Is Galois Theory a Powerful Tool for Understanding Field Extensions?

[fishturtle1]

Homework Statement

84i) If $B$ and $C$ be subfields of a field $E$, then their $\textbf{compositum}$ $B \lor C$ is the intersection of all the subfields of $E$ containing $B$ and $C$. Prove that if $\alpha_1, \dots, \alpha_n \in E$, then $F(\alpha_1) \lor \dots \lor F(\alpha_n) = F(\alpha_1, \dots, \alpha_n)$.

84ii) Prove that any splitting field $K/F$ containing $B$ (as in exercise 83) has the form $K = B_1 \lor \dots \lor B_r$, where each $B_i$ is isomorphic to $B$ via an isomorphism that fixes $F$. (Hint: If $Gal(K/F) = \lbrace \sigma_1, \dots, \sigma_r \rbrace$, then define $B_i = \sigma_i(B)$.

Relevant Equations

Definition: Let $E/F$ be a field extension and $\alpha_1, \dots, \alpha_n \in E$. Then $F(\alpha_1, \dots, \alpha_n)$ is the intersection of all subfields containing $F$ and $\alpha_1, \dots, \alpha_n$.

Definition: $Gal(E/F)$ is the group of automorphisms of $E$ that fix $F$.

Exercise 83 (Since it is referred to in 84ii): Let $B/F$ be a finite extension. Prove that there is an extension $K/B$ so that $K/F$ is a splitting field for some polynomial $f(x) \in F[x]$.

Proof of 84i): We assume that $E/F$ is a field extension. For each $i$, $F(\alpha_i)$ is the smallest subfield of $E$ containing $F$ and $\alpha_i$. Let $F'$ be a subfield of $E$ containing $F$ and $\alpha_1, \dots, \alpha_n$. Then $F'$ containing $F(\alpha_i)$ for all $i$. This implies $F'$ contains $F(\alpha_1) \lor \dots \lor F(\alpha_n)$. This shows $F(\alpha_1) \lor \dots \lor F(\alpha_n)$ is the smallest subfield of $E$ containing $F$ and $\alpha_1, \dots, \alpha_n$. We can conclude $F(\alpha_1) \lor \dots \lor F(\alpha_n) = F(\alpha_1, \dots, \alpha_n)$. []

Proof of 84ii) Let $f(x) \in F[x]$ be a polynomial whose splitting field is $K/F$. Denote the roots of $f(x)$ as $\alpha_1, \dots, \alpha_n$. Let $Gal(K/F) = \lbrace \sigma_1, \dots, \sigma_r \rbrace$ and define $B_i = \sigma_i(B)$ for all $i$. We know $K = F(\alpha_1, \dots, \alpha_n)$. By part i), it is enough to show $r = n$ and $\sigma_i(B) = F(\alpha_i)$. ... But I'm not sure how to proceed. I know the $\sigma_i$'s permute the roots of $f(x)$ but I'm not sure if i can use that here.
Does this work as a counterexample? Consider $B = F = \mathbb{Q}$ and $E = \mathbb{Q}(\sqrt{2})$. Then $B/F$ is a finite extension. We see $E/F$ is a splitting field for $f(x) = x^2 - 2 \in \mathbb{Q}[x]$ and $Gal(E/F) = \lbrace id, \sigma \rbrace$ where $\sigma: \mathbb{Q}(\sqrt{2}) \mapsto \mathbb{Q}(\sqrt{2})$ is defined by $a\sqrt{2} + b \mapsto a(-\sqrt{2}) + b$ for all $a, b \in \mathbb{Q}$.

Then $id(B) \lor \sigma(B) = B \lor B$ since $id$ and $\sigma$ fix $B$. This implies $E \neq id(B) \lor \sigma(B)$.

 

[Office_Shredder]

It's a little unclear to me what B is even supposed to be, can you post exercise 83?

I think your answer to the first part is correct.

 

[fishturtle1]

It's a little unclear to me what B is even supposed to be, can you post exercise 83?

I think your answer to the first part is correct.

Thanks for the reply! Exercise 83: Let $B/F$ be a finite extension. Prove that there is an extension $K/B$ so that $K/F$ is a splitting field of some polynomial $f(x) \in F[x]$. (Hint: Since $B/F$ is finite, it is algebraic, and there are element $\alpha_1, \dots, \alpha_n$ with $B = F(\alpha_1, \dots, \alpha_n)$. If $p_i(x) \in F[x]$ is the irreducible polynomial of $\alpha_i$, take $K$ to be the splitting field of $f(x) = p_1(x) \cdot\dots\cdot p_n(x).$)

 

[Office_Shredder]

I think you should focus on the f(x) they give you in 83. B is generated by something like one root of $p_1,...,p_n$ for each $p_i$ . $K$is generated by all the roots of all the $p_i$. So you want to create some fields that have the other roots of these polynomials and then compositum them together and show you get K.

 

[fishturtle1]

We use the following lemma from the textbook:

Lemma 50. Let $\sigma : F \to F'$ be an isomorphism of fields, let $\sigma^* : F[x] \to F'[x]$, defined by $\sum r_ix^i \mapsto \sum \sigma(r_i)x^i$, be the corresponding isomorphism of rings, let $p(x) \in F[x]$ be irreducible, and let $p^*(x) = \sigma^*(p(x)) \in F'[x]$. If $\beta$ is a root of $p(x)$ and $\beta'$ is a root of $p^*(x)$, then there is a unique isomorphism $\hat{\sigma} : F(\beta) \to F'(\beta)$ extending $\sigma$ with $\hat{\sigma}(\beta) = \beta'$.

