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7

Given two finite fields $E_1$ and $E_2$ that are both extensions of a field $\mathbb{K}$, we have that $$E_1\cong E_2 \text{ as fields} \iff \lvert E_1\rvert = \lvert E_2\rvert \iff E_1\cong E_2\text{ as \mathbb{K}-vector spaces}$$ so you can't find any examples with $\mathbb{K}$ finite. Hint: Try using the field $\mathbb{K}=\mathbb{F}_p(T)$. For ...

6

Let $F$ be a finite field, and $X$ an infinite set, let $\hat X$ be the set of all finite strings of things in $X$ where two strings are equal if they differ only by their order (e.g. $x_1x_3x_2=x_1x_2x_3$). Then let $A$ be the set of formal sums $\{\sum_{i=1}^n f_ix_i\mid n\in\Bbb N,\ f_i\in F\ \forall i,\ x_i\in \hat X\ \forall i\}$. Then $A$ is a ...

4

No, there is no such example, because if the polynomial ring $R[x]$ is a PID, then $R$ is a field, so that $R[x]$ is also Euclidean. For references see here.

3

On the question of whether choice is required: as Gregory Grant's answer shows, the following is a theorem of ZF: $$\text{Suppose \kappa is a cardinality. Then there is a field F with \vert F\vert \ge \kappa - in particular, \kappa injects into F.}$$ Moreover, if $\kappa$ is equinumerous with $\kappa^{<\omega}$ (the cardinality of finite ...

3

There are many more than that. As $\pi$ is transcendental, $\mathbf Q(\pi)\simeq \mathbf Q(x)$, and it is known that $$\operatorname{Aut}(\mathbf Q(x))=\mathbf{PGL}_2(\mathbf Q),$$ the projective linear group of order $2$ over $\mathbf Q$ which is the set of homographic transformations: $$x\mapsto \frac{ax+b}{cx+d}, \quad ad-bc\neq 0.$$ Counter-example: ...

3

Hints: If $\mathbb{F}$ is a finite field then $\mathbb{F}^{\times}:=\mathbb{F}\setminus\{0\}$ is a cyclic group with respect to the multiplication in $\mathbb{F}$ $a+a+a=0$ for any $a\in\mathbb{F}$

3

If $\sigma(t)=-t$, then $\sigma(t^k) = (-1)^k t^k$, so a polynomial $\sum a_n t^n$ is fixed by $\sigma$ iff $a_i=-a_i$ when $i$ is odd. So a rational function is fixed iff both numerator are fixed or both are negated. So it should be the field generated by quotients of even polynomials and quotients of odd polynomials.

2

The answer given by Spooky is excellent. Note that the fixed field, as described, is simply $\mathbb{R}(t^2)$. Regarding uniqueness of $\sigma$. Since $\mathbb{R}(t)$ is generated by the single element $t$, once we choose $\sigma(t)$ we determine $\sigma$ completely.

2

Say your field isomorphism is $f: K \to L$. Take a non-constant polynomial with coefficients in $L$, apply $f^{-1}$ to get a non-constant polynomial with coefficients in $K$. This has a root, since $K$ is algebraically closed; call it $x$. Then you can easily check that $f(x)$ is a root of your original polynomial.

2

Hint: The primitive $n$th roots of unity are $e^{2\pi i k/n}$ for $(k,n) = 1$, while the primitive $2n$th roots of unity are $e^{2\pi i \ell/2n}$ for $(\ell,2n)=1$. If $(k,n) = 1$ then $(2k+n,2n)=1$. Therefore $$\Phi_{2n}(x) = \prod_{(k,n)=1} (x-e^{2\pi i(2k+n)/2n}) = \prod_{(k,n)=1} (x+e^{2\pi i k/n}) = \prod_{(k,n)=1} (-x-e^{2\pi i k/n}) = \Phi_n(-x).$$

