How to solve Exercise 2? How does embedding, I suppose in meaning of embedding defined before in the post linked as order-preserving ring homomorphism, $e:\mathbb{Q}\,{\rightarrow}\,K$, imply that for all ordered fields $K$ there exists a subfield of $K$ isomorphic to $\mathbb{Q}$?
124060
I'm going through the first chapter in a text on real analysis, which contains preliminaries on ordered fields, the real numbers, etc. Supposedly I had learned about such things already, in calculus, but I thought it wouldn't hurt to go over it again.  Up until the paragraph which is the subject of my question, everything is thoroughly proven and examples are provided. However the following excerpt just goes through facts, whose proof I cannot conceive:      Although $\mathbb Q$ is an archimedean field, these properties cannot be used to define $\mathbb Q$ since $\\$ there are many Archimedean ordered fields. What distinguishes $\mathbb Q$ from the other $\\$Archimedean ordered fields is that $(\mathbb Q,&lt;)$ is the $smallest$ ordered field in the following sense: $\\$ if $(X,\prec)$ is an ordered field, then $X$ contains a sub-field which is (field) isomorphic to $\mathbb Q$.$\\$ Furthermore such an isomorphism preserves the order relation.   I've concatenated the paragraph a bit, so the text isn't idem from the book (if anyone wishes to know it's Phillips, An Introduction to Analysis and Integration Theory, Dover Publications). Also $\mathbb Q$ contains no proper subfields, and this can be verified by the apparently easy-to-prove fact that, if a field has characteristic zero, then it contains a subfield isomorphic to $\mathbb Q$. Then it remains to prove the preservation of order.  I'm guessing that some concepts from abstract algebra or field theory would easily suffice, but at the moment such topics are a bit over my head, so all I can think of doing is actually constructing the isomorphism $\phi:\mathbb Q\to\hat{\mathbb Q}$, where $\hat{\mathbb Q}$ is a certain subfield of  $X$ in such a way that field operations are preserved, but I really can't come up with anything. So the question is: how do I construct this isomorphism, or, if there's a better way of proving this, how is it done? Thanks for any help.  Edit: Assume the existence of $(\mathbb Q,&lt;)$ as an archimedean ordered field.
6082
I was reading an example about describing homomorphisms and I'm having a bit of trouble with this one.  Show there is no group homomorphism $f : \mathbb{Z}_{10} \to \mathbb{Z}_{25}$ such that $f(1) = 3$. The first step of the proof is  $|1| = 10$ in $\mathbb{Z}_{10}$, so $|f(1)| \in\{ 1, 2, 5, 10\}$.  Does this mean that $|f(1)|$ can be $1$, $2$, $5$ or $10$? I see that $|1| = 10$ since $1$ is a generator of $\mathbb{Z}_{10}$, and that $|f(1)|$ seems to be a factor of $10$. If $f(1)$ is the image of $1$ under $f$, how would I find the order of $f(1)$?  Thanks for your help.
122229
I was preparing for an area exam in analysis and came across a problem in the book Real Analysis by Haaser &amp; Sullivan. From p.34 Q 2.4.3, If the field F is isomorphic to the subset S' of F', show that S' is a subfield of F'. I would appreciate any hints on how to solve this problem as I'm stuck, but that's not my actual question.  I understand that for finite fields this implies that two sets of the same cardinality must have the same field structure, if any exists. The classification of finite fields answers the above question in a constructive manner.  What got me curious is the infinite case. Even in the finite case it's surprising to me that the field axioms are so "restrictive", in a sense, that alternate field structures are simply not possible on sets of equal cardinality. I then started looking for examples of fields with characteristic zero while thinking about this problem. I didn't find many. So far, I listed the rationals, algebraic numbers, real numbers, complex numbers and the p-adic fields. What are other examples? Is there an analogous classification for fields of characteristic zero?
75584
Let me ask Prob.F6 in Pinter's algebra book.  'Let $p$ be a prime and $f$ be a homomorphism from $G$ to $H$ where $G$ and $H$ are groups.  If range of $f$ has an element of order $p$, then $G$ has an element of order $p$.'   If $f$ is injective, life gets easier. But without the injectivity of the homomorphism, i feel lost... Would somebody help please?
