New answers tagged

0

This follows automatically from one of the equivalent definitions of normal extension: $\;K/E\;$ is normal iff every injection $\;K\to\overline E\;$ is in fact an $\;E\,-$ automorphism of $\;K\;$ , with $\;\overline E\;$ a fixed algebraic closure of $\;E\;$ . Thus, Gal$\,(K/\Bbb Q)\;$ is the set of all $\;\Bbb Q\,-$ automorphism of $\;K\;$, and since the ...


2

Here's another example. First, note that $x^3-2$ is irreducible over $\mathbb{Q}$, by Eisenstein's Criterion with $p=2$. This polynomial has the following roots: $$\sqrt[3]{2},\ \zeta_3\sqrt[3]{2},\ \zeta_3^2\sqrt[3]{2},$$ where $\zeta_3 = -\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}$ is a primitive 3rd root of unity. Note that since each of the above numbers is a ...


8

No, take $\mathbb{Q}(x)$ and $\mathbb{Q}(y)$ where $x$ is any transcendental number and $y$ is any transcendental number not in $\mathbb{Q}(x)$ (which exists since $\mathbb{Q}(x)$ is countable and transcendental numbers are uncountable) : they are both isomorphic to $\mathbb{Q}(X)$ but they are not equal. PS : all subfields of $\mathbb{C}$ contain ...


2

By the Rational Root Theorem, the only possible rational roots are $\pm 1, \pm 3$, but substituting shows that none of these are roots. Hence (since $\deg f < 4$), $f$ is irreducible, so $\operatorname{Gal}(f)$ is transitive and thus $3 \mid \#\operatorname{Gal}(f) = [T : \Bbb Q]$. On the other hand, the discriminant of $f$ is $$\Delta_f = 4(18)^3 - ...


2

If you denote $\alpha$ the real root of $f(x)$, you obtain the minimal polynomial of the complex roots of $f(x)$ dividing it by $x-\alpha$. It will have degree $2$ (actually it is $x^2+\alpha x+\alpha^2+18$), hence the degree of the splitting field over $\mathbf Q$ is $6$.


4

The converse is false: let $k=\mathbb{Q}$, $K=\mathbb{Q}(\sqrt{2})$, and $L=\mathbb{Q}(\sqrt{3})$. Then $\mathrm{Gal}(K/k)\simeq \mathrm{Gal}(L/k)\simeq \mathbb{Z}/2\mathbb{Z}$, but $K$ and $L$ are not isomorphic, since $K$ contains a square root of $2$ but $L$ doesn't.


1

The usual meaning of “having a multiple root” includes “in some extension field”. I can only interpret “has a linear factor” assuming “in $F[x]$” or it would make no sense. I don't have Dummit-Foote available, so I guess that the exercise has some context, because the statement can be false in case the field has characteristic $3$. A polynomial $f(x)\in ...


1

If the cubic $f$ has a multiple root in some extension field $E$, and $F$ does not have characteristic $3$, then the cubic and its derivative have a common root in the extension field, so they are not relatively prime in that field. It follows $f$ and $f'$ are not relatively prime in $F$. If $f'$ divides $f$, then $f$ has a linear factor. If $f'$ does not ...


1

A complex number $z$ is constructible if there is a series $$\Bbb Q \subset K_1, \quad K_1 \subset K_2, \quad \ldots, \quad K_{n - 1} \subset K_n$$ of quadratic field extensions such that $z$ is in the last field, $K_n$. In particular, $$\omega = e^{2 \pi i / 3} = -\tfrac{1}{2} + \tfrac{1}{2} \cdot i \sqrt{3}$$ is in $\Bbb Q(i \sqrt{3})$, but $i \sqrt{3}$ ...


1

$\frac{-1+i\sqrt 3}{2}$ is constructible because we only need to adjoin square roots to obtain that number. Specifically, we only need $i$ (the square root of $-1$) and $\sqrt 3$ (the square root of, well, $3$). In fact, the product $i\sqrt 3$ is by itself a square root of the rational ("obviously constructible" would suffice) number $-3$. The rest is just ...


1

It can be shown that rings of algebraic integers algebraic numbers which are roots of a monic polynomials are so-called Dedekind domains – a generalisation of both PIDs and UFDs. They have several characterisations, most notably that ideals have a decomposition as a product of prime ideals, and this decomposition is unique, up to the order of the factors. ...


