Give an example of fields $k\subseteq K\subseteq L $, and $l\subseteq L$, for which $l/k$ and $L/K$ are algebraic, $k$ is algebraically closed in $K$, and $lK = L$, but $l$ is not algebraically closed in $L$.   This question has already been posted here , but has not yet had an answer, the truth is that I can not find such an example, could someone please help me? Thank you very much.
48704
This is problem 3.18 from Patrick Morandi's Field and Galois Theory.     Give an example of fields $k \subseteq K \subseteq L$ and $l \subseteq L$ for which $l/k$ and $L/K$ are algebraic, $k$ is algebraically closed in $K$, and $lK = L$, but $l$ is not algebraically closed in $L$.   This question has been asked twice, namely  and , though neither had any answers, and only some inconclusive comments. Normally I wouldn't ask the question again, but after struggling with it for weeks at this point, I emailed Morandi, and he gave me some hints.  He told me that he remembers finding an example, but that he isn't sure. He told me that it would need to be over characteristic $p$ and that the extensions (at least some of them I suppose) could not be separable. Moreover, he told me that $K = k(x)(a)$, where $a$ is a root of an irreducible polynomial $f(x,y) \in k(x,y)$ (I'm not 100% what the root of a two variable polynomial is). Finally, he said that $l$ is an algebraic extension of $k$, where $f$ becomes reducible over $l$. Still however, I can't come up with anything. I was hoping that with these extra hints, someone would be able to figure it out.  He did point out that there is a chance that there is an error in the question, but I haven't managed to prove that such an example cannot exist.
107053
Can you find an example of non-isomorphic fields which embed inside each other?   Most probably we can't but I am looking for extraordinary answer...
98032
I am trying to prove that if $L/K$ is an algebraic extension and if $\alpha \in L$, then    $\alpha$ is separable over $K$ if $\mathrm{char}(K)=0$. This is clear because $K$ is perfect which in turn implies that $L/K$ is seperable .  Now if $\mathrm{char}(K)=p$ is prime, then the statement is: $\alpha$ is separable if and only if $K(\alpha) =K(\alpha^p)$.  This problem is from a past examination and looks very interesting, but somehow I am not able to connect what's happening with the extension of $K$ with $\alpha$ and $\alpha^p$ .   I need some help here.    This somehow makes me think that the minimal polynomial of $\alpha$ and $\alpha^p$ have different roots and $\alpha$ is given by some root of minimal polynomial of $\alpha^p$ .   Note that I am using the definition of separability as follows :      An extension $L/K$ is called separable if for every $\alpha \in L$ the minimal polynomial of $\alpha$ has distinct roots in $L$ .    Thanks for helping.
124213
Let $K$ be a field and $f,g$ irreducible polynomials in $K[X]$, is there a nice iff condition for $K[X]/(f)\cong K[X]/(g)$?  ($\cong$ denotes an isomorphism that is the identity on restriction to $K$).  Thoughts: It is sufficient that they are $K^\times$ multiples of each other. I'd hoped this was necessary but it isn't as $\mathbb{Q}[X]/(X^2-2)\cong\mathbb{Q}(\sqrt2)=\mathbb{Q}(\sqrt2+1)\cong\mathbb{Q}[X]/(X^2-2X-1)$ with the middle two fields viewed as subfields of $\mathbb{C}$. It is necessary that they have the same degree. Please let me know if there are any other simple necessary conditions. Thanks!
70911
Consider following two statements from same book of Cohn: Basic Algebra.  Proposition : If $k$ is any field and $\mathcal{F}$ is a set of polynomials over $k$, then any two minimal splitting fields for $\mathcal{F}$ over $k$ are isomorphic. Theorem: If $k$ is any field then $k$ has algebraic closure and any two algebraic closures are isomorphic.  My question is, if we take $\mathcal{F}$ to be set of all the monic irreducible polynomials over $k$, then by 7.3.2, doesn't it imply that any two algebraic closures are isomorphic? Further, just after Theorem 7.3.2, the author gives also the existence of minimal splitting field for $\mathcal{F}$ (it is not visible in link). Then doesn't it imply that algebraic closure of any field exists and is unique up to isomorphism? In other words, I was thinking that Theorem 7.3.2 and existence of minimal splitting field actually gives proof of theorem 11.8.3 above. Why it was differently stated and proved I didn't understand. Are these two theorems really different?
38028
Let $K$ be a subfield of $\mathbb{C}$ not contained in $\mathbb{R}$. Show that $K$ is dense in $\mathbb{C}$.     completely stuck on it. can I get some help please.
