In paragraph V.1 of Algebra proposition 1.7 Lang claims that the class of algebraic extensions is distinguished. I know that if $F/k$ and $E/F$ are algebraic extensions than so is the $E/k$ - that is easy to prove. However, the second required property is that if $E/k$ is algebraic and $F/k$ is arbitrary, then (assuming the compositum is defined) $EF/F$ is algebraic. Lang more or less skips the proof of this, only saying that "an element remains algebraic under lifting, and hence does the extension."  However, there are no finiteness conditions here and no element given in advance. Although $EF$ is generated over $F$ by a set of elements which are algebraic over $F$ (namely, all the elements of $E$), there are no facts known (up to this point in the book at least) about the infinitely generated extensions.  What am I missing?  Thank you.
123274
Suppose $K$ is a field, and $K\subset E$ is an algebraic extension, and $K\subset F$ is any extension.  Suppose that $L$ is some field which contains both $F$ and $E$ as subfields.  I want to show that $FE$ (the smallest subfield of $L$ containing both $F$ and $E$) is algebraic over $F$.  Lang mentions that the result is true because an element remains algebraic under lifting, but I can't prove the claim with this.  I interpret his comment to mean that if $x\in E$ is algebraic over $K$, then $x\in FE$ is algebraic over $F$.  This is certainly true.    But this doesn't seem enough if I start with any element of $FE$.
66994
Fraleigh(7ed) Theorem33.12. Let $p$ be a prime and let $n\in\mathbb{Z}^+$. If $E$ and $E&#39;$ are fields of order $p^n$, then $E \simeq E&#39;$.   Proof in the text: Both $E$ and $E&#39;$ have $\mathbb{Z}_p$ as prime field, up to isomorphism. By Corollary 33.6(A finite extension $E$ of a finite field $F$ is a simple extension of $F$), $E$ is a simple extension of $\mathbb{Z}_p$ of degree $n$, so there exists an irreducible polynomial $f(x)$ of degree $n$ in $\mathbb{Z}_p[x]$ such that $E\simeq \mathbb{Z}_p[x]/ \langle f(x) \rangle$. Because the elements of $E$ are zeros of $x^{p^n}-x$, *we see that $f(x)$ is a factor of $x^{p^n}-x$ in $\mathbb{Z}_p[x]$*. Because $E&#39;$ also consists of zeros of  $x^{p^n}-x$, we see that $E&#39;$ also contains zeros of irreducible $f(x)$ in $\mathbb{Z}_p[x]$. Thus, because $E&#39;$ also contains exactly $p^n$ elements, $E&#39;$ is also isomorphic to $E\simeq \mathbb{Z}_p[x]/\langle f(x) \rangle$.  I don't know why $f(x)$ divides $x^{p^n}-x$. $E$ has a zero $\alpha$ of $f(x)$, but it doesn't need to have all the zeros of $f(x)$. So $f(x)=(x-\alpha)g(x)$ in $E[x]$ and $g(x)$ need not be splitted into linear factors. How can $f(x)$ divide $x^{p^n}-x$?
5969
If $[L:K]$ is finite, then $[L:K]$ is algebraic.  The author's proof is to simply state that given $[L:K] = n$, then ${1, x, x^2, ... , x^n}$, for any $x$ in $L$, would contain $n+1$ elements, so it must be linearly dependent over $K$.  First of all, why must ${1, x, x^2, ... , x^n}$ be linearly dependent over $K$? I do undertsand that if $L=K(a_1, ... , a_m)$, and we pick an non-algebraic $a_i$, then ${1, a_i, a_i^2, ... , a_i^n}$ would be, by definition, linearly dependent over $K$. But how can we know that this works with all the elements in $L$?  Second of all, even if we find that $c_0+ c_1x+c_2x^2 + ... + c_{m-1}≠0$, for some $x$ in $L$, $c_i$ in $K$, why would this mean that $[L:K]=∞$?    I asked this question a couple days ago, though I didn't explain my doubt in such detail as in this post. Anyway, the replies asked if I understood the meaning of being 'linearly dependent over a field $F$'. My understanding is that a set ${b_1, ... , b_n}$ is linearly independent over $F$ if each element in $F$ can be written as $f_1b_1+...+f_nb_n$ (so that  the different combinations of $b_i$ form the whole $F$) and that the only way $f_1b_1+...+f_nb_n=0$ is for all the $b_i$ to be $0$ (so that we make sure all the elements in the set are needed to map create $F$).  I guess they asked me that question because is rather obvious why the author's proof works. Yet I don't know why, but it isn't that obvious to me.    I would really appreciate any help/thoughts.
