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0

The original statement is true. We have that if $f$ is an irreducible cubic, and then $\mathbb{Q}(\alpha): \mathbb{Q}$ is radical. Then there exist, from the definition, $\beta_1, \ldots, \beta_n$ with $$ \mathbb{Q}(\alpha) = \mathbb{Q}(\beta_1, \ldots , \beta_n), $$ and for each $i$ we may assume there is some prime $p_i$ such that $$ \beta_i^{p_i} \in ...


0

Let $f(x) = x^q - x$. To show that $f$ splits in $\mathbb{F}_q$, we wish to demonstrate the existence of $q$ roots, perhaps counting multiplicity. Now if $x$ is a root of $f$, then notice it must satisfy $x^q = x$. Obviously this holds for $0$, so let's turn our focus to the invertible elements. The multiplicative group $\mathbb{F}_q^\times$ has order ...


0

Hint: If $x$ is a non-zero element of $F_q$, then $x^{q-1}=1$.


2

Hint: Show that as $m (x)$ is irreducible then through the homomorphism $\varphi : K[x] \to \Omega$ defined as $\varphi(f)(x) = f(a)$ as $a \in \Omega$ is such that $m (a) = 0$ you have that $$\frac{K[x]}{\langle m(x) \rangle} \simeq K[a]$$ (Use the First Isomorphism Theorem) Show that for any $f(x) \in K[x]$, $f(a)$ can be written uniquely as ...


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.


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 $F$. ...


0

As other have pointed out the key to finding the splitting field and the Galois group is to prove that $f(x)=x^4+x-1$ is irreducible in $\Bbb{F}_3[x]$. Because this is degree four, we can no longer test this simply by checking the absence of zeros in the prime field. David Wheeler described a good way of handling a quartic by testing the possibility of it ...


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

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 $F(a)=\{p(a):p(x)\in F[x]\}$


2

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 ...


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 ...


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 ...


0

To the implied question about finite automorphism groups and infinite extensions: $\mathbb{Aut}(\mathbb{R}/\mathbb{Q})$ is trivial, but $\mathbb{R}$ is clearly not a finite extension of $\mathbb{Q}$. (To see this note that in $\mathbb{R}$, $x > y$ iff there is a $z$ with $x - y = z^2$. Hence an automorphism of $\mathbb{R}$ must be order-preserving, from ...


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: ...


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 $\mathbb{Q}$. ...


0

If $n$ is prime, you get a lot more than a domain. You get a field of characteristic $p$. Moreover, every finite field of characteristic $p$ must have $p^n$ elements for some $n$ and $\mathbb{Z}/p\mathbb{Z}$ is its prime field. In this way one can apply linear algebra and dimension theory for vector spaces. ADDED 1: $\mathbb{Z}/p\mathbb{Z}$ has been used ...


0

The assertion is the same as $\mathbf Z/n\mathbf Z$ has no zero-divisor iff $n$ is prime. Having no zero-divisor means that if $ab\in n\mathbf Z$, then $a$ or $b\in n\mathbf Z\,$; in other words, if $n$ divides $ab$, then $n$ divides $a$ or $b$. This is exactly Euclid's lemma. Some details: suppose $ \mathbf Z/n\mathbf Z$ is a domain, and $n$ divides a ...


0

The claim that there are fields of every infinite cardinality depends essentially on "choicy methods". Without the Axiom of Choice, it is consistent that there may exist an amorphous set, that is, an infinite set that does not have two disjoint infinite subsets. Such a set cannot be given a field structure. An amorphous field cannot contain any nonzero ...


0

You can always arrange for a fraction $r(t) = \frac{P(t)}{Q(t)}$ to have an even denominator $r(t) = \frac{P(t) Q(-t)}{Q(t)Q(-t)}= \frac{P_1(t)}{Q_1(t)}$. An even polynomial like $Q_1(t)$ is invariant under $\sigma$. So if $r(t)$ is invariant, $P_1(t)$ must also be even. One may need to do some reductions to get an invariant fraction as a quotient of two ...


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 ...


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.


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.


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 ...


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 ...


0

If $\kappa$ is a finite cardinality, this is only possible if $\kappa =p^n$ for some prime $p$ and whole number $n$. If $\kappa$ is infinite this is possible by Zorn Lemma. Indeed pick a set $A$ or cardinality $\kappa$ [we need this to be sure that we work inside a set, otherwise we might run into the set of all sets]. Define $$\mathcal F := \{ (B, +, ...


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 ...


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 $\cos(\frac{2\pi}{n})$. ...


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 ...


0

We have $3$ proposed roots, and a cubic equation. We calculate: $(x - a_1)(x - a_2)(x - a_3) = (x - a_1)(x - (a_1^2 - 2))(x - (a_1^3 - 3a_1))$ $= x^3 - (a_1 + a_1^2 - 2 + a_1^3 - 3a_1)x^2 + (a_1(a_1^2 - 2) + a_1(a_1^3 - 3a_1) + (a_1^2 - 2)(a_1^3 - 3a_1))x - a_1(a_1^2 - 2)(a_1^3 - 3a_1)$ $= x^3 - ((a_1^3 + a_1^2 - 2a_1 - 1) - 1)x^2 + (a_1^5 + a_1^4 - ...