2

In a Dedekind domain every ideal is (in a unique way) the product of prime ideals. (The product of two ideal is the ideal generated by all products of elements.) The ring of integers of an algebraic number field is a Dedekind domain. The prime ideal decomposition of $p$ is the factorization of the principal ideal generated by $p$ into prime ideals of this ...


1

In these cases, considering the conjugate can help. The conjugate of $a+b\sqrt{2}$ is $a-b\sqrt{2}$. Now, if $(a+b\sqrt{2})(c+d\sqrt{2})=0$, also $$ (a+b\sqrt{2})(a-b\sqrt{2})(c+d\sqrt{2})(c-d\sqrt{2})=0 $$ and therefore $$ (a^2-2b^2)(c^2-2d^2)=0 $$ Since the integers form a domain, we conclude $a^2-2b^2=0$ or $c^2-2d^2=0$. The irrationality of $\sqrt{2}$ ...


1

Hint Suppose we have $$(a + b \sqrt{2})(c + d \sqrt{2}) = 0$$ from some $a, b, c, d \in \Bbb Z$. Expanding gives $$(ac + 2 bd) + (ad + bc) \sqrt{2} = 0,$$ and since $\sqrt{2}$ is irrational, the coefficients must vanish separately $$ac + 2bd = ad + bc = 0 .$$ Substituting $0$ for any of the four parameters gives quickly that $a = b = c = d = 0$, so we may ...


2

The main proof you mention is the easiest and the best since it generalizes very well. If you really want something more intrinsic, note $$(a+b\sqrt 2)(c+d\sqrt 2)=0\implies (a^2-2b^2)(c^2-2d^2)=0$$ But then if so, either $a^2=2b^2$ or $c^2=2d^2$, WLOG assume the former. Then $a$ is even, but then if the prime factorization of $a$ is $2^kp_1^{e_1}\ldots ...


0

No there is not: the archimedean place is defined inherently by the ordering, which is not an algebraic property. In fact, this can easily be seen because equivalent absolute values generate the same topology, and you know that the standard topology on $\Bbb R$ is generated by it's total ordering (this is in general how one can define topologies on totally ...


5

I have no idea how to answer question 1, but in the presence of the axiom of choice, everything is very simple. If $K$ is an algebraically closed field of characteristic $0$, then the additive group of $K$ is isomorphic to a direct sum of $|K|$ copies of $\mathbb{Q}$ (this is trivial when $K$ is uncountable; when $K$ is countable note that it must have ...


2

I don't think your approach is right. However, you can construct a homomorphism on $\def\qq{\mathbb{Q}}$$\qq(\sqrt{n})$ that fixes $\qq$ and maps $\sqrt{n}$ to $-\sqrt{n}$. You need to prove that it works, but then after that you are more or less done since homomorphism fixes polynomials over $\qq$. The other way is as Darij said in a comment, namely that ...


1

Leading off with a few easy (trivial) observations with a view of extracting a more precise range for $|S|$ where something good might happen. Undoubtedly you knew about these arguments. An averaging argument: shows that there exists $t\neq0$ such that $$ \left|S\cap (S+t)\right|\ge\left\lceil\frac{|S|^2-|S|}{2^n-1}\right\rceil. $$ Proof. Consider the sum ...


2

The key fact is that 21 is not prime. You will want to prove that if $f(x)$ whose degree is a multiple of 21 has no roots in $F$, then in fact there is a polynomial of degree 3 or 7 which has no roots in $F$. This violates the hypothesis, so all polynomials must actually split (so $F$ is algebraically closed). Try thinking about the Galois group of $f$ ...


0

Eisenstein's criterion and Gauss's Lemma are useful here. By Gauss' Lemma, $X^p-t$ is irreducible in $\mathbb F_p(t)[X]$ if and only if it is irreducible in $\mathbb F_p[t][X]$. But the ideal $(t) \trianglelefteq \mathbb F_p[t]$ is prime ; this is proved by the fact that if the product of two polynomials in $t$ (with coefficients in $\mathbb F_p$) is ...


1

Because it only has one root, and none of them are in $\Bbb F_p(t)$. Recall for any root $\zeta$ we have $\zeta^p=t$, but then since $\Bbb F_p(t)[x]$ is a UFD, it means that $(x-\zeta)^p=x^p-t$ has just the one root. So if it is reducible, it reduces all the way, and in fact there is an element of $\Bbb F_p(t)$ such that $\left(\displaystyle{q(t)\over ...