15412
I was having some trouble finding an explicit isomorphism between $\mathbb{Z}[i]/(7)$ and $\mathbb{Z}[\sqrt{-2}]/(7)$.  $\textbf{What I have noticed is}:$   7 is a prime element in $\mathbb{Z}[i]$ so $(7)$ is a maximal ideal in $\mathbb{Z}[i]$ and $\mathbb{Z}[i]/(7)$ is  a field.  7 is also a prime element in $\mathbb{Z}[\sqrt{-2}]$   $\textbf{What I have been trying to do is this}$   Find a surjective homomorphism between $\mathbb{Z}[i]/(7) \rightarrow \mathbb{Z}[\sqrt{-2}]/(7)$. Since $\mathbb{Z}[i]/(7)$ was a field, the kernel of this homomorphism will be either the whole ring or just $0$. In the latter case it would be a isomorphism.    I am having trouble finding this surjective homomorphism:  I have noticed that $\bar{{i}}^{2}=-1$ so the image of $\bar{{i}}$ must be sent to something whose square is $-1$. Any help would be appreciated. I may be missing some obvious insight.
37211
I'm trying to find the ring homomorphisms from $\mathbb{Z}_n$ to $\mathbb{Z}_m$.    What I think is the answer is two cases:   1) If $m \gt n $ there is no ring homomorphism.  2) Otherwise, the homomorphism sends each element $x$  in $\mathbb{Z}_n$ to $(x \bmod m)$ in $\mathbb{Z}_m$.  Is this correct?  EDIT: Part of the ring homomorphism definition is that 1 is sent to 1
11027
Is there a proper subfield $K\subset \mathbb R$ such that $[\mathbb R:K]$ is finite? Here $[\mathbb R:K]$ means the dimension of $\mathbb R$ as a $K$-vector space.  What I have tried:   If we can find a finite subgroup $G\subset Gal (\mathbb C/\mathbb Q)$ such that $G$ contains the complex conjugation, it will be done by letting $K$ be the fixed field of $G$. But I don't know whether such a group exists. Maybe we can start with finding a suitable subgroup of $Gal(\bar{\mathbb Q}/\mathbb Q)$ and then lift it to $Gal(\mathbb C/\mathbb Q)$, where $\bar{\mathbb Q}$ denotes the algebraic closure of $\mathbb Q$. By isomorphism extension theorem, we can find many automorphisms of $\mathbb C$, none of them carries $\mathbb R$ to itself except for the identity. This is because $Gal(\mathbb R/\mathbb Q)$ is the trivial group. For example, now suppose $\{x_\alpha\}\subset \mathbb R$ is a transcendence basis over $\mathbb Q$. Let $\sigma$ be a permutation of $\{x_\alpha\}$, then by the isomorphism extension theorem, $\sigma$ extends to an automorphism of $\mathbb C$, which we still denote by $\sigma$. Then $L=\sigma(\mathbb R)$ is a copy of $\mathbb R$ and $\mathbb R$ is algebraic over $K=\mathbb R\cap L$. How large can $K$ be? Is it possible that $[\mathbb R:K]$ is finite?   Does anyone has some ideas?  Thanks!
62097
Let $\phi\colon R \to R'$ be a ring homomorphism. Prove that if $R$ is a field then either $\phi$ is an isomorphism or $\phi(r) = 0$ for all $r \in R$.   I am stuck on this problem and don't know where to begin. I feel like I'm very weak in writing proofs.
117529
If we are given a homomorphism $g$ between a field $k$ and an algebraically closed field $\Omega$, and a field $k'$ which is a finite algebraic extension of $k$, how do we extend $g$ to a homomorphism $g'$ from $k'$ to $\Omega$?
12423
This is part of the exercise, I'm stuck with it.  $A$ is a commutative ring with unit.  1) Suppose we have an homomorphism $\phi : A^{m} \to A^{n}$ surjective. Is true that $m \geq n $ ?  2) Suppose we have an homomorphism $\phi : A^{m} \to A^{n}$ injective. Is true that $m \leq n $ ?
112363
This is a question I made up, but couldn't solve even after some days' thought. Also if any terminology is unclear or nonstandard, please complain.  Given groups $G$ and $H$, we say that $G$ can be embedded in $H$ if there exists an injective homomorphism $\varphi : G \to H$. (Note that the image $\varphi(G)$ is then isomorphic to $G$.) I am interested in the situation where a pair of groups $G$ and $H$ can be embedded in each other. Of course, this is guaranteed to be the case when $G \cong H$. But is the converse true? More precisely:     Q1. Do there exist non-isomorphic groups $G$ and $H$ such that each of them can be embedded in the other?   I am interested in this because, in my mind, this question is analogous to the  in set theory. Of course, this view could be too naive or useless. Oh well.  The only "progress" I could make is to create another question. Let $\varphi_G:G \to H$ and $\varphi_H:H \to G$ be a pair of embeddings as in the question. Then the homomorphism $\varphi := \varphi_H \circ \varphi_G : G \to G$ is also injective; i.e., it is an embedding. I can show that the image of this map ($K := \varphi(G)$) is a proper subgroup of $G$ unless $G \cong H$. This leads me to another question:      Q2. Does there exists a group $G$ that is isomorphic to a proper subgroup of itself?   If the answer to this is negative, then so is the case for Q1. Though both of these seem "obviously false", I cannot prove them. Nor can I construct a counterexample. Any suggestions?   Some remarks:   Nothing is inherently special about groups here. I suppose one could ask the same question for rings, fields, or other structures; I focused on this specific question for clarity. I tried to search through Wikipedia and Google books, but I cannot figure out the answer or where I can find the answer.  I have no idea as to how easy or difficult these questions are. If they are trivial/easy (say, the level of a standard undergrad homework exercise), then please give me hints rather than a complete solution :-).
