If $E$ and $F$ are two fields and $f : E \to F$ and $g : F \to E$ be field homomorphisms. Is it true that $E$ and $F$ are isomorphic?  We know that both $f$ and $g$ are injective which prompts the question.  EDIT: As pointed out in comments and other posts, this is not true in general. Are there any conditions on $E$, $F$ which makes this result true? Algebraically closed, characteristic zero, positive etc.?
120534
Are there any nice examples of structures (groups, modules, rings, fields) $A$ and $B$ such that there are embeddings $A → B → A$ while $A \not\cong B$? I would especially like to see an example for modules $A$, $B$. Or is it even true that the existence of such embeddings implies $A \cong B$?  Background: I’m correcting exercises and I wanted to give a counterexample to a failing argument. (Well, I’m not certain it fails, but I’m pretty sure it does and it’s not sufficiently justified at least.)
40420
My question has been asked for groups I think, but there are no free fields, and fields seem more rigid than groups, so I feel like here the answer may be positive.  The question is the following : suppose we have field morphisms $L\to K\to L$ (I don't know if this is the used convention but I take my morphisms to have $f(1)=1$; and so they are embeddings), must we then have $L\simeq K$ ?   Now obviously the first thing is that, as in Cantor-Bernstein's theorem, the morphism $L\to K$ need not be itself the isomorphism (consider the field $\mathbb{Q}(X_n)_{n\in \mathbb{N}}$ which has plenty of non surjective embeddings into itself). So one would have to construct the isomorphism, not prove that the given morphisms are isomorphisms.  More generally, the question that naturally arises from this is : given a class of universal algebras (of one given type), under what conditions do we have "$A$ can be embedded in $B$ and conversely $\implies A\simeq B$" ?   The answer for the class of fields is the answer to my first question, and for the class of groups, the answer is no (see the question "When can two groups be embedded in eachother ?")
14834
I have two ring homomorphisms $f,g\colon \mathbb{Q}\to X$. I know that $f=g$ on the integers. How can I show that $f$ and $g$ agree on the rationals?   My attempt:  let $x,y \in \mathbb{Z}$, $y\neq 0$.  $$\begin{align*} f(x/y)&amp;=f(x)f(1/y) &amp;&amp;\text{because }f\text{ is a ring homomorphism}\\  &amp;=g(x)f(1/y) &amp;&amp;\text{because }f\text{ and }g\text{ agree on the integers} \end{align*}$$  I do not know how to rewrite $f(1/y)$ so that I can replace it with $g(1/y)$. Any hints will be much appreciated. Thanks.
82712
Is my proof ok?  Let $f:G\to G^{\prime}$ be a group homomorphism and let $H\lt G$. $Im(H) = \{f(x):x\in H\}$. To show that $Im(H)$ is a group, it suffices to show that $f(x)f(y)^{-1}\in Im(H)$. $$f(x)f(y)^{-1}=f(xy^{-1})\in Im(H)\text{ because } xy^{-1}\in H$$  $Ker(f)=\{x:f(x)=e^{\prime}\}$ ($e^{\prime}$ is the identity in $G^{\prime}$). $$f(xy^{-1})=f(x)f(y)^{-1}=e^{\prime}(e^{\prime})^{-1}=e^{\prime}\in Ker(f)$$.
26441
A question that may sound very trivial, apologies beforehand. I am wondering why $( \mathbb{Q}_{&gt;0} , \times )$ is not isomorphic to $( \mathbb{Q} , + )$. I can see for the case when $( \mathbb{Q} , \times )$, not required to be positive, one can argue the group contains elements with order 2 (namely all negatives). In the case of the requirement for all rationals to be positive this argument does not fly. What trivial fact am I missing here?
122001
Here we have $F$ a field and $f: \mathbb{Z} \to F$ a ring epimorphism. We are to prove that $F$ is a finite field with non-zero characteristic.   I know that since $f$ is an epimorphism, we have that $\forall x\in F$, $\exists y \in \mathbb{Z}$ such that $f(x) = y$. (As an epimorphism is a surjective ring homomorphism).   I am unsure about two things: Am I to prove that F is finite, or just prove that it has a non-zero characteristic. Either way, I could use some guidance. Thank you :)   PS This is for HW, so I only require a good hint about how to approach the problem.
