12

Let us consider a field $K$ and let us say we fix an algebraic closure $K^{\text{alg}}$ of it. Now, $K^{\text{alg}}$ might contain elements that are not separable (over $K$), that is their minimal polynomial is not a separable polynomial (i.e., it has roots of multiplicity greater $1$ in $K^{\text{alg}}$ or equivalently it is not co-prime with its ...


5

Try this approach: Consider the minimal $F$-polynomial for $x$, call it $H(T)$. It’s $F$-irreducible, and so of form $h(T^q)$ with $q=p^s$ for an $s\ge0$ and $h(T)$ an $F$-irreducible polynomial that’s separable. If $s>0$, then $x$ is not separable over $F$ (and so $K$ is not separable over $F$), but $x^q$ is separable over $F$, as a root of the ...


5

This may be using a pile-driver to crack a peanut, but here’s how I’ve always thought of this problem: Since $L$ is finite and separable over $K$, the Galois closure $\mathscr L$ of $L$ over $K$ is also finite over $K$. You can construct $\mathscr L$ by taking the minimal polynomial $f$ of your primitive element and adjoin all roots of $f$. In other words, $...


5

Every subfield of $\overline{\mathbb{F}_p}$ is perfect. You can use that to produce many examples of the sort that you are after: the absolute Galois group of $\mathbb{F}_p$ is isomorphic to the profinite completion $\hat{\mathbb{Z}}$ of $\mathbb{Z}$, and that group has many infinite quotients (it is isomorphic to $\prod_l \mathbb{Z}_l$, where the product ...


5

Take your favorite infinite non-separably closed field $F$ of characteristic $p>0$ (e. g. $F = \Bbb F_p(X)$; the field $\Bbb F_{p^2}(X)$ is a separable non-trivial extension of $F$). Let $\overline F$ be an algebraic closure of $F$ and consider the perfect closure $F^{\rm perf}$ inside $\overline F$ given by $$ F^{\rm perf} := \{x\in \overline F \mid x^{p^...


4

Suppose that $L/K$ is a finite separable extension, and $E/K$ is an arbitrary extension. Then $\mathrm{Hom}_K(L,E)$ consists of at most $[L:K]$ elements, because a field homomorphisms $\sigma :L\rightarrow E$ that fixes $K$ is uniquely determined by the image $\sigma (x)$ of a primitive element of $L/K$, which in turn must be a root of the minimal polynomial ...


3

$\Bbb F_p(\sqrt[p]{x}) / \Bbb F_p(x)$ is not separable. A field of characteristic $p$ is perfect iff the Frobenius endomorphism $x \mapsto x^p$ is surjective. In the case of $\Bbb F_p(x)$, $x$ cannot be contained in the image of the Frobenius endomorphism for degree reasons.


3

Using your example, let $K = \mathbb{F}_2(t), L = \mathbb{F}_2(t^{1/6})$. Consider the equation $$x^3 - t = 0$$ then it has a root in $L$, namely $t^{1/3}$. Then all the roots of this equation will be $t^{1/3}, \zeta t^{1/3}, \zeta^2 t^{1/3}$, where $\zeta$ is a primitive third root of unity. If $\zeta t^{1/3} \in L$, we must have $\zeta \in \mathbb{F}_2$, ...


3

Let $A={\bf Q}(\root4\of2)$, $B={\bf Q}(\sqrt2)$, $C=D={\bf Q}$, then $A/B$, $B/C$, $C/D$ are all normal, but $A/D$ isn't.


3

You can't find non-separable elements in $\mathbb Q, \mathbb R$ or $ \mathbb C$ because these are all fields of characteristic zero. You should instead think about infinite fields of characteristic $p$. For example, consider the field extension $\mathbf F_p (t): \mathbf F_p(t^p)$, and consider the element $t \in \mathbf F_p(t)$. Its minimal polynomial over ...


3

I would highly recommend to take another characterisation of a normal field extension, namely the 2. point on your list. The result immediately follows from $\operatorname{Hom}_F(K,\overline K) = \operatorname{Hom}_L(K,\overline K)$, because this equality tells you that normality of $K/F$ and $K/L$ are the same. The reason for this equality is very easy: ...


3

Take the characteristic of our fields to be $p$. Let $\alpha\in K$. As $K/L$ is normal, then $f(\alpha)=0$ where $f$ is monic with has coefficients in $L$, and $f$ splits into linear factors in $K$. Then $f(X)=X^m +a_1X^{m-1}+\cdots+a_m$. As $L/F$ is purely inseparable, then for each $i$, $a_i^{p^{m_i}}\in K$ for some $m_i$. Let $m$ be the maximum of the $...


3

$\gamma=w+i$ with $w=\sqrt[4]2$ is a good guess (and something similar, i.e. in rare cases maybe with an additional factor, always works, as per the proof of primitive element theorem). If you write down powers of $\gamma$ and simplify, you will note that they all have the form $$\gamma^k=a+bw+cw^2+dw^3+ei+fwi+gw^2i+hw^3i$$ with the eight coefficients $a,\...


3

As you say, any char $0$ field is perfect. Apart from that, any finite field is perfect, and any algebraic extenstion of a perfect field is perfect. So, you can easily find infinite algebraic extensions of $\Bbb F_p$ for some $p$, apart from the algebraic closure, and it will be perfect. Say the quadratic closure of $\Bbb F_p$, for instance (the "...


3

This is not always true as shows the answer of this question for $X^n-2$, the order of the Galois group is $n\phi(n)$ or $n\phi(n)/2$. https://mathoverflow.net/questions/143739/galois-group-of-xn-2


3

Since the characteristic is zero, the extension is separable. It is the splitting field of $X^8-2$ as you have remarked.


