Let $K$ be a field, and let $\alpha$ be transcendental over $K$ and algebraic over $K(\beta)$. We have a Hasse diagram of field extensions     Now, by reduction to absurdity $\beta$ must be transcendental over $K$ as well (this is asked but I'm not sure if it's relevant to the question). The problem is to prove that $\beta$ must be algebraic over $K(\alpha)$.  The first thing I tried is to take whatever polynomial annihilates $\alpha$: $$a_0 + a_1\alpha\cdots + a_n\alpha^n = 0$$ with $a_0,\dots,a_n\in K(\beta)$. If $\beta$ were algebraic over $K$, then I could rewrite this as $$\sum\nolimits_1^m\lambda_{i,0}\beta^i + \left(\sum\nolimits_1^m\lambda_{i,1}\beta^i\right)\alpha + \cdots + \left(\sum\nolimits_1^m\lambda_{i,n}\beta^i\right)\alpha^n$$ And this would of course give a polynomial that annihilates $\beta$ with coefficients in $K(\alpha)$ (aside from the fact that $K(\alpha,\beta)/K(\alpha)$ would be algebraic by mulitplicity of degree). The thing is, when it comes to transcendental extensions, I'm just not sure that $$\left\{\beta^k\right\}_{k\in\Bbb N}$$ is still a $K$-basis for $K(\beta)$. If this were true then I suppose my question is how to prove this. If not, what other method might I try?  I also wonder whether or not this is a special case of $$E,L\supset K \,\wedge\, F/E \text{ algebraic }\stackrel ?\implies F/L\text{ algebraic }$$ My guess is no, but I also can't think of an counterexample because I only know of simple transcendental extensions (where it would be true by what I originally wanted to prove).
124109
Let $E$ be an extension of field $F$, and let $\alpha, \beta \in E$. Suppose $\alpha$ is transcendental over $F$ but algebraic over $F(\beta)$.   Show that $\beta$ is algebraic over $F(\alpha)$.  Okay, first questions: What does the notation $F(\alpha)$ and $F(\beta)$ mean? And being transcendental means it solves no equations with rational coefficients, but what does it mean for a field?
3394
Suppose $E/F$ is a field extension and $\alpha, \beta \in E$ are algebraic over $F$. Then it is not too hard to see that when $\alpha$ is nonzero, $1/\alpha$ is also algebraic. If $a_0 + a_1\alpha + \cdots + a_n \alpha^n = 0$, then dividing by $\alpha^{n}$ gives $$a_0\frac{1}{\alpha^n} + a_1\frac{1}{\alpha^{n-1}} + \cdots + a_n = 0.$$     Is there a similar elementary way to show that $\alpha + \beta$ and $\alpha \beta$ are also algebraic (i.e. finding an explicit formula for a polynomial that has $\alpha + \beta$ or $\alpha\beta$ as its root)?    The only proof I know for this fact is the one where you show that $F(\alpha, \beta) / F$ is a finite field extension and thus an algebraic extension.
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.
105931
Considering the irreducible (by condition) polynomial $f=X^4-2aX^2+b\in\mathbb Q[X]$ and the field extension $K=\mathbb Q(\sqrt{a^2-b})$ of $\mathbb Q$, I want to show the following result:    For $[L:\mathbb Q]=4$ we have $\sqrt b\in K$ where $L$ is the splitting field of $f$.   My thoughts so far: The four roots of $f$ are $\pm\sqrt{a\pm\sqrt{a^2-b}}$. If $L$ is of degree $4$ it must be primitive since $f$ is irreducible, so $L=\mathbb Q\left(\sqrt{a+\sqrt{a^2-b}}\right)$. Then, by multiplying the two roots $\sqrt{a+\sqrt{a^{2}-b}}\cdot\sqrt{a-\sqrt{a^{2}-b}}=\sqrt{b}$ we have that $\sqrt{b}\in L$. Now I want to look at $$\mathbb Q\subset K\subset K(\sqrt b)\subset L$$ and differentiate some cases. The case where $[L:K]=1$ is trivial and the case $[L:K]=4$ can't happen because if we square any root and subtract $a$, we get $\pm\sqrt{a^2-b}\in K$ which means $[L:K]\leq 2$. So let's look at the case where $[L:K]=2$. As I see it, we get two subcases:   For $[L:K(\sqrt b)]=2$ we have $K(\sqrt b)=K$ which finishes the proof. For $[L:K(\sqrt b)]=1$ we have $K(\sqrt b)=L$. This means that $L$ is also the splitting field of $(x^2-b)(x^2-(a^2-b))$. Is this a contradiction? How do I finish this case?