26776
Let $E|F$ be an algebraic field extension and a ring $K$ such that $F\subseteq K\subseteq E$. It is true that $K$ is a field?
67393
Is $\mathbb{Z}[x]/(x^2+1)$  isomorphic to  $\mathbb{Z}[i]$?   My attempt is that try to define a mapping $g$ from $\mathbb{Z}[x]$ to $\mathbb{Z}[i]$ by $g(f(x))= f(i)$, for $f(x)\in\mathbb{Z}[x]$. If it is possible then $\ker g$ is $(x^2+1)$? Am I on the right track? Please Help.
95808
I recently came to the topic of field extensions in my abstract algebra course, and there has been a slight issue which has been bothering me that I was hoping I might be able to clear up.  We have defined an extension field for a field F to be a field E such that $F \subseteq E$ and that $F$ is a field under the operations of $E$ restricted to $F$.   Sounds easy enough and I realize that we have been using objects like this for a long time. For example we know that $\mathbb{C}$ is an extension field of $\mathbb{R}$.  Something that has been bothering me a little bit though is that we have started proving theorems where we need to construct extension fields, but these extension fields don't seem to contain the original field $F$ but rather an isomorphic copy of $F$.  For example if our field was $F$, then $F[x]/(p(x))$ is a field if $p(x)$ is irreducible, which contains a subfield isomorphic to $F$. It seems strange that in the theorems (Gallian's Text) that $F[x]/(p(x))$ is considered an extension field for $F$ even though it doesn't really contain $F$ as a subset, but rather another set which is isomorphic.  I don't think I would have normally though this as being much of a problem, but I remember that earlier in the text Gallian seems to mention that even when structures are isomorphic and that they behave essentially the same, that we need to keep in mind that they are not exactly the same.  If this distinction does matter, why not make the definition of an extension field just say that $E$ is an extension field of $F$ if $E$ has a subfield isomorphic to $F$? This would seem to include all cases. Is this largely a historical issue related to how mathematicians thought about isomorphic structures in the past?
68159
Let $f$: $Z_m \rightarrow G$, $f(n)$ = $a^n$. Show that f is a function and an homomorphism.  Doing a review on the whole semester, and a little refresh on this topic would be great.   By definition I know $G$ has the same order of $Z_m$, but not sure how to attack this problem with ease.   I also know that cyclic groups of the same order are isomorphic.   (I cannot use the fact that cyclic groups of the same order are isomorphic, that's why I ask for another approach, because I can't use theorems or lemmas that have not been discussed in class.)
86729
What can be the general statements made, when two finite groups can be isomorphic to each other: $\mathbb{Z}_{pq}$ and $\mathbb{Z}_{p} \times \mathbb{Z}_{q}$  Say  $$\mathbb{Z}_{pq} \simeq\mathbb{Z}_{p} \times \mathbb{Z}_{q},$$  what are the conditions for $p$ and $q$ and their factorization relations?  How do we show this in the most elegant way? And the intuitive way?
113811
Let $\phi : C \to D$ be an injective ring homomorphism such that $\phi(1)=1$, where $1$ denotes the constant function $1$ and $C$, $D$ are the rings of continuous, respectively differentiable functions on $\mathbb{R}$. If $f \in C ,g \in D$ are such that $\phi(f)=g$, and $t_0\in \mathbb R$, is it true that $\phi (f-g(t_0))=g-g(t_0)$ ? This is used on page 944, . And why if the image of $\phi $ consists of only constant functions, then $\phi$ is not injective ? Please help. Thanks in advance.
40530
Consider $f:A \to B$ and $g: B \to C$. If I know that both $f$ and $g$ are injective, then can I state that $$g  \circ f:A \to C$$ is injective?
9546
Let's say $f$ is an isomorphism $f:G  \rightarrow G'$, where $G$ and $G'$ are multiplicative groups.  Then for $x\in G$, if $f(x) = x' \in G'$. Do we have always have $\text{ord}(x) = \text{ord}(x')$? Why?