74088
I am trying to prove this statement in the characteristic $p&gt;0$ case.     Every algebraic extension of a perfect field is separable and perfect.   This is stated as a corollary of Proposition V.6.11 in Lang.     Let $k$ be a field and let $K$ be a normal extension of $k$. Let $G$ be the group of automorphisms of $K$ that fix $k$. Let $K^G$ be the fixed field of $G$ acting on $K$, and $K_0$ the maximal separable subextension of $K$. Then $K^G$ is purely inseparable over $k$, $K$ is separable over $K^G$, $K_0\cap K^G =k$  and $K_0K^G=K$.   Lang doesn't write many details in the proof except that every finite extension is contained in a normal extension.   This is what I have so far. Suppose, $k=k^p$. Let $E$ be an algebraic extension of $k$. Let $\alpha \in E$. Now, $\alpha$ is contained in the field, say $K$ generated by the roots of the minimal polynomial of $\alpha$ over $k$ (and is hence normal as it's the splitting field of the minimal polynomial of $\alpha$). At this point, I don't know how to proceed. I am tempted to use the Frobenius map, but it does not fix every element of $k$ although it does stabilize it in this perfect case. Any hints will be appreciated. Thanks.
92871
What is the difference, if there is any, between $F(x)$ and $F[x]$, where $F$ is a field and $x$ is an element of a field extension $E/F$ that is not in $F$? I have some guesses, but I was never sure about this. For example, for one specific usage in a textbook, I see $\mathbb{Q}[i]$ and $\mathbb{Q}(\pi)$. And maybe a more general question: what happens when $F$ is just a ring? For example, for the ring of Gaussian integers, the notation I usually see is $\mathbb{Z}[i]$, with brackets instead of open parentheses. I guess I never properly learned the notation.
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?
73097
If $F$ is a field, then $F[x]$ is a principal ideal domain   By a previous theorem, we know that $F[x]$ is an integral domain. Now, let $I$ be an ideal in $F[x]$. If $I = \{0\}$, then $I = \langle 0 \rangle$. If $I \neq \{0\}$, then among all the elements of $I$, let $g(x)$ be one of minimum degree. We will show that $I = \langle g(x) \rangle$. Since $g(x) \in I$, we have $\langle g(x) \rangle \subseteq I$. Now let $f(x) \in I$. Then, by the division algorithm, we may write $f(x) = g(x)q(x) + r(x)$, where $r(x) =0$ or deg $r(x) \lt$ deg $g(x)$. Since $r(x) = f(x) - g(x)q(x) \in I$, the minimality of deg $g(x)$ implies that the latter condition cannot hold. So, $r(x) = 0$ and, therefore, $f(x) \in \langle g(x) \rangle$. This shows that $ I \subseteq g(x)$.  My question is: What is a polynomial of minimum degree in this context?  And how does this minimality imply that deg $r(x) \lt$ deg $g(x)$ cannot hold?
61639
I read that if $F$ is the field of algebraic numbers over $\mathbb{Q}$, then every polynomial in $F[x]$ splits over $F$. That's awesome! Nevertheless, I don't fully understand why it is true. Can you throw some ideas about why this is true?
16317
Let $X$ be transcendental over a field $F$, and let $E$ be a subfield of $F(X)$ properly containing $F$. Prove that $X$ is algebraic over $E$.   Could we maybe use also the following??   Let $f(x) \in E \Rightarrow f(x) \in F(X) \Rightarrow f(x)=\frac{h(X)}{g(X)}$, where $g(X) \neq 0, h,g \in F[x]$.   Or doesn't this help ??
