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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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