- #1

- 96

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**Proposition:**

Every extension of degree 2 is a root field.

**Proof:**

Let F be a field. Let p(x) ε F[x] and suppose p(x) has degree n. Then p(x) has n roots, say c

_{1},c

_{2},...,c

_{n}. Let E be the extension of F that contains the aforementioned roots, and suppose [E:F]=2. Now, we know F(c

_{1},c

_{2},...,c

_{n}) is the root field of p(x) over F. Previously, we have shown that if an extension over a field is of a degree that is prime, then there is no proper field between the field and its extension*. The extension E clearly contains F, and by our hypothesis c

_{1},c

_{2},...,c

_{n}ε E. Thus, F(c

_{1},c

_{2},...,c

_{n}) is subset of E. Then since,

F(c

_{1},c

_{2},...,c

_{n}) ≠ F, F(c

_{1},c

_{2},...,c

_{n}) = E. Hence, our extension is indeed a root field. QED

*I simply used the theorem that is analogous to LaGrange's Theorem for Finite Groups, but for fields!

Now, this is my "second version" of the proof. I am going back and trying to prove a few exercises different way. I am not sure if I set up my assumptions correctly. Any criticism would be very helpful! Thanks! :)