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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
Generating functions and recursions smelling from 1000 km
Assassino9931   13
N 8 minutes ago by watery
Source: IMC 2022 Day 1 Problem 3
Let $p$ be a prime number. A flea is staying at point $0$ of the real line. At each minute,
the flea has three possibilities: to stay at its position, or to move by $1$ to the left or to the right.
After $p-1$ minutes, it wants to be at $0$ again. Denote by $f(p)$ the number of its strategies to do this
(for example, $f(3) = 3$: it may either stay at $0$ for the entire time, or go to the left and then to the
right, or go to the right and then to the left). Find $f(p)$ modulo $p$.
13 replies
Assassino9931
Aug 5, 2022
watery
8 minutes ago
more rafinament limits integrals
teomihai   0
2 hours ago
Let$ f:[0,1]->R$ continously function with $f'$ bounded
Find $$\lim_{n \rightarrow \infty}n(n\int_{0}^{1}x^{n}f(x)dx-f(1))$$
0 replies
teomihai
2 hours ago
0 replies
Differentiation Marathon!
LawofCosine   188
N 3 hours ago by vaishnav
Hello, everybody!

This is a differentiation marathon. It is just like an ordinary marathon, where you can post problems and provide solutions to the problem posted by the previous user. You can only post differentiation problems (not including integration and differential equations) and please don't make it too hard!

Have fun!

(Sorry about the bad english)
188 replies
LawofCosine
Feb 1, 2025
vaishnav
3 hours ago
Derivative of Normalization Map has null space of dimension 1
myth17   2
N 4 hours ago by myth17
Let $f(\vec{x}) = \frac{\vec{x}}{||\vec{x}||}$ be defined on $\mathbb{R}^n \setminus \{\vec{0}\}$. Show that the dimension of the kernel of $Df_{\vec{x}}$ for any $\vec{x} \in \mathbb{R}^n \setminus \{\vec{0}\}$ is $1$.
2 replies
myth17
Yesterday at 5:16 PM
myth17
4 hours ago
No more topics!
find the isomorphism
nguyenalex   14
N Mar 14, 2025 by Royrik123456
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 $\bar{\sigma}$?
14 replies
nguyenalex
Mar 12, 2025
Royrik123456
Mar 14, 2025
find the isomorphism
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nguyenalex
152 posts
#1
Y by
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
112 posts
#2
Y by
I believe in this case your extended automorphism $\overline{\sigma}$ is basically the conjugation automorphism.
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ILOVEMYFAMILY
641 posts
#3
Y by
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
112 posts
#4
Y by
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
641 posts
#5
Y by
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
1724 posts
#6
Y by
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
112 posts
#7
Y by
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.
Z K Y
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Royrik123456
112 posts
#8
Y by
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
641 posts
#9
Y by
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
641 posts
#10
Y by
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$\bar{\sigma}$.
Z K Y
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Royrik123456
112 posts
#11
Y by
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
641 posts
#12
Y by
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 $\bar{\sigma}$.
This post has been edited 1 time. Last edited by ILOVEMYFAMILY, Mar 14, 2025, 7:05 AM
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GreenKeeper
1679 posts
#13
Y by
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
112 posts
#14
Y by
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
112 posts
#15
Y by
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.
This post has been edited 1 time. Last edited by Royrik123456, Mar 14, 2025, 1:51 PM
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