# Field extension question

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)##.

Last edited:

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

fishturtle1
It's a little unclear to me what B is even supposed to be, can you post exercise 83?

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).##)

Staff Emeritus
Gold Member
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
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##. []

Staff Emeritus
Gold Member
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.