has no other roots. For now let's assume that these four are all distinct, and consider other cases later. Multiplying four expressions (x-), we observe a polynomial Q with rational coefficients. (This is tiresome to do by hand; we check with WolframAlpha.) This Q must divide P, because all of its roots are of multiplicity 1, and are also roots of P. Given that P is irreducible over ℚ, P must be the same as Q up to a rational coefficient, and therefore cannot have any other roots but these four. Some of the four observed roots of P might be the same; e.g. if b=c=0, d≠0, then among the 4 expressions there are only 2 distinct elements: a+d√6 and a-d√6. In this case we will multiply only distinct roots to acquire a product polynomial. There are some cases to consider (variant where a≠0 and exactly two of b,c,d are 0; variant where a≠0 and b=c=d=0; variant where all four are 0), but in the end, we always collect several distinct roots of P, multiply their (x-), observe that the product is a polynomial with rational coefficients that divides P, and conclude that P has no other roots.
Thanks for helping us get to the roots of the concepts!
Great video, thanks!
6:54 But how do we make sure that EVERY irreducible polynomial in F[t] splits over E or is irreducible over E? There will be infinite of those. What if one such polynomial has only one root in E.
The polynomial p and F[t] are designated to be 'measured' against F and E. How this is all done was detailed in previous videos in the playlist. This is a definition, so there isn't a justified proof. Normal Extension is encapsulating this structure for invocation at a future time.
If there is only one root in E, and the polyn p only has one root, I'd say that E is a normal Extension of F.
I'm not quite convinced by your "every" argument. You've shown that every polynomial with sqrt(3)-sqrt(2) as a root has all its roots in E. But what about a polynomial having some other element of E as a root? It's not obvious to me that it will also have all its roots in E.
This is a continuation of an example from a previous video 302.S3a, .S3b, .S3c, where your questions are answered.
Why can we conclude in the final example that every polynomial having p as a factor, splits completely??? We only know that p splits completely in this example...if f(x) = p(x) q(x) in E and p splits completely in E, can we say anything about splitting q in E?
Remember that we are talking about irreducible polynomials here. In the example p(x) = x^4 -10x^2 + 1, and it is irreducible over the rationals. Therefore, any polynomial, lets say f(x), that contains sqrt(3) -sqrt(2) would have to be in the form f(x) = p(x)q(x) for some polynomial q(x). Therefore, it is reducible. Since we are only concerning ourselves with irreducible polynomials we cannot say anything about q(x) in this case. Hope that helps.
@@matthewmcgilvary3141 What if the polynomial has some root in E not equal to sqrt(3)-sqrt(2). How do we know it splits completely?
Agreed, how do we know there isn't an irreducible polynomial that doesn't have sqrt(3)-sqrr(2) as a root, but that does have some other root in the field extension? Seems that something is missing in the proof.
superb
Re: normality of ℚ/t⁴-10t²+1. Here's a sketch of a full proof.
Let's for simplicity assume without proof that any element of this extension field can be presented as a+b√2+c√3+d√6, where a,b,c,d are rational. (This is true, but slightly annoying to demonstrate.)
Consider a polynomial P over ℚ that's irreducible over ℚ, but has a+b√2+c√3+d√6 as a root. No polynomial over ℚ can distinguish √2 and -√2, nor √3 and -√3; therefore, P must also have as roots the following three elements of ℚ/t⁴-10t²+1: a-b√2+c√3-d√6, a+b√2-c√3-d√6, a-b√2-c√3+d√6.
Now, we want to show that P has no other roots. For now let's assume that these four are all distinct, and consider other cases later.
Multiplying four expressions (x-), we observe a polynomial Q with rational coefficients. (This is tiresome to do by hand; we check with WolframAlpha.) This Q must divide P, because all of its roots are of multiplicity 1, and are also roots of P. Given that P is irreducible over ℚ, P must be the same as Q up to a rational coefficient, and therefore cannot have any other roots but these four.
Some of the four observed roots of P might be the same; e.g. if b=c=0, d≠0, then among the 4 expressions there are only 2 distinct elements: a+d√6 and a-d√6. In this case we will multiply only distinct roots to acquire a product polynomial. There are some cases to consider (variant where a≠0 and exactly two of b,c,d are 0; variant where a≠0 and b=c=d=0; variant where all four are 0), but in the end, we always collect several distinct roots of P, multiply their (x-), observe that the product is a polynomial with rational coefficients that divides P, and conclude that P has no other roots.