2

If the Galois group is readily available, as it is in this case, it makes sense to use the fact that a polynomial with distinct roots is irreducible if and only if the Galois group is transitive on its roots. So in this case you have just to check that the orbit of the obvious candidate $\alpha = \sqrt[4]{2} + i$ under the Galois group has length $8$. (...


2

The extension $E/K$ need not be separable. Here is the example I learned from a note by J. Lipman. Consider the rational function field $F=\mathbb{F}_2(y,z)$ and the extension $E=F(x)$, where $x$ is a root of $$f(t)=t^4+yt^2+z\in F[t].$$ If $E/K$ was separable, we would have $f=g^2$, for $g\in K[t]$. We have $g=t^2+\sqrt{y}t+\sqrt{z}$, which means that $\...


2

Hint: How do you get the field $\mathbb{F}_{p^2}$ from the field $\mathbb{F}_p$? The addition of the transcendent variable $t$ does not change things much. Second hint: How does separability relate to minimal polynomials?


2

The group $G=Gal(E/F)$ is finite. By Galois theory $K$ is the fixed field of the subgroup $G_1 = Gal(E/K)$ of $G$, and $L$ is the fixed field of the subgroup $G_2 = Gal(E/L)$ of $G$. So the fixed field of $(G_1, G_2)$ ( the subgroup of $G$ generated by $G_1$, $G_2$) is $K\cap L$. Now we use the following fact: If $H$ is a finite group of automorphisms of a ...


2

No, consider $L_1=\mathbb Q$ and $L_2=\mathbb Q(\sqrt 3)$ which aren't equal but certainly exist inside $L_3=\mathbb Q({}^4\!\!\!\sqrt 3)$. Here with $w={}^4\!\!\!\sqrt 3$ we have $L_1(w)=L_2(w)=L_3$ but not $L_1=L_2$. Edit: since you want the fields to contain $\mathbb C$, we can replace the number with any variable $t$. Consider $\mathbb C(t^4) \subset ...


2

Here is a counterexample. Take $L_1=\mathbb{Q}$, $L_2=\mathbb{Q}(i)$ and $w=\zeta_8=e^{2\pi i/8}$. Then $L_3=\mathbb{Q}(\zeta_8)$, and $L_1(w)=L_2(w)=L_3$. On the other hand, $L_1\neq L_2$.


2

I'm going to follow the usual notation and write $\overline{F}$ for an algebraic closure of $F$ containing $E$, rather than $F'$. You have probably misread what Jacobson wrote. He is writing "Let $F^{1/q}$ (respectively $D^{1/q}$) be the subfield (resp. subring) of elements $v$ of $\overline{F}$ such that $v^{1/q} \in F$ (resp. $D$)." He omits the word "...


2

An inseparable extension of degree $2$ is $L=F(\sqrt a)$ in characteristic $2$. As $\sqrt a$ is the unique square root of $a$ then any $F$-automorphism fixes $\sqrt a$ and so is trivial on $F$. The classic example is the rational function field $L=\Bbb F_2(t)$ and $F=\Bbb F_2(t^2)$. A separable extension of degree $2$ in characteristic $2$ is an Artin-...


2

Suppose that $[E:k]_s = n < \infty$. Then, for every $\alpha \in E_s$ we have that $[k(\alpha):k] \leq n$. So, the set $$\mathcal{A} = \{\, [k(\alpha):k] \mid \alpha \in E_s \,\} \subset \Bbb{N}$$ is bounded above by $n$. Let $m = \max(\mathcal{A})$ and let $\alpha_0 \in E_s$ be an element for which the maximum is attained. We claim that $E_s = k(\...


2

A purely transcendental extension $F(x)$ is never perfect when $F$ is a field of characteristic $p \neq 0$; the polynomial $f(t) = t^p - x$ is an irreducible over $F(x)$, but has repeated roots. One way to see this is that if we let $F(y)$ be another purely transcendental extension, then $F(x) \cong F(y^p)$. Since $F(y)$ is generated over $F(x)$ by a root ...


2

$\DeclareMathOperator{Gal}{Gal}$ If $f\in F[x]$ is of degree $n$, denote $K$ as its splitting field, one of its root $\alpha$. Note that $\Gal(K/F) \cong S_n$ implies $K/F$ is separable, $f$ is irreducible over $F$ and $[K:F] = n!$, hence $[F(\alpha):F] = n \implies [K:F(\alpha)] = (n-1)!$, so $|\Gal(K/F(\alpha))|= | S_{n-1}|$. Since $\Gal(K/F(\alpha))$ ...


2

First, remember the classic example of a non-separable extension, $\mathbb{F}_p(\sqrt[p]{X})/\mathbb{F}_p(X)$. It's non-separable because the minimal polynomial for $\sqrt[p]{X}$ has repeated roots: $T^p-X=(T-\sqrt[p]{X})^p$. Observe that the $p$th power of any element of $\mathbb{F}_p(\sqrt[p]{X})$ is in $\mathbb{F}_p(X)$, because the Frobenius map is a ...


2

Let $f\in K[X]$ be the minimal polynomial of $\alpha$ over $K$. Let $K\subset K_1 \subset K(\alpha)$ be an intermediate extension. Let $f_1$ be the minimal polynomial of $\alpha$ over $K_1$. Then $f_1\in K_1[X]$ is a factor of $f$. Let's notice that $K_1$ is the field $K_1'$ generated over $K$ by the coefficients of $f_1$. Indeed, the extensions $K(\alpha)...


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