Proof of 84ii): From exercise 83, we have $B = F(\alpha_1, \dots, \alpha_n)$ and $K/F$ is the splitting field of $p_1(x)\cdot\dots\cdot p_n(x)$, where $p_i(x)$ is the minimal polynomial of $\alpha_i$ over $F$. We note that each $p_i(x)$ is irreducible over $F$.

Fix $p_i(x)$. Let $\beta$ be a root of $p_i(x)$. We will construct a $\gamma \in Gal(E/F)$ such that $\hat{\sigma}(\alpha_i) = \beta$. Define $\sigma : F \to F$ to be the identity isomorphism. Using the notation in Lemma 50, $\alpha_i$ is a root of $p_i(x)$ and $\beta$ is a root of $p^*(x)$. By Lemma 50, there is a unique isomorphism $\hat{\sigma} : F(\alpha_i) \to F(\beta)$ extending $\sigma$ with $\hat{\sigma}(\alpha_i) = \beta$. So, we have found an isomorphism that fixes $F$ and maps $\alpha \mapsto \beta$. Repeatedly extending $\hat{\sigma}$ using Lemma 50 gives an automorphism $\gamma : E \to E$ such that $\gamma(\alpha_i) = \beta$ and $\gamma$ fixes $F$ ( i think ? ). We can conclude that there exists $\gamma \in Gal(E/F)$ such that $\gamma(\alpha_i) = \beta$ where $\beta$ is any root of $p_i(x)$.

Next, we write $Gal(E/F) = \lbrace \gamma_1, \gamma_2, \dots, \gamma_r \rbrace$ and define $B_i = \gamma_i(B)$. Denote the roots of $f(x)$ as $\alpha_1, \dots, \alpha_N$. By exercise 84i) and possibly reordering some terms, we have
$$B_1 \lor \dots \lor B_r = B(\alpha_1) \lor \dots \lor B(\alpha_N) = B(\alpha_1, \dots, \alpha_N)$$

We see $E \subset B(\alpha_1, \dots, \alpha_N)$. On the other hand, $E$ extends $B$ and contains $\alpha_1, \dots, \alpha_N$. So, $B(\alpha_1, \dots, \alpha_N) \subset E$. We may conclude $E = B(\alpha_1, \dots, \alpha_n) = B_1 \lor \dots \lor B_r$. []

 

[Office_Shredder]

What is $E$ supposed to be?

 

[fishturtle1]

We use the following lemma from the textbook:

Lemma 50. Let $\sigma : F \to F'$ be an isomorphism of fields, let $\sigma^* : F[x] \to F'[x]$, defined by $\sum r_ix^i \mapsto \sum \sigma(r_i)x^i$, be the corresponding isomorphism of rings, let $p(x) \in F[x]$ be irreducible, and let $p^*(x) = \sigma^*(p(x)) \in F'[x]$. If $\beta$ is a root of $p(x)$ and $\beta'$ is a root of $p^*(x)$, then there is a unique isomorphism $\hat{\sigma} : F(\beta) \to F'(\beta)$ extending $\sigma$ with $\hat{\sigma}(\beta) = \beta'$.

Proof of 84ii): From exercise 83, we have $B = F(\alpha_1, \dots, \alpha_n)$ and $K/F$ is the splitting field of $p_1(x)\cdot\dots\cdot p_n(x)$, where $p_i(x)$ is the minimal polynomial of $\alpha_i$ over $F$. We note that each $p_i(x)$ is irreducible over $F$.

Fix $p_i(x)$. Let $\beta$ be a root of $p_i(x)$. We will construct a $\gamma \in Gal(K/F)$ such that $\gamma(\alpha_i) = \beta$. Define $\sigma : F \to F$ to be the identity isomorphism. Using the notation in Lemma 50, $\alpha_i$ is a root of $p_i(x)$ and $\beta$ is a root of $p_i^*(x)$. By Lemma 50, there is a unique isomorphism $\hat{\sigma} : F(\alpha_i) \to F(\beta)$ extending $\sigma$ with $\hat{\sigma}(\alpha_i) = \beta$. So, we have found an isomorphism that fixes $F$ and maps $\alpha_i \mapsto \beta$. Repeatedly extending $\hat{\sigma}$ using Lemma 50 gives an automorphism $\gamma : K \to K$ such that $\gamma(\alpha_i) = \beta$ and $\gamma$ fixes $F$ ( i think ? ). We can conclude that there exists $\gamma \in Gal(K/F)$ such that $\gamma(\alpha_i) = \beta$ where $\beta$ is any root of $p_i(x)$.

Next, we write $Gal(K/F) = \lbrace \gamma_1, \gamma_2, \dots, \gamma_r \rbrace$ and define $B_i = \gamma_i(B)$. Denote the roots of $f(x)$ as $\alpha_1, \dots, \alpha_N$. By exercise 84i) and possibly reordering some terms, we have
$$B_1 \lor \dots \lor B_r = B(\alpha_1) \lor \dots \lor B(\alpha_N) = B(\alpha_1, \dots, \alpha_N)$$

We see $K \subset B(\alpha_1, \dots, \alpha_N)$. On the other hand, $K$ extends $B$ and contains $\alpha_1, \dots, \alpha_N$. So, $B(\alpha_1, \dots, \alpha_N) \subset K$. We may conclude $K = B(\alpha_1, \dots, \alpha_N) = B_1 \lor \dots \lor B_r$. []

Sorry! The $E$'s should have been $K$'s. I have edited in the above.