117941
Let L/K be a field extension with [L:K] = p, where p is prime. Show that L/K is a simple extension.  I tried using the Tower Law but to no avail. I'd be grateful if somebody can assist me with a hint!
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 ?
60060
It is true that all finite field extensions are algebraic. It is not true however that all algebraic extensions are finite.   In lectures we were given the example of the field extension  $$\mathbb{A}=\{\alpha \in \mathbb{C}: \alpha \text{ is algebraic over } \mathbb{Q} \}$$  I want to prove to myself that this is indeed a non-finite algebraic extension.   I have shown that $\mathbb{A}$ is a subfield of $\mathbb{C}$. Now I consider roots of the equation $x^n-2$ for $n \geq 1$. Clearly such roots are in $\mathbb{A}$ by definition. I think that if I show that for a given $k$, the root of $x^k-2$ (say $\psi_k$) does not lie in $\mathbb{Q}(\psi_1,...,\psi_{k-1})$, then I have proven the extension is not finitie (since this will hold for all $n$).  Can someone please confirm this reasoning and suggest how the proof could be finished.
118804
I am having difficulty with this proof. I have previously proved that for an automorphism on the extension field $K$ of $F$ that when restricted to $F$ is the identity automorphism must take roots of $f(x)\in F[x]$ in $E$ to roots of $f(x)$ in $E$. I was thinking of using the minimal polynomial $x^{3}-2\in\mathbb{Q}[x]$ and the result of previous problem I solved, but I suspect there are automorphisms that don't fix the subfield, so the result can't be used or I need to prove that any automorphism of an extension field must fix the subfield that is being extended or always sends roots in $E$ to roots in $E$.  I am not asking for a solution to this, but maybe a hint or pointer in the right direction. This problem is introduced in the text prior to Galois theory, so please keep that in consideration.
66507
How does one go about proving that the sums and products of two algebraic numbers over a field $F$ (say $a,b\in K$, where $K/F$ is a field extension) is also algebraic?   If we call $f_a$ and $f_b$ the min. poly's of $a$ and $b$, then I'm assuming the answer involves such polynomials. Perhaps looking at their roots in splitting fields for both of them? And finding a "big" splitting field, constructed from those two other ones?      In particular, I'd like a way of explicitly constructing the minimal polynomials $\ f_{ab}$ of $ab$ and $f_{a+b}$ of $a+b$.     I read somewhere that $g(x)=\Pi_j\Pi_i (x-\alpha_i\beta_j)$ works for $ab$, where the $\alpha_i$ and $\beta_j$ are the roots of $f_a$ and $f_b$, respectively, but I do not know why $g(x)\in F[x]$.  Similar remarks for $a+b$
79364
Suppose $K$ is a field and $F$ is an algebraic extension of some degree $n=(F:K)$. It is stated that the field of rational functions $F(x)$ is in fact an algebraic extension of the field $K(x)$ and moreover $(F(x):K(x))=n$.    How do I approach this exercise? I'm new in this area so any help would be greatly appreciated!
24760
Let $K$ be a finite extension of $\mathbf{Q}_p$, i.e., a $p$-adic field. (Is this standard terminology?)  Why is (or why isn't) an algebraic closure $\overline{K}$ complete?  Maybe this holds more generally:  Let $K$ be a complete Hausdorff discrete valuation field. Then, why is $\overline{K}$ complete?  I think I can show that finite extensions of complete discrete valuation fields are complete.
98625
Let $K$ be an algebraic extension field (not necessarily finite) of $\mathbb{Q}$. Let $\mathscr{O}_K$ be the integral closure of $\mathbb{Z}$ in $K$.  Then, is $\mathscr{O}_K$ Noetherian?  If $K$ is a finite extension of $\mathbb{Q}$, then $\mathscr{O}_K$ can be shown to be Noetherian. However, if we do not assume that finiteness condition, is it still true?  (It can be shown that $\mathscr{O}_K$ is an integrally closed domain of Krull-dimension $1$. Hence, if it is Noetherian, it is a Dedekind domain.)