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Since $a^5=(1+\sqrt[3]{7})^{7\cdot 13} /((\sqrt[3]{7}-7)^3+77)^{5\cdot 13}$, it's clear that $a^5\in\mathbb{Q}(\sqrt[3]{7})$. Since the degree of $\mathbb{Q}(\sqrt[3]{7})$ over $\mathbb{Q}$, 3, is prime, there are no intermediate fields between $\mathbb{Q}(\sqrt[3]{7})$ and $\mathbb{Q}$. It follows that unless $a^5\in\mathbb{Q}$, $a^5$ will generate ...

2

The zero ideal is in fact maximal in $M_2(\mathbb{R})$, as it is for $M_n(R)$ for any $n>1$ and ring $R$ (see Why is the ring of matrices over a field simple?). I will give the proof for $M_n(F)$, $F$ a field and $n>1$ since vector space theory makes it a bit more elegant. To prove the zero ideal is maximal, we take any non-zero ideal $\mathfrak{A}$ ...

1

You're one the right track, but you need to follow through! Substituting $x=\alpha$ and using the fact that $p(\alpha)=q(\alpha)=0$, you can conclude that $r(\alpha) = 0$. Thus, you have found a polynomial of strictly smaller degree that also claims $\alpha$ as a root. Now, apply the division algorithm to $q(x)$ and $r(x)$ as in the Euclidean algorithm. ...

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Based on the link that Gregory Grant gave, here is my solution. ~~~ Let $B \subseteq S$ be a transcendence basis for $F(S)$ over $F$. Then $F(S)/F(B)$ is algebraic and finitely generated, and hence finite. Given any field $M$ such that $F(S)/M/F$:   If $[M:F] > [F(S):F(B)]$:     Let field $N$ be such that $M/N/F$ and $\infty > ... 0 As its mentioned that q divides order of A. That means, order of A is a natural number(i.e. finite). You cannot deduce that domain of characteristic q(not equal to 0) is finite. There are fields of characteristic q(not equal to 0), which are infinite. Consider, for example, algebraic closure of Z/pZ, for p a prime number. Now, that the integral domain given ... 1 Moos is right: if you have a finite domain, it makes the most sense to solve is using Lagrange's theorem. Even if you haven't covered it, you certainly will at some point, and this is a great opportunity. You will find a proof in any group theory chapter, or in learnmore's post. Since the characteristic of a ring with identity is the same as the additive ... 0 Note the following segment from the fundamental theorem of galois theory Let$L$be an intermediate field of$E/Q$, then$L/Q$is galois$\iff Gal(E/L)\trianglelefteq Gal(E/Q)$That is,$L/Q$is a galois extension when the galois group for$E/L$is a normal subgroup of the galois group for$E/Q$. Since$E/Q$is galois,$|Gal(E/Q)|=[E:Q]=p^2$. Any group ... 0 Equivalently: If$L$is an integral domain, finite-dimensional over$K$, is$L$a field? The answer is yes: If$a\in L$is nonzero, then$x\mapsto ax$is a linear transformation of finite-dimensional$K$-vector spaces. It is injective, hence surjective, and it follows that$a$is a unit. 0 We have a homomorphism$K \hookrightarrow L \twoheadrightarrow L/p$. Any homomorphism between fields is automatically a field extension. So$p$must be maximal, then this is a field extension. In the finite-dimensional case, this is always the case. A field extension is nothing else but an algebra over a field, which also happens to be a field itself. 0 If$L_i = K(x_i)$, then$L_i=K(x_i+c)$for any$c \in K$(and since$char(K)=0$, we have enough$c$'s to choose from) yields another primitive element with another minimal polynomial. So you can assume$A = L_1 \times \dotsc \times L_n$with$L_i = K(x_i)$, the$x_i$having distinct minimal polynomials$f_i \in K[T]$. In particular they are pairwise ... 0 Let$I$denote the integral domain since char ($I)=q$;$q$is the least positive integer such that$q.a=0\forall a\in I;(0) $denotes zero element of$I$Assuming$o(I)=n$we have$n.a=0$To prove$q$divides$n$; let$n=pq+r$where$0\leq r<q$So$na=p(qa)+ra\implies 0=0+ra\forall a\in I\implies ra=0\forall a$But$q$is the least positive integer ... 1 Yes,$A$should be finite to make such a conclusion. For the proof of the statement, you should look at the underlying abelian group und use Lagrange's theorem. Characteristic$q$implies that$A$contains a copy of$\mathbb Z/q\mathbb Z$. 1 Let$\;f(x)=\sum_{k=0}^n a_nx^n\in\Bbb Q[x]\;$, and$\;f(\pi)\in\Bbb Q)\pi)\;$. Then, $$\sigma(f(\pi))=f(-\pi)\implies f(\pi)\in\Bbb Q(\pi)^\sigma\iff a_r=0\;\;\text{for all odd}\;\;0\le r\le n$$ and from here follows the claim. 4 Call the fixed field$F$. (1) The extension$\mathbf{Q}(\pi)/F$is generated by$\pi$which is quadratic algebraic over$F$(as it is the square root of the element$\pi^2\in F$). Also$\mathbf{Q}(\pi)/\mathbf{Q}(\pi^2)$is quadratic for the same reason. Now you can see that$F/\mathbf{Q}(\pi^2) $is of degree 1. 5 If you know a little Galois theory, then one way of going about this is to start by listing the intermediate fields between$\mathbb{Q}$and$\mathbb{Q}[\sqrt{2}, \sqrt{3}]$, of which there are only three:$\mathbb{Q}[\sqrt{2}]$,$\mathbb{Q}[\sqrt{3}]$, and$\mathbb{Q}[\sqrt{6}]$. (Galois theory guarantees this list is exhaustive.) If$\sqrt{5} \in ...