2

As commented, the question is missing an essential piece of information, the ground field. To get a somewhat non-trivial question, the ground field should probably be $\mathbb Q(\pi)$. Now the following reasoning works: Any $a\in\mathbb Q(\pi)$ has the form $a = \frac{f(\pi)}{g(\pi)}$ with $f,g\in \mathbb Q[x]$, $g\neq 0$. If $a^3 = \pi$, then $f(\pi)^3 - ... 2 Suppose that your regular polygon has a vertice on the$x$-axis. Then the first vertice counted counter clock wise has coordinates$(\cos(\frac{2\pi}{n}),\sin(\frac{2\pi}{n}))$. Hence if you know the construction of the polygon, by projecting the first vertice on the$x$axis, which can be done with a ruler and a compass, you can get$\cos(\frac{2\pi}{n})$. ... 2 No! Consider a set$I$of indices and the field of rational functions$K=k(X_i|i\in I)$over an arbitrary field$k$. The extension field$K\subset L=k(\sqrt X_i|i\in I)$is algebraic (since it is generated by algebraic elements), of degree$[L:K]\geq \operatorname {card} I$because the elements$\sqrt X_i$are linearly independent over$K$. Thus by taking ... 2 Your first question should be: is this polynomial irreducible in$\Bbb F_3[x]$? It's clear it has no root in$\Bbb F_3$, but this is not enough, we must check for possible quadratic factors. So, suppose (by way of contradiction) we had:$x^4 + x - 1 = (x^2 + ax + b)(x^2 + cx + d)$with$a,b,c,d \in \Bbb F_3$. Then$a + c = 0$(since our polynomial has no ... 1 Are you familar with Kronocker? The polynomial is irreducible over the finite field. Kronocker gives you a field which contains a root of your polynomial. The quotient is generated by$\{1,x,x^2,x^3\}$then how many elements does it have? Does this new field contain the other$3$roots? How does the isomorphic copy look? (knowing what the basis is for the ... 1 Here's an example.$K=\mathbb C$,$F=\mathbb Q$,$a=\sqrt[3]{2}.$Then$F(a)$is going to be the smallest field containing$F$and$a$, from your definition. So here, as$\mathbb Q(\sqrt[3]{2}) \supseteq \mathbb Q$,$\mathbb Q(\sqrt[3]{2}) \ni a$, we have$\mathbb Q(\sqrt[3]{2}) \supseteq \{a+b\sqrt[3]{2}|a, b \in \mathbb Q\}$. But this isn't yet a field, as ... 1 The answer by Kaj is good. But I'll add one thing: This only works if$a,b$are algebraic over$F$. Same goes for$F(a)=\{p(a):p(x)\in F[x]\}$1 You are correct in your first hunch. If we have fields$F \subset K$and an element$a \in K$, then by definition,$F(a)$is defined to be the smallest subfield of$K$that contains$a$. However, if$a$is algebraic over$F$, then$F(a) = \{p(a) \ | \ p(x) \in F[x] \}$. For a proof that these are equivalent statements when$a$is algebraic, see my post ... 1 One way to describe the algebraic closure is that it is in some sense a "maximal" algebraic extension: it's an algebraic extension into which every other extension embeds. So it seems to me like the following question is a more basic one that should be answered first: What's an algebraic extension of commutative rings? There are various ways to answer ... 1 Sorry, I'm not going to use blackboard bold, but here goes. Let$p\in E$,$p\not \in K$since$K\subsetneq E$, then since$E\subseteq F$, we have$p=\frac{f(u)}{g(u)}$, for$f$and$g$polynomials in$K[x]$. Then$pg(x)-f(x)\in E[x]$has$u$as a root since$pg(u)-f(u)=f(u)-f(u)=0$, note that this polynomial is necessarily nonzero since$p\not\in K$. Thus ... 1 This is actually not true;$\mathrm {Aut}(\mathbb{Q}(\pi)/\mathbb Q)$is infinite. In general, if we have a field of rational functions$\mathbb{Q}(x_1,x_2,\ldots,x_n)$in indeterminates$x_1,x_2,\ldots,x_n$(for which$\mathbb{Q}(\pi)$is with$n=1$) the automorphism group of this field over$\mathbb{Q}$contains the general linear group over$\mathbb{Q}$. ... 1 Two points: One, Galois closure is a relative concept, that is not defined for a filed, but for a given extension of foields. Second, it is not something maximal. To the contrary it is something minimal. Given an extension of fields$F\subset E$if it is not Galois, then the smallest extension of$F$that containing$E$and that is a Galois extn of$F$is ... 1 Actually any finite subgroup of the multiplicative group of a field (whether the field itself is finite or not) is cyclic. In the present case, $$\mathbf F_9^{\times}\simeq \mathbf Z/8\mathbf Z$$ 1 The one thing you know for sure about$H$and$K$is that they each contain$F$as a subfield (up to isomorphism). What they are saying is that$\mu$fixes$F$, just like the other answers say. Given some$x \in F \subset K$, its image under$\mu$is$x \in F \subset H$. 1 It means that$\nu|_F$is the identity map on$F$. In other words,$\nu(x)=x$for all$x\in F$. 1 Yes. The elements of$E$algebraic over$F$form a subfield of$E$, call it$L$, that contains$F$. Since$E=F(\{\alpha\})$where$\{\alpha\}$is a set of algebraic elements,$E$is contained in$L$. Thus$E=L$is algebraic over$F$. 1 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 ... 1 I think you are getting confused because the book uses the symbol$a$for two different purposes. Let's rephrase like this: Let$F$be a field, and let$K$be an extension of$F$. If there is some$b\in K$such that$K=F(b)$, and the minimal polynomial of$b$over$F$has degree$2$then we say that$K$is a quadratic extension of$F\$. ...

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