0

Another example of uncountable-degree algebraic extension: Let the small field be $k=\Bbb R(t)$, rational functions in one variable over the reals. Then the irrationalities $\sqrt{t-\alpha}$, for $\alpha\in\Bbb R$, are clearly $k$-linearly independent, and uncountable in number, so the compositum of all the quadratic extensions $\Bbb R(\sqrt{t-\alpha})$ over ...


0

Note that be assumption, there exist an element $a\in S_1$ and $b\in S_2$ which are algebraically dependent. Also, by assumption, we know that two elements $a_1,a_2\in S_1$ are independent, and the same for $S_2$. Thus, if $a\in S_1\cap S_2$, then for every element $c\in S_1$, $a$ and $c$ are independent, as they are both in $S_1$, and the same is true of ...


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 ...


0

We can also employ a basic inductive argument. Suppose the proposition holds for all positive integers $n<m$. We have $$\left(\frac{1}{x}\right)^m-1=\prod_{d|m}\Phi_d\left(\frac{1}{x}\right)=\left(\prod_{\stackrel{1<d<m}{d|m}}\Phi_d\left(\frac{1}{x}\right)\right)\cdot\Phi_m\left(\frac{1}{x}\right)\left(\frac{1}{x}-1\right).$$ Multiplying by the ...


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 ...


0

Suppose we start from $$\Phi_n(x) = \prod_{d|n} (x^d-1)^{\mu(n/d)}.$$ With $n$ odd we get $$\Phi_{2n}(x) = \prod_{d|2n} (x^d-1)^{\mu(2n/d)} = \prod_{d|n} (x^d-1)^{\mu(2n/d)} \prod_{d|n} (x^{2d}-1)^{\mu(2n/d/2)} \\ = \prod_{d|n} (x^d-1)^{-\mu(n/d)} \prod_{d|n} (x^{2d}-1)^{\mu(n/d)} \\ = \prod_{d|n} (x^d+1)^{\mu(n/d)}.$$ This is $$(-1)^{\sum_{d|n} ...


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 $a\in\mathbb{F}$


0

I think I've got an answer. the hint hasn't totally sunken in it but I think I have made a little progress. seeing we have $(−1)^k=1$ then it must be that 2∣k and we have $(ξ)^k=1$ and that $ξ$ is a primitive nth root so its powers are nth roots so then n∣k but GCD(2,n)=1 so 2n=αk for some α i.e. 2n∣k so I can conclude that (−1)(ξ)=−ξ is a primitive 2nth ...


0

This is true. The multiplicative group of $\Bbb{F}_q$ is cyclic of order $$ q-1=\ell^2-1=(\ell-1)(\ell+1). $$ If $\ell$ is odd (or, equivalently, if $p$ is an odd prime) then $\ell+1$ is even. Therefore $2(\ell-1)\mid q-1$. This means that in $\Bbb{F}_q^*$ there is an element of order $2(\ell-1)$. Let $a$ be such an element. Then we know that ...


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 $$ \Phi_{2n}(x) = \prod_{(k,n)=1} (x-e^{2\pi i(2k+n)/2n}) = \prod_{(k,n)=1} (x+e^{2\pi i k/n}) = \prod_{(k,n)=1} (-x-e^{2\pi i k/n}) = \Phi_n(-x). $$


0

Put another way: $K$ is a vector space over the field $H$. It must be finite-dimensional, since $K$ is finite, and a basis is a subset of $K$. So say the dimension of $K$ as a vector space over $H$ is $r$. Then $K$ is isomorphic to $H^r$ (which is also a $k$-dimensional vector space over $H$). This may be the part you don't get: that any two ...


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 ...


0

If $p$ is the characteristic, then $X^p-a^p=(X-a)^p$, and if there were any nontrivial $F$-factorization, one of the factors would be $g=(X-a)^m$ for $1\le m<p$. But the $m-1$-degree term of $g$ is $-maX^{m-1}$, implying that $ma\in F$, and thus that $a\in F$. If $p$ is prime to the characteristic, then the $p$-th roots of unity are $p$ in number, and ...


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 - ...


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.


0

The computation is required by the definition of the homomorphism: basically it is defined first for positive integers, then extended to negative integers. So for formulae involving two integers $m$ and $n$, one has to examine the different combinations. You meet the same problem in mid school on extending the exponent notation from natural numbers to ...


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 $F$.


0

Here is an elementary argument that does not even need the structure theorem of abelian groups: Let $n$ be the exponent of $H$, that is, the smallest $n$ such that $h^n=1$ for all $h\in H$. By Lagrange's theorem, $n\le 8$. If $n<8$, then equation $x^n=1$ would have $8$ solutions, and this cannot happen in a field. Hence, $n=8$. If all elements have ...


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, $$\mathbf F_9^{\times}\simeq \mathbf Z/8\mathbf Z$$



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