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Every nonempty subset of $\,S = \{\sqrt2,\sqrt3,\sqrt 5\}\,$ has product $\not\in \Bbb Q\,$ thus $\,[\Bbb Q(\sqrt2,\sqrt3,\sqrt 5):\Bbb Q] = 8\,$ by this answer. Said informally, multiplicative independence implies linear independence (the reason for such becomes clearer when one studies Galois theory of Kummer extensions). One can also give a direct ...

1

One way to do this is: $$n|m\implies X^n-1|X^m-1 \tag 1$$ to prove it take $m=nt$ then : $$X^m-1=(\color{#0a0}{X^n})^t-1=(\color{#0a0}{X^n}-1)\left(\sum_{i=0}^{t-1}(\color{#0a0}{X^n})^i\right) \tag 2$$ Or you can see this as a consequence of $Y-1|Y^t-1$ for $Y=X^n$. Another result which follows from this is that if you replace $X$ by a specific number in ...

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If $k\mid n$, then every root of $x^{p^k-1}-1$ is also a root of $x^{p^n-1}-1$, so the field $\mathbb{F}_{p^k}$ is a subfield of $\mathbb{F}_{p^n}$, by recalling that, in an algebraic closure of $\mathbb{F}_p$, the field with $p^r$ elements is uniquely determined as the set of roots of $x^{p^r}-x$. Let's see the converse. If $\mathbb{F}_{p^k}$ is a subfield ...

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Some ideas: Prove that $\;w\;$ is a multiple root of a non-zero polynomial $\;f\;$ iff $\;w\;$ is also a root of its derivative $\;f'\;$. Deduce that if $\;f\;$ is irreducible (over some field), then this happens iff $\;f'=0\;$, and thus over a field of characteristic $\;p>0\;$ this can happen iff all the non-zero coefficients of $\;f\;$ correspond to ...

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Extended hints: The polynomial is irreducible, because Eisenstein with $p=$___ says so. The discriminant of this polynomial is $81$, which is a ________. Therefore the Galois group is __________. The first step in solving a cubic is to get rid of the quadratic term with the aid of a linear substitution. Here this amounts to calculating that $$... 1 Your solution is fine. It does follow in a similar way that the cube cannot be quadrupled. But, for example, this method would not work to show that the cube cannot be duplicated 8-fold, since \sqrt[3]{8} is not degree three. 1 The question means that if we have a finite fied K then there exists a prime p such that K is isomorphic to F_p. First of all,the characteristic of K must be a prime p_1 hence the subfield:$$K_{p_1}=\{1_k,2.1_k,\cdots,(p-1).1_k\} $$Is isomorphic to F_{p_1}. 1 Take 1 \in K. Then  1+1, (1+1)^{-1} , etc. are in K. If K is infinite, this gives you a copy of the rationals. If not, you get a finite field, which is \mathbb Z_p 0 It says that, if you take finite field F, it contains a preime field (isomorphic to \mathbf Z/p\mathbf Z also denoted \mathbf F_p in the context of finite fields) for some p. You simply have to map \mathbf Z to F by sending n to n\cdot 1 and consider the kernel of this homomorphism. 0 The non-zero elements of \mathbf F_{q^n} are a cyclic group of order q^n-1=2m. So, if y^2=r, we have r^m=y^{2m}=1. Conversely, suppose  r^m=1, and let \zeta be a generator of \mathbf F_{q^n}^{\times}. We can write r=\zeta^k for some k. Since a generator has order 2m, the hypothesis implies:$$\zeta^{km}=1\iff km\equiv 0\mod 2m \iff ...

