finite/algebraic field extensions and minimal polynomial 1) To show: $L/K$ field extension is algebraic iff every subring with $K\subset R\subset L$ is a field.
My answer: I can write $R=\bigcup\limits_{\alpha\in R}K[\alpha]$ with $K[\alpha]$ (field as $L/K$ algebraic) is the image of $f\mapsto f(\alpha)$. Now $\alpha\in R \Rightarrow\alpha\in K[\alpha]\Rightarrow \alpha^{-1}\in K[\alpha]\Rightarrow \alpha^{-1}\in R$. The other way $K[\alpha]$ is always a field for every $\alpha$ and because of $K[\alpha]\cong K[X]/(\ker(\varphi))$ it follows that the kernel isn't zero. This should be fine right? Is there a "better way"?
2) $\alpha\in\mathbb C$ and $\alpha^3+2\alpha-1=0$. What is the minimal polynomial from $\alpha$ and $\alpha^2+\alpha$.
Now the polynomial is irreducible in $\mathbb F_3[X]$ so its irreducible in $\mathbb Q[X]$, so the given is the minimal polynomial of $\alpha$ over $K$. Now I took the basis $(\alpha^0,\alpha^1,\alpha^2)$ and made the matrix to this basis of the endomorphism $(\alpha^2+\alpha)$ by looking how the image of $(1,0,0), (0,1,0), (0, 0, 1)$ looks like: $(0,1,1), (1,-2,1), (1,-1-2)$ (if the degree of $\alpha$ was greater than $2$ I substracted the equation given by a needed facotr of $\alpha^n$). So with this I've got $-x^3-4x^2-3x+4$ for the characterisc polynomial of the found matrix, which is the minimal polynomial of $\alpha^2+\alpha$ as it has no roots in $\mathbb F_3[X]$, so its irreducible in $\mathbb Q[X]$. Now can I compute this polynomial somehow directer? My way doesn't seem very efficient (did I do any mistakes?). And is there some way I could argue that $[K[\alpha^2+\alpha]:K]=[K[\alpha]:K]$ so i wouldn't have to check the irreducibility of the calculated polynomial (I can't think of an argument for $[K[\alpha^2+\alpha]:K]\geq[K[\alpha]:K]$). I guess it could be only $1$ or $3$, as it has to divide $3$, so I can rule out the one case as else $\alpha$ would be algebraic. But is there a more direct argument?
3) $L/K$ fieldextension $x\in L$ transcentdental over $K$. Now I should show that for $n\geq 1$ $x^n$ is transcendental over $K$ and $[K(x):K(x^n)]=n$, so I only need to show the second statement. I cant seem to get ahead with this one. A tip would be very helpful!
4) $L/K$ field extension $\alpha\in L$ algebraic over $K$. To show is for $n\geq 1:[K(\alpha^n):K]\geq(1/n)[K(\alpha):K]$. I guess for this one I need the idea of the last one.
 A: There are many questions here.
1) Is fine. You can also say that, if $\alpha^{-1}=a_0+a_1\alpha+\dots+a_n\alpha^n\in K[\alpha]$, then $a_n\alpha^{n+1}+\dots+a_1\alpha^2+a_0\alpha-1=0$, so $\alpha$ is algebraic.
2) Irreducibility of $X^3+2X-1$ is fine. Since $\beta=\alpha^2+\alpha\in\mathbb{Q}[\alpha]$, the degree of $\beta$ is either $1$ or $3$. Since the elements $1$, $\alpha$, $\alpha^2$ form a basis of $\mathbb{Q}[\alpha]$ over $\mathbb{Q}$, we can't have $\alpha^2+\alpha+r=0$ for any $r\in\mathbb{Q}$, so $\beta\notin\mathbb{Q}$ and so…
We have $\beta^2=\alpha^4+2\alpha^3+\alpha^2=\alpha(-2\alpha+1)+2(-2\alpha+1)+\alpha^2$, so
$$
\beta^2=-\alpha^2-3\alpha+2
$$
Compute $\beta^3$ and find a linear combination of $1$, $\beta$, $\beta^2$ and $\beta^3$ which is equal to $0$.
3) Consider the polynomial $f(t)=t^n-x^n$ in $K(x^n)[t]$; then $x$ is a root of $f(t)$, so $x$ has degree at most $n$ over $K(x^n)$. Now show that the degree must be at least $n$.
4) Yes, the idea is similar to 3. The polynomial $t^n-\alpha^n$ in $K(\alpha^n)[t]$ has $\alpha$ as a root, so $[K(\alpha):K(\alpha^n)]\le n$. Now use
$$
[K(\alpha):K(\alpha^n)][K(\alpha^n):K]=[K(\alpha):K]
$$
