find the isomorphism

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nguyenalex

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nguyenalex

#1 h Mar 12, 2025, 3:58 PM

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I have the following exercise:

Let $E$ be an algebraic extension of $K$ , and let $F$ be an algebraic closure of $K$ containing $E$ . Prove that if $\sigma : E \to F$ is an embedding such that $\sigma(c) = c$ for all $c \in K$ , then $\sigma$ extends to an automorphism of $F$ .

My attempt:

Theorem (*): Suppose that $E$ is an algebraic extension of the field $K$ , $F$ is an algebraically closed field, and $\sigma: K \to F$ is an embedding. Then, there exists an embedding $\tau: E \to F$ that extends $\sigma$ . Moreover, if $E$ is an algebraic closure of $K$ and $F$ is an algebraic extension of $\sigma(K)$ , then $\tau$ is an isomorphism.

Back to our main problem:

Since $K \subset E$ and $F$ is an algebraic extension of $K$ , it follows that $F$ is an algebraic extension of $E$ . Assume that there exists an embedding $\sigma : E \to F$ such that $\sigma(c) = c$ for all $c \in K$ . By Theorem (*), there exists an embedding $\tau : F \to F$ that extends $\sigma$ . Since $F$ is algebraically closed, $\tau(F)$ is also an algebraically closed field.

Furthermore, because $\sigma(c) = c$ for all $c \in K$ and $\tau$ is an extension of $\sigma$ , we have
$K = \sigma(K) \subset K \subset \sigma(E) \subset \tau(F) \subset F.$
This implies that $F$ is an algebraic extension of $\tau(F)$ . We conclude that $F = \tau(F)$ , meaning that $\tau$ is an automorphism. (Finished!!)

Let choose $F = A$ be the field of algebraic numbers, $K=\mathbb{Q}$ . Consider the embedding $\sigma: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{2}) \subset A$ defined by
$a + b\sqrt{2} \mapsto a - b\sqrt{2}.$ Then, according to the exercise above, $\sigma$ extends to an isomorphism
$\bar{\sigma}: A \to A.$ How should we interpret $\bar{\sigma}$ ?

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Royrik123456

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Royrik123456

#2 h Mar 12, 2025, 8:30 PM

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I believe in this case your extended automorphism $\overline{\sigma}$ is basically the conjugation automorphism.

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ILOVEMYFAMILY

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ILOVEMYFAMILY

#3 h Mar 13, 2025, 12:00 AM

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Royrik123456 wrote:

I believe in this case your extended automorphism $\overline{\sigma}$ is basically the conjugation automorphism.

Can you prove it?

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Royrik123456

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Royrik123456

#4 h Mar 13, 2025, 12:34 AM

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Possibly I can but will give it a try later (I dont particularly think it will be hard). But honestly I dont think it requires proof, because of two reasons.
1. The conjugation automorphism when restricted to $\mathbb{Q}(\sqrt{2})$ actually is the map he mentioned.
2. He never mentioned the extension is unique.

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ILOVEMYFAMILY

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ILOVEMYFAMILY

#5 h Mar 13, 2025, 5:06 AM

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Royrik123456 wrote:

Possibly I can but will give it a try later (I dont particularly think it will be hard). But honestly I dont think it requires proof, because of two reasons.
1. The conjugation automorphism when restricted to $\mathbb{Q}(\sqrt{2})$ actually is the map he mentioned.
2. He never mentioned the extension is unique.

You should check it. I doubt that your answer is incorrect

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Tip_pay

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Tip_pay

#6 h Mar 13, 2025, 6:35 AM

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nguyenalex wrote:

I have the following exercise:

Let $E$ be an algebraic extension of $K$ , and let $F$ be an algebraic closure of $K$ containing $E$ . Prove that if $\sigma : E \to F$ is an embedding such that $\sigma(c) = c$ for all $c \in K$ , then $\sigma$ extends to an automorphism of $F$ .