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The relation $\beta=f(\alpha)/g(\alpha)$ can be written as $$f(\alpha)-\beta g(\alpha)=0$$ which means that $\alpha$ satisfies the polynomial $$f(X)-\beta g(X)$$ which has coefficients in $K(\beta)$. Since $\beta\ne0$ and $g(X)\ne0$, this polynomial is nonzero, hence $\alpha$ is algebraic over $K(\beta)$.

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You've got the right idea, but I think the way you are going about things might be a bit difficult. Instead, consider $F(\beta)$ as an intermediate subfield between $F$ and $F(\alpha)$. If $[F(\beta):F]$ and $[F(\alpha):F(\beta)]$ are both finite, then by the multiplicativity of field extension degree, $[F(\alpha):F]$ is finite, a contradiction. Hence, if we ...

2

It is obvious that $F(1+a^{-1})\subset F(a)$. And since $$\frac1{(1+a^{-1})-1}=a$$ we have that $F(a)\subset F(1+a^{-1})$. They have the same index because they are the same extension.

1

$\mathbb Q[\sqrt{2}]/\mathbb Q$ and $\mathbb Q[\sqrt[4]{2}]/\mathbb Q[\sqrt{2}]$ are normal, but $\mathbb Q[\sqrt[4]{2}]/\mathbb Q$ is not.

1

Your first implication is not correct it should be: $$a=\sqrt{a}\cdot \sqrt{a}\implies a\in \mathbb{Q}(\sqrt{a})\implies \mathbb{Q}(a)\subset \mathbb{Q}(\sqrt{a})\$$ in the second inclusion: $$\sqrt{a}=1+a\implies \sqrt a\in \mathbb{Q}(a)\implies \mathbb{Q}(\sqrt a)\subset \mathbb{Q}(a)\$$

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$(a+1)^2 = a^2+2a+1 = a \,\,\,$

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Fact: Consider two polynomials $f$ and $p$ over $\Bbb{Q}$, with $p$ irreducible. It can be proved that if $f$ and $p$ share a root, then $p$ divides $f$. How does this help? Suppose that $\alpha$ is a root of an irreducible polynomial $f \in \Bbb{Q}[X]$ of degree $n$ and that $[\Bbb{Q}(\alpha):\Bbb{Q}] = m < n$. Since $[\Bbb{Q}(\alpha):\Bbb{Q}]$ is ...

3

The polynomial is irreducible over the rationals, because its possible rational roots are to be found among $\pm1$, $\pm2$, $\pm4$ and $\pm8$. A direct check shows these numbers are not roots. Since the polynomial has degree $3$, reducibility over $\mathbb{Q}$ coincides with having a rational root. So, if $\alpha$ is a root of the polynomial, $f$ is its ...

0

edit: It seems that Thomas is asking for is a proof that such a vector space has infinitely many subspaces. The following answer rather regards the existence of infinite subspaces. (First note that a vector space $V$ has infinite subspaces iff it is infinite. Maybe what you meant is infinite proper subspaces.) You couldn't find a counter example because ...

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Take $\sigma=id: F=\mathbb C\to \bar F=\mathbb C$ and choose $E$ and $\alpha$ any way you like: $\sigma$ won't extend to $\mathbb C(\alpha)$.

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Any example will do, such as $F=\mathbb Q$, $\alpha=\pi$. Since $\overline F$ is algebraic over $F$, $\tau(\alpha)$ would have to be algebraic, say $f(\tau(\alpha))=0$ with $f\in F[X]$. Since $f(\alpha)\ne 0$, $\tau$ fails to be monomorphic.