96113
My questions are:     $1.$ Is there a field morphism $\Bbb R(X) \hookrightarrow \Bbb R$ ?      $2.$ If the answer to $1.$ is "yes", are $\Bbb R$ and $\Bbb R(X)$ isomorphic as fields?   $ $    $ $  For $1.$, here are some comments:   This is true for $\Bbb C$, see . The key point here is the fact that $\Bbb C$ is algebraically closed, of cardinality $2^{\aleph_0}$, and of characteristic $0$. The same argument doesn't work for $\Bbb R$. We can actually find a subfield $F$ of $\Bbb R$ such that $F \cong F(X) \cong F(X,Y) \cong \cdots$, see .  that $\Bbb Q(X)$ does not embed in $\Bbb Q$ (as a field).  This is equivalent to: Is there an injective ring morphism $\Bbb R[X] \hookrightarrow \Bbb R$ ? (By taking the fraction fields).   As for 2. :   They are isomorphic as abelian groups, and actually as $\Bbb Q$-vector spaces. They are not isomorphic as $\Bbb R$-algebras, because they haven't the same transcendence degree over $\Bbb R$. Let's try to see how to construct such a field isomorphism. We could try$^{[1]}$ to find a subset $B$ of $\Bbb R$ such that $\Bbb R = \Bbb Q(B)$ and $C \subsetneq B \implies \Bbb R \neq \Bbb Q(C)$.  The set $B$ must contain transcendental elements, pick such an $x_0 \in B$.  Then $\Bbb R = \Bbb Q(B) = \Bbb Q(B \setminus \{x_0\})(x_0)$ and we could try$^{[2]}$ to find a field isomorphism $\Bbb Q(B \setminus \{x_0\}) \cong \Bbb Q(B)$, so that $\Bbb Q(B \setminus \{x_0\})(x_0) \cong \Bbb Q(B)(x_0) \cong \Bbb R(X)$.  For ${[1]}$, I tried to use Zorn's lemma on $\mathscr B = \{B \subset \Bbb R \mid \Bbb R = \Bbb Q(B)\}$ and the partial order $B_1 ≤ B_2 \iff B_1 \supset B_2$. But this is not clear to me why this should be an inductive set.  For ${[2]}$, my idea was to define $f : \Bbb Q(B \setminus \{x_0\}) \cong \Bbb Q(B)$ by choosing a bijection $b : B \setminus \{x_0\} \to B$ (these sets should be uncountable), and then trying to extend $f\vert_{B \setminus \{x_0\}} = b$ to a field morphism.     Edit for $^{[1]}$: in  article (MINIMAL GENERATING SETS OF GROUPS, RINGS, AND FIELDS, by LORENZ HALBEISEN, MARTIN HAMILTON, AND PAVEL RUZICKA) the theorem 2.4. proves that $\Bbb R$ has no minimal generating set as field (or $\Bbb Q$-algebra, i.e. as ring). The argument uses Artin-Schreier theorem. Minimal generating sets of modules are   .  I think that $B$ is a minimal generating set of a subfield $K \subset \Bbb R$ over $\Bbb Q$ if and only if any element of $B$ cannot be written as a rational fraction of other elements of $B$. Under these conditions and using (transfinite) induction, we can probably prove (and this was my first idea) that any function $f: B \to \Bbb C$ such that $b$ and $f(b)$ are conjugate over $\Bbb Q$ (i.e. have the same minimal polynomial) extends to a (unique) field morphism $\sigma : K = \Bbb Q(B) \to \Bbb C$.  Actually, the collection $\mathscr B$ defined above doesn't have to be an inductive set, at least if we replace $\Bbb R$ by $K =\Bbb Q(\sqrt 2, \sqrt[4]{2},\sqrt[8]{2},\dots)$, because the descending chain $\left(E_n = \{\sqrt[2^m]{2} \mid m \geq n\}\right)_{n \geq 1}$ satisfy $K = \Bbb Q(E_n)$, but $K \neq \Bbb Q\left(\bigcap_{n≥1} E_n\right)=\Bbb Q$, so the chain has no upper bound w.r.t. $≤$ in $\mathscr B$.
50393
It is well-known that is $E$ is a simple finite extension over $F$, then it has finite many intermediate fields.   Actually Minle's notes says that if an finite extension has finite intermediate fields, then it is simple! I am curious about this fact, could someone give a proof? Thanks!
98612
Let $F$ be a field. When happens that the additive group of $F$ is isomorphic to the multiplicative group?  It is easily to work out that $F$ must have characteristic $0$, but then what?