Consider the set of all pairs $(L, \tau)$ , where $L$ is an intermediate field such that $E \subseteq L \subseteq F$ , and $\tau : L \to F$ is a $K$ -embedding extending $\sigma$ , that is, $\tau|_E = \sigma$ . This set is nonempty, since the pair $(E, \sigma)$ satisfies the conditions. We define a partial order $(L_1, \tau_1) \leq (L_2, \tau_2)$ if $L_1 \subseteq L_2$ and $\tau_2|_{L_1} = \tau_1$ . Given a chain ${(L_i, \tau_i)}$ , the union $L = \bigcup L_i$ is a field, and we can define $\tau : L \to F$ as $\tau(x) = \tau_i(x)$ if $x \in L_i$ . Since the chain is constraint-compatible, $\tau$ is a well-defined $K$ -embedding extending $\sigma$ . Thus, every chain has an upper bound. By Zorn's lemma, there exists a maximal element $(M, \rho)$ , where $M \subseteq F$ and $\rho : M \to F$ is a $K$ -embedding such that $\rho|_E = \sigma$

Suppose that $M \neq F$ , i.e., there exists $\alpha \in F \setminus M$ . Since $F$ is algebraic over $K$ and $K \subseteq M$ , the element $\alpha$ is algebraic over $M$ . Let $p(x) \in M[x]$ be the minimal polynomial of $\alpha$ over $M$ , it is irreducible over $M$ . Applying $\rho$ to the coefficients of $p(x)$ , we obtain the polynomial $\rho(p)(x) \in \rho(M)[x]$ . Since $F$ is algebraically closed, $\rho(p)(x)$ has a root $\beta \in F$ , that is, $\rho(p)(\beta) = 0$ . Since $\rho : M \to \rho(M)$ is an isomorphism and $p(x)$ is irreducible over $M$ , the polynomial $\rho(p)(x)$ is irreducible over $\rho(M)$ . Therefore, there is an embedding $\tau : M(\alpha) \to F$ such that $\tau|_M = \rho$ and $\tau(\alpha) = \beta$ (e.g., $M(\alpha) \cong \rho(M)(\beta)$ ). This means that $\rho$ extends to $\tau$ on a larger field $M(\alpha)$ , where $M \subsetneq M(\alpha) \subseteq F$ , which contradicts

It remains to show that $\tau': F \to F$ is an automorphism. Since $\tau'$ is an embedding, it is injective. The image of $\tau'(F)$ is an algebraically closed subfield of $F$ containing $K$ . But $F$ is the minimal algebraic closure, so $\tau'(F) = F$

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Royrik123456

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Royrik123456

#7 h Mar 13, 2025, 8:58 AM

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ILOVEMYFAMILY wrote:

Royrik123456 wrote:

Possibly I can but will give it a try later (I dont particularly think it will be hard). But honestly I dont think it requires proof, because of two reasons.
1. The conjugation automorphism when restricted to $\mathbb{Q}(\sqrt{2})$ actually is the map he mentioned.
2. He never mentioned the extension is unique.

You should check it. I doubt that your answer is incorrect

I don't know. But the conjugation automorphism when restricted here, gives the mapping he wants. I dont aee why that cane be incorrect.

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Royrik123456

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Royrik123456

#8 h Mar 13, 2025, 9:05 AM

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So as I am given to understand the author of this question is not interested in the proof but a specific case of this statement..

This post has been edited 1 time. Last edited by Royrik123456, Mar 13, 2025, 9:06 AM

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ILOVEMYFAMILY

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ILOVEMYFAMILY

#9 h Mar 13, 2025, 9:43 AM

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Tip_pay wrote:

nguyenalex wrote:

I have the following exercise:

Let $E$ be an algebraic extension of $K$ , and let $F$ be an algebraic closure of $K$ containing $E$ . Prove that if $\sigma : E \to F$ is an embedding such that $\sigma(c) = c$ for all $c \in K$ , then $\sigma$ extends to an automorphism of $F$ .

Consider the set of all pairs $(L, \tau)$ , where $L$ is an intermediate field such that $E \subseteq L \subseteq F$ , and $\tau : L \to F$ is a $K$ -embedding extending $\sigma$ , that is, $\tau|_E = \sigma$ . This set is nonempty, since the pair $(E, \sigma)$ satisfies the conditions. We define a partial order $(L_1, \tau_1) \leq (L_2, \tau_2)$ if $L_1 \subseteq L_2$ and $\tau_2|_{L_1} = \tau_1$ . Given a chain ${(L_i, \tau_i)}$ , the union $L = \bigcup L_i$ is a field, and we can define $\tau : L \to F$ as $\tau(x) = \tau_i(x)$ if $x \in L_i$ . Since the chain is constraint-compatible, $\tau$ is a well-defined $K$ -embedding extending $\sigma$ . Thus, every chain has an upper bound. By Zorn's lemma, there exists a maximal element $(M, \rho)$ , where $M \subseteq F$ and $\rho : M \to F$ is a $K$ -embedding such that $\rho|_E = \sigma$