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yes It must have two linearly independent elements $x$ and $y.$ For any $\theta \in (0,1)$ consider the set spanned by $\theta x + (1-\theta) y.$ This will be a different subspace for each $\theta.$

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Another way to see the result is as follows: You found the irreducible polynomial fo r $\alpha=\sqrt{1+\sqrt 2}$ essentially by noticing that if you square $\alpha$, then subtract $1$, you obtain a anumber that, when squared, gives $2$: $f(X)=(X^2-1)^2-2=X^4-2X^2-1$. But it is of course the case for all roots of $f$ that squaring, subtracting $1$, and ...

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Note that finding the conjugates of your given root is equivalent to factoring the minimal polynomial into linear terms $(X-\alpha_1)(X-\alpha_2)(X-\alpha_3)(X-\alpha_4)$. Using the quadratic formula, you can rewrite your minimal polynomial as $(X^2-1-\sqrt{2})(X^2-1+\sqrt{2})$, which is then easy to factor into linear terms.

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You can use what is called a “primitive element” of K for what you can take an element x such that the images of x by the 20 Q-automorfisms of K (splitting field) be distinct. You can verify that the simple sum of your elements satisfy this condition.

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As Andreas Caranti pointed out, there do exist finite extensions that are splitting fields and yet are not separable. I think now that the best way to make sense of the terminology is to abandon the idea of a "finite normal extension." Leave the definition of a normal extension $E$ of $F$ as a splitting field over $F$, then based on this, define a Galois ...

1

Since $[K:F]$ is finite, $a$ is a root of a polynomial in $F[X]$, and this polynomial belongs to $E[X]$, too.

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The statements agree when the extension is finite as you point out. However, the notion of degree is not meaningful in the context of infinite extensions. Thus, the second characterization does not exist for infinite extensions.

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Yes. Look at the union of all fields $\mathbb F_{p^{\ell^n}}$, where $p$ and $\ell$ are two fixed prime numbers. This is an infinite field, and all elements are roots of unity.

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I think the algebraic closure of $\mathbb{F}_p$ does the job for $p$ a prime number. Take $x\in\overline{\mathbb{F}_p}$ such that $x\neq 0$ then $\mathbb{F}_p[x]$ is a finite field and hence $x$ is of $\underline{\text{multiplicative}}$ finite order in $\mathbb{F}_p[x]$ and hence in $\overline{\mathbb{F}_p}$. Of course, the last thing to know is that ...

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Let $F$ be a field of characteristic $p$. Let $f = (1 + x)^p \in F[x]$. We want to show that $f = 1 + x^p$. Take the formal derivative: $f' = p(x+1)^{p-1} = 0$ Now we know that $f$ has degree $p$, and its derivative is $0$, so $f$ must be in the form $A + Bx^p$ with $A$, $B \in F$. $f(0) = 1$ so $A = 1$. A product of monic polynomials is always monic so ...

1

If you look at Zagier's one-sentence-proof of Fermat's theorem on sums of two squares, you will notice that every prime of the form $p=4k+1$ can be written as the sum of two squares: from $$a^2+b^2 = p$$ it follows that $-1$ is a quadratic residue $\!\!\pmod{p}$. Obviously, the last sentence also follows from the fact that the Legendre symbol satisfies: ...

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In this case $F[i]=F(i)$. This is because $F(i)$ is the field of fractions of $F[i]$, which is already a field. You can see this becuase the characteristic polynomial of $i$ is irreducible in any subfield or $\mathbb{R}$, or more explicitly by computing $(a+bi)\frac{1}{a^2+b^2}(a-bi)=1$, so that the inverse of any nonzero element exists.

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You are missing the primitive element theorem. That is the critical argument. To be more precise: a separable extension of degree $n$ would have a primitive element, whose minimal polynomial would be irreducible of degree $n$, contradiction!

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Take $\mathbb Q(x)/\mathbb Q(x^k)$. This is not a normal extension for any $k \geq 3$. The normal closure is $\mathbb Q(x,\zeta)$ with $\zeta = \exp(\frac{2\pi i}{k})$ a primitive $k$-th root of unity.

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