# 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
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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.