Suppose that $M \neq F$ , i.e., there exists $\alpha \in F \setminus M$ . Since $F$ is algebraic over $K$ and $K \subseteq M$ , the element $\alpha$ is algebraic over $M$ . Let $p(x) \in M[x]$ be the minimal polynomial of $\alpha$ over $M$ , it is irreducible over $M$ . Applying $\rho$ to the coefficients of $p(x)$ , we obtain the polynomial $\rho(p)(x) \in \rho(M)[x]$ . Since $F$ is algebraically closed, $\rho(p)(x)$ has a root $\beta \in F$ , that is, $\rho(p)(\beta) = 0$ . Since $\rho : M \to \rho(M)$ is an isomorphism and $p(x)$ is irreducible over $M$ , the polynomial $\rho(p)(x)$ is irreducible over $\rho(M)$ . Therefore, there is an embedding $\tau : M(\alpha) \to F$ such that $\tau|_M = \rho$ and $\tau(\alpha) = \beta$ (e.g., $M(\alpha) \cong \rho(M)(\beta)$ ). This means that $\rho$ extends to $\tau$ on a larger field $M(\alpha)$ , where $M \subsetneq M(\alpha) \subseteq F$ , which contradicts

It remains to show that $\tau': F \to F$ is an automorphism. Since $\tau'$ is an embedding, it is injective. The image of $\tau'(F)$ is an algebraically closed subfield of $F$ containing $K$ . But $F$ is the minimal algebraic closure, so $\tau'(F) = F$

Thanks, but I already solved it. I'm stuck at finding such $\bar{\sigma}$ .

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ILOVEMYFAMILY

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ILOVEMYFAMILY

#10 h Mar 13, 2025, 9:44 AM

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Royrik123456 wrote:

So as I am given to understand the author of this question is not interested in the proof but a specific case of this statement..

Yes. But I can find such $\bar{\sigma}$ .

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Royrik123456

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Royrik123456

#11 h Mar 13, 2025, 11:38 AM

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ILOVEMYFAMILY wrote:

Royrik123456 wrote:

So as I am given to understand the author of this question is not interested in the proof but a specific case of this statement..

Yes. But I can find such $\bar{\sigma}$ .

Exactly. And I found one that works.

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ILOVEMYFAMILY

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ILOVEMYFAMILY

#12 h Mar 14, 2025, 7:04 AM

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Royrik123456 wrote:

ILOVEMYFAMILY wrote:

Royrik123456 wrote:

So as I am given to understand the author of this question is not interested in the proof but a specific case of this statement..

Yes. But I can find such $\bar{\sigma}$ .

Exactly. And I found one that works.

The complex conjugation does not work as it fixes real number hence in particular does not extend the particular $\bar{\sigma}$ .

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GreenKeeper

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GreenKeeper

#13 h Mar 14, 2025, 8:39 AM

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The proof of the existence of $\bar{\sigma}$ is purely existential, by the Zorn lemma argument above. And it's highly non-unique, in fact there's $2^{\aleph_0}$ of them, and none of them is gonna have a "closed form" in any reasonable sense.

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Royrik123456

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Royrik123456

#14 h Mar 14, 2025, 1:47 PM

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ILOVEMYFAMILY wrote:

The complex conjugation does not work as it fixes real number hence in particular does not extend the particular $\bar{\sigma}$ .

I see now why you maybe confused. I never meant complex conjugation. In fact complex conjugation can never be meant because $\mathbb{Q}(\sqrt{2})$ is a totally real field. What I mean't by the map is Galois conjugation. You send it to one of it's Galois conjugates. And I believe the map is not unique.

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Royrik123456

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Royrik123456

#15 h Mar 14, 2025, 1:49 PM

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GreenKeeper wrote:

The proof of the existence of $\bar{\sigma}$ is purely existential, by the Zorn lemma argument above. And it's highly non-unique, in fact there's $2^{\aleph_0}$ of them, and none of them is gonna have a "closed form" in any reasonable sense.

exactly. The proof doesn't give a closed form as such as it is not a constructive proof. But sending to one of it's Galois conjugates works as a reasonable extension I believe in the explicit example he gave.

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