34495
I have recently started studying algebraic field extensions and I got to know that algebraic closures $\overline{F}$ of finite fields $F$ are infinite. Therefore, I've asked myself the following question and couldn't come up with an answer.  Suppose $F$ was a finite field, i.e. $|F| &lt; \infty$. Is it possible to construct an algebraic extension $F'$ of $F$ sucht that $|F'| = \infty$ and $F' \subsetneq \overline{F}$? Meaning, that $F'$ is not algebraically closed.
121570
I am trying to work through Hartshorne's book and while working through one of the exercises, I need to show the following:  Let $k$ be an algebraically closed field. Let $Y \subseteq A^3$ be the set $Y=\{(t,t^2,t^3)\;|\;t \in k\}$. Show that $Y$ is an affine variety.  The rest of the problem I have completed. I'm just not sure how to show that this set is closed and irreducible. I assume that they intend for $A^3$ to have the Zariski topology.
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?
2679
For which $c$ will the set $K(c) = \{a + b\sqrt{c} : a, b \in \mathbb{Q}\}$ be a field? I know for example, that $K(\frac{2}{3})$ will be one, I am just wondering what the most general result of this type is.
116277
Let $\mathbb{F}, \mathbb{F'}$ be fields, $\mathbb{F} \subset \mathbb{F'}$, and $\alpha \in \mathbb{F'}$. Assume $\mathbb{F(\alpha)}$ is finitely generated algebra over $\mathbb{F}$.   Is this true: $(\mathbb{F}(\alpha):\mathbb{F}) &lt; \infty$ ?   First i thought it's a consequence of this theorem:   But we don't know if $\alpha$ is algebraic or not.
70370
Let $ K/ L /F $ be fields. If $K / L$ is Galois and $ L / F $ is Galois, then $ K / F$ is Galois. We mentioned this very quickly in today's class without justifying. But I have trouble seeing this. Any help is appreciated.  Our definition for $ K / F$ being Galois is the $F$-automorphism group $G = G( K / F) : = \{ \sigma \in aut (G) : \sigma_F = id_F \}$ fixes only $F$, i.e. $ \{ x \in K : \sigma (x) = x \text { for all } \sigma \in G  \} = F$.  But we also showed that $ K /F$ is Galois if and only if $ K/F$ is normal and separable.
53896
I need help proving  this stunning result.  Let me state it this way.  Let  $\mathbb L$ be a field and $\mathbb K$ be a field with $\mathbb K \subset \mathbb L$  Let $A$ be the set of algebraic numbers ie $A=\{a \in \mathbb L, \exists P \in \mathbb K [X], P(a)=0 \}$  I managed to prove that $A$ is a ring with the following reasoning:  If $a,b \in A$ then $ \mathbb K[a]$ is a ring.  And $ \mathbb K \subset \mathbb K[a] \subset \mathbb K[a][b] $  Some finite-dimension argument(ask details if needed) proves that $\mathbb K[a][b]$ has finite dimension as a vector space over the field $\mathbb K$  Reminding that $\mathbb K[a+b] \subset \mathbb K[a][b]$ and $\mathbb K[ab] \subset \mathbb K[a][b]$ proves that they both have finite dimension, hence $a$ and $b$ algebraic.  How should I proceed with inverses ?
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?
118561
An algebraic closure of a field K is an algebraic extension of K that is algebraically closed. It is one of many closures in mathematics. But why is it a countable set?
115087
Let $K/F$ be a field extension, I know that if $K/F$ is a finite extension then a simple argument from linear algebra shows that since every homomorphism of fields from $K$ to $K$ that fixes $F$ is 1-1 it is also onto, i.e. an automorphism of $K$.  Can someone please give an example that this is not the case if $K/F$ is not a finite extension ?
24276
I'm interested in understanding compositum of general fields better. Assume we have $\Omega/K/F$ and $\Omega/L/F$ field extensions, and consider the composite $KL$.   It seems to me that every $a \in KL$, exists $n_1,n_2 \in \mathbb{N}$, $ \left \lbrace k_i \right \rbrace_{i=1}^{n_1}, \left \lbrace k'_i \right \rbrace_{i=1}^{n_2} \subseteq K$ and $ \left \lbrace l_i \right \rbrace_{i=1}^{n_1}, \left \lbrace l'_i \right \rbrace_{i=1}^{n_2} \subseteq L$ such that: $$ a = \frac{\sum_{i=1}^{n_1} k_i l_i}{\sum_{j=1}^{n_2} k'_j l'_j} $$  (That is since the family of all such sums is closed under addition, multiplication, and distribution because of the abelian nature of fields, therefore all numbers of this form are subfields that contain both K and L)  Is it a correct way to refer to a compositum of fields? When is it correct that the general item of an extension is $a = \sum_{i=1}^{n} k_i l_i$?