73590
I am currently working to find all ring homomorphisms $\phi : \mathbb{Q}(\sqrt{2})\rightarrow \mathbb{Q}(\sqrt{2})$.   My work so far: Since $\mathbb{Q}(\sqrt{2})$ is a field extension of $\mathbb{Q}$, a field, it is obviously a field as well. Therefore, its only ideals are $\{0\}$ and itself. Since every homomorphism's kernel is an ideal, we have that either $Ker(\phi)=\{0\}$ or $\mathbb{Q}(\sqrt{2})$.   In the latter case, we have the trivial homomorphism. In the former case, we have that the homomorphism is injective (and therefore bijective).   My question: I think it is true that in the former case $\phi(1)=1$, but am not sure why. If this is the case, can I prove inductively that $\phi(x)=x, \forall x\in \mathbb{Q}(\sqrt{2})$?  Thank you in advance for your help.
35224
I have seen that some authors are defing the number field as a subfield of $ \mathbb C$ which is a finite extension of the rational numbers $ \mathbb Q $, while some others without referering to complex numbers $ \mathbb C$ . I think we don't need $ K$ to be a subfield of $ \mathbb C$ in the definition. So, my question is the follwing:     Is it neceserily to define $K$ as a subfield of $ \mathbb C$ or not ?   And if no why ? Is it true that if we omit this in the definition, that then $K$ will turn out to be a subfield of $ \mathbb C$?   I came up with this question, when I saw that in order to define infinite primes in a number field then these are determined by the embeddings $ K \to \mathbb C $  Any idea would be really appreciated.  Thank you in advance.
19687
Let $G$ be a finite group of order $n\ge2$. Which of the following statements are true? a. There always exists an injective homomorphism from $G$ into $S_n$. b. There always exists an injective homomorphism from $G$ into $S_m$ for some $m\lt n$. c. There always exists an injective homomorphism from $G$ into $GL_n(R)$.  I think b. is right (don't know how to prove though). I also think that c. is false. But not sure.
106228
This is related to this .  I just didn't want a prolonged discussion in the comments.  Let $\phi: G \to G'$ be a homomorphism.  Let $G$ be a finite group.  Let $K \leq G$ be the kernel of $\phi$.  Let $I \leq G'$ be the image of $\phi$.  Let $H' \leq G'$.  Find a formula relating the order of $\phi^{-1}(H')$ in terms of $H', I, K$.    Attempt at a solution: $|\phi^{-1}(H')|=|H'\cap I|\cdot |K \cap \phi^{-1}(H')|$.  My justification for this is as follows.  $\phi^{-1}(H')/(K \cap \phi^{-1}(H'))\cong H' \cap I$
113212
True or false. (Prove or give a counterexample.) Let $G$ be a group. Then $|g| = |\phi(g)|$, for all homomorphisms $\phi: G \to G$ and all $g \in G$. Solution. False. $\phi: \mathbb Z_{10} \to \mathbb Z_{12}$ defined by $\phi(x)=0$ for all $x \in \mathbb Z_{10}$ is a counterexample. This function is a homomorphism because $\phi(x+y) = 0 = 0+0 = \phi(x) + \phi(y)$ for all $x, y \in \mathbb Z$ (this function is discussed in problem 3 in assignment 7 as the function sending $[1]$ to $[0]$). The order of $g=1$ is infinity. The order of $\phi(1)=0$ is one.  The problem is that he said $G \to G$ is equivalent to $\mathbb Z_{10} \to \mathbb Z_{12}$, which is think is not right. Explain please? EDIT: Please look at this test and give me your honest opinion,
113060
How many homomorphism are there from $(\mathbb Q^+ , \,\cdot\,)$ to $(\mathbb Z_n , +)$?   Could anyone give a hint how to proceed? I couldn't even start