123992
I am having some trouble understanding Proposition 5.15 in Introduction to Commutative Algebra by Atiyah and Macdonald.      Let $A\subset B$ be integral domains, $A$ integrally closed, and $x\in B$ be integral over an ideal $\mathfrak{a}$ of $A$. Then $x$ is algebraic over the field of fractions $K$ of $A$, and if its minimal polynomial over $K$ is $t^n+a_1t^{n-1} + \cdots + a_n$, then the $a_1, a_2, \ldots , a_n$ all lie in the radical of $\mathfrak{a}$.   The proof of the proposition states that the coefficients of the minimal polynomial of $x$ are polynomials in the $x_i$ (the conjugates of $x$). I don't understand this, and am struggling with the rest of the proof as well. Thanks for any help.
86188
Let $E$ be an extension field of $F$ and $\alpha \in E$. Then $\alpha$ is transcendental over $F$ if and only if $F(\alpha)$ is isomorphic to $F(x)$, the field of fractions of $F[x]$.   This was a theorem in an abstract algebra textbook with a very brief proof. Can someone please explain why this theorem holds? I'm having difficulty grasping the concepts at hand.  Thanks!
94121
Suppose $a$ and $b$ are complex numbers and both transcendental over $\mathbb Q$. I am wondering why $ab$ and $a+b$ can not both be algebraic.  Thanks for any help.
47939
If we have an algebraic number $\alpha$ with (complex) absolute value $1$, it does not follow that $\alpha$ is a root of unity (i.e., that $\alpha^n = 1$ for some $n$).  For example, $(3/5 + 4/5 i)$ is not a root of unity.  But if we assume that $\alpha$ is an algebraic integer with absolute value $1$, does it follow that $\alpha$ is a root of unity?    I know that if all conjugates of $\alpha$ have absolute value $1$, then $\alpha$ is a root of unity by the argument below:  The minimal polynomial of $\alpha$ over $\mathbb{Z}$ is $\prod_{i=1}^d (x-\alpha_i)$, where the $\alpha_i$ are just the conjugates of $\alpha$.  Then $\prod_{i=1}^d (x-\alpha_i^n)$ is a polynomial over $\mathbb{Z}$ with $\alpha^n$ as a root.  It also has degree $d$, and all roots have absolute value $1$.  But there can only be finitely many such polynomials (since the coefficients are integers with bounded size), so we get that $\alpha^n=\sigma(\alpha)$ for some Galois conjugation $\sigma$. If $\sigma^m(\alpha) = \alpha$, then $\alpha^{n^m} = \alpha$.  Thus $\alpha^{n^m - 1} = 1$.
96369
Is it true that if $x_1, \dots, x_n \in \Bbb C$ are algebraically independent over $\mathbb Q$, then they are algebraically independent over $\overline{\mathbb Q}$?  I know how to prove it for $n=1$, by the contrapositive. If $P(x)=0$ and $P \neq 0$ has coefficients $a_0,...,a_n$ in $\overline{\mathbb Q}$, then $x$ is algebraic over $\mathbb Q(a_0,...,a_n)$ which is itself algebraic over $\mathbb Q$, therefore $x$ is algebraic over $\mathbb Q$, so it satisfies some relation $Q(x)=0$ where $Q \neq 0$ is a polynomial with rational coefficients.  For $n \geq 2$, I tried induction, but I was not sure how to proceed.  Thank you for your help!
