Great question! In this problem, we are trying to find the splitting field over Q. As you point out, we aren't able to get the roots of the polynomial (complex or real in this case) by simply looking in Q. We have to enlarge out scope. And that is precisely what the splitting field is all about: find the smallest field extension of Q that contains the roots. Now, you might wonder why the Q is mentioned at all if we are just going to include the roots anyway. The reason is that the splitting field can change depending on your base field. If we asked for the splitting field over R, then, since the fourth root of 2 is in R, we would obtain R(i) = C, a much bigger splitting field than the one obtained over Q. Similarly, of we asked for the splitting field over C, then as all of the roots lie in C, the answer would just be C again. A more interesting question is to ask what the splitting is if we view the polynomial over a finite field, such as Z/3Z (the integers modulo 3).
@@Justicewarrior795 Ah, I see. In that case, the answer is still no. You can see this in a couple of different ways. The first is that you can't use the field operations to get ∜2 from ∜2.i and so Q(∜2.i) ⊆ Q(∜2, i), but the inclusion is proper. A second way is to use the extension degrees. Since ∜2.i is a root of the irreducible polynomial x^4 - 2, we have [Q(∜2.i):Q] = 4. On the other hand, [Q(∜2, i): Q] = [Q(∜2)(i) : Q(∜2)][Q(∜2) : Q] = 2.4 = 8, where the latter degree follows from ∜2 being a root of x^4 - 2.
@@Justicewarrior795 It is possible to achieve the splitting field with a single element, however. This follows from the Primitive Element Theorem. In this case, Q(∜2, i) = Q(∜2 + i).
It's important to know what Q(a, b) means. It is the smallest field containing Q, a, and b. So if I write Q(a,b) = Q(a,c), then I'm not saying that b = c. Instead, I am saying that the smallest field containing Q, a, and b equals the smallest field containing Q, a, and c. Immediately after I replace 4√2i by i, I explain how I can use 4√2 and 4√2i to get i using field operations, as well as how to use 4√2 and i to get 4√2i (again using field operations. This is enough to show that Q(4√2, i) = Q(4√2, 4√2i).
Can I get a general definition of the topic?!
why are complex roots allowed if it's over Q? Thank you
Great question! In this problem, we are trying to find the splitting field over Q. As you point out, we aren't able to get the roots of the polynomial (complex or real in this case) by simply looking in Q. We have to enlarge out scope. And that is precisely what the splitting field is all about: find the smallest field extension of Q that contains the roots.
Now, you might wonder why the Q is mentioned at all if we are just going to include the roots anyway. The reason is that the splitting field can change depending on your base field. If we asked for the splitting field over R, then, since the fourth root of 2 is in R, we would obtain R(i) = C, a much bigger splitting field than the one obtained over Q.
Similarly, of we asked for the splitting field over C, then as all of the roots lie in C, the answer would just be C again.
A more interesting question is to ask what the splitting is if we view the polynomial over a finite field, such as Z/3Z (the integers modulo 3).
@@AdamGlesser thank you a lot! It became a lot clearer to me now
What is the tools used to explain this video
Inform
When I recorded this, I was just using the screen recorder on my iPad and the app Notability.
Shouldn't it just be Q(sqrt(2)*i) ?
That would be a splitting field for x^2 + 2 over Q, but not for x^4 - 2.
@@AdamGlesser my bad I meant 4th root of 2 *i . In other words can't the splitting field be achieved by only adjoining one element?
@@Justicewarrior795 Ah, I see. In that case, the answer is still no. You can see this in a couple of different ways. The first is that you can't use the field operations to get ∜2 from ∜2.i and so Q(∜2.i) ⊆ Q(∜2, i), but the inclusion is proper. A second way is to use the extension degrees. Since ∜2.i is a root of the irreducible polynomial x^4 - 2, we have [Q(∜2.i):Q] = 4. On the other hand, [Q(∜2, i): Q] = [Q(∜2)(i) : Q(∜2)][Q(∜2) : Q] = 2.4 = 8, where the latter degree follows from ∜2 being a root of x^4 - 2.
@@AdamGlesser ok I see, thank you so much.
@@Justicewarrior795 It is possible to achieve the splitting field with a single element, however. This follows from the Primitive Element Theorem. In this case, Q(∜2, i) = Q(∜2 + i).
Why 4√2i replace by i
It's important to know what Q(a, b) means. It is the smallest field containing Q, a, and b. So if I write Q(a,b) = Q(a,c), then I'm not saying that b = c. Instead, I am saying that the smallest field containing Q, a, and b equals the smallest field containing Q, a, and c. Immediately after I replace 4√2i by i, I explain how I can use 4√2 and 4√2i to get i using field operations, as well as how to use 4√2 and i to get 4√2i (again using field operations. This is enough to show that Q(4√2, i) = Q(4√2, 4√2i).
@@AdamGlesser thank you sir...its very helpful to me....
r/videosthatendtoosoon!!!