33205
Consider a field extension $L/K$ and $a,b \in L$ algebraic over $K$. Is it true that in this setting $K[a,b]$ is already a field?   I know that $K[a]=K(a)$ if and only if $a$ is algebraic over $K$.
54082
When I'm self-studying  Parick Morandi's book Field and Galois Theory,I came across some problems,which I can't work out fully. Let $K$, $L$, be two extension fields of base field $F$. If  $K/F$ and $L/F$ are both Galois extension ,how to proof that their composite $KL$ is also Galois over $F$. And why we have the same conclusion when "Galois "is replaced by "normal","separable","purely inseparable",or "algebraic"? I worked on these problems and searched this website  for two days but could only proof the cases for "separable","purely inseparable",and "algebraic".I have no idea how to proof the "normal" and "Galois" cases. In addition ,why the degree inequality  holds for $$ [KL:F]\leq[K:F][L:F]$$ If you know how to proof these ,please tell me.I am here waiting for the answers.Thank you!
59989
Let     $$ \mathbb{Z}^{'}_{\mathbb{C}}=\{ z \in \mathbb{C} | \exists f \in \mathbb{Z}[X] \text{ monic such that } f(z)=0\} $$   be the integral closure of $ \mathbb{Z} $ in $ \mathbb{C} $. Prove that $ \mathbb{Z}^{'}_{\mathbb{C}} $ is not finitely generated as a $ \mathbb{Z} $-module.  My attempt: Suppose $ \mathbb{Z}^{'}_{\mathbb{C}}=\mathbb{Z}\alpha_{1}+...+\mathbb{Z}\alpha_{n} $ for some $ \{ \alpha_{i}\}_{i=\overline{1,n}} \subset \mathbb{Z}^{'}_{\mathbb{C}} $. I know that given $ \alpha , \beta \in \mathbb{Z}^{'}_{\mathbb{C}} $ with minimal polynomials of degrees $ m $ and $ n $ respectively, then we can construct a monic polynomial with integral coefficients $ P(z)=\prod(z-(\alpha_{i}+\beta_{j})) $ where $ \alpha_{i},\beta_{j} $ are the conjugates of $\alpha $ , $\beta $ and therefore the minimal polynomial of $ \alpha + \beta $ has degree $ \leq mn $.   I was thinking that then the degrees of the minimal polynomials of the elements of $ \mathbb{Z}\alpha_{1}+...\mathbb{Z}\alpha_{n} $ could be somehow bounded by a certain constant, whereas $ \mathbb{Z}^{'}_{\mathbb{C}}$ contains elements whose degrees of the minimal polynomials are arbitrarily large, take $ 2^{\frac{1}{n}} $ for example,$ n\geq 2$. I think this works when we are asked to prove that the set of all algebraic numbers $ \overline{\mathbb{Q}} $ is not a finitely dimensional $\mathbb{Q}$-vector space, but I don't know how to proceed in my case.  I would really appreciate any ideas, hints or solutions. Thank you very much!
53357
Of course this is a particular case of the closure under addition and multiplication of algebraic numbers (which proof would also be appreciated), but I would prefer a simple transformation on the already known polynomial giving $a$ as a root in order to obtain a polynomial giving $ia$ as a root. Thank you in advance for your help.
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 ??
79758
Let $F$ be a subfield of a field $K$, $a$ an element of $K$.  Prove that $a$ is algebraic over $F$ iff $F[a]=F(a)$.  For the first direction I've tried this: $a$ is algebraic over $F$ $\Rightarrow$ there is a polynomail $f$ over $F$ such that $f(a)=0$ $\Rightarrow$ $1,a,a^2,...,a^m$ are linearly dependent $\Rightarrow$ $\deg(F[a])&lt;m$.    but it contains $f(a)$ so they are equal..  is that correct? about the other direction.. i simply have no clue
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?
72333
Show the algebraic closure $\bar{\mathbb{Q}}$  is not even finitely generated over the field $\mathbb{Q}$   I'm not sure how to go about this..