1

Given an ordered ring $(A,<)$ (which is an integral domain) there is a unique order $<'$ on $Frac(A)$ which coïncides with $<$ on $A$, because given $a,c \in A, \ b,d > 0 \in A$, $\frac{a}{b} <' \frac{c}{d} \longleftrightarrow ad <' bc$. Moreover, there is a universal property which is basically that of the fraction field along with ...


1

If $f(x)=g(x^2)$, then the splitting fields satisfy $E_f\supseteq E_g\supseteq\mathbb Q$. Since $E_g/\mathbb Q$ is also Galois, you have that $H=\textrm{Gal}(E_f/E_g)$ is normal in $G_f$ and $G_f/H\cong G_g$. In the situation you discuss this gives you that $G_f/\mathbb Z_2=\mathbb Z_2$, but that's as far as you can go without taking the specifics of the ...


4

You should be thinking geometrically. The field $K=\Bbb Q(\sqrt3,\sqrt5\,)$ is of degree four over $\Bbb Q$, as the comments have pointed out. There are only four proper subfields, namely $\Bbb Q$, $\Bbb Q(\sqrt3\,)$, $\Bbb Q(\sqrt5\,)$, and $\Bbb Q(\sqrt{15}\,)$. Each of the fields involving a square root is two-dimensional. So here you are, in a ...


1

No, the restriction of $|\cdot|$ to $\mathbb{R}$ need not be trivial. For instance, the valuation on $\mathbb{Q}_p$ extends to a valuation on its algebraic closure $\overline{\mathbb{Q}_p}$. We can choose a field isomorphism $\mathbb{C}\to\overline{\mathbb{Q}_p}$ to then get a valuation on $\mathbb{C}$ whose restriction to $\mathbb{Q}$ is the $p$-adic ...


0

Yes, the method of proof works. It is easy to see that the only 3rd root of unity in $\Bbb{F}_5$ is $1$. To complete the proof, all you need is the following fact: Claim: Let $F$ be a field. Then, the only elements of $F(t)$ that are algebraic over $F$ are those in $F$. I will sketch the proof. Each element $\alpha\in F(t)$ is of the form ...


2

The prime subfield will be the subfield obtained by taking the additive subgroup generated by $1$ and then throwing in multiplicative inverses. If $K$ has characteristic $0$, then the additive subgroup generated by $1$ will be isormorphic to $\mathbb{Z}$ and so adding inverses gives that the prime subfield is isomorphic to $\mathbb{Q}$. If $K$ has ...


1

Basically your question is : for any $P\in \mathbb{Q}[X]$ of degree $n$, is it true that $x^{n+1}>P(x)$ ? Now if $P=\sum a_k X^k$, then $P(x)\leqslant (\sum |a_k|)\cdot x^n$ since $x^k\leqslant x^n$ for all $k\leqslant n$. Then since $x>(\sum |a_k|)$, you get $x^{n+1}>(\sum |a_k|)\cdot x^n$, so $x^{n+1}> P(x)$.


0

Seems to me the splitting field of $x^4+2x^2+2$ over the rationals is ${\bf Q}(i,\sqrt2)$, from which it's easy to work out that the Galois group is Klein-4.


2

The splitting field of $f$ is the field $\mathbf Q(\zeta,\sqrt[6]2)$, where $\zeta$ is a primitive $n$th root of unity. The minimal polynomial of $\zeta$ is the cyclotomic polynomial $\;\Phi_6(x)=x^2-x+1$. so that $$[\mathbf Q(\zeta:\mathbf Q]=2, \enspace[\mathbf Q(\sqrt[6]2:\mathbf Q]=6.$$ Furthermore, these extensions are linearly independent and thus ...


0

The splitting field of $f$ is $\mathbb{Q}(e^{i \pi/3},\sqrt[6]{2})$, which is not the same as $\mathbb{Q}(e^{i \pi/3}\sqrt[6]{2})$. It has degree $12$ since $e^{i \pi/3}$ has degree $2$ and $\sqrt[6]{2}$ is real.


2

The answer will depend on whether or not $\alpha \in F$. Hint: a polynomial of degree $2$ or $3$ is irreducible over a given field if and only if it has no roots.


4

This is because the prime subfield is generated as a field by $1$. Since you have no choice but to send $1$ to itself, the prime subfield remains fixed as well.


1

Note that $\gamma$ also is a root of $(x-1)p(x)=x^p-1$. Then for any $k$, $\gamma^k$ is also a root of $x^p-1$, for $(\gamma^k)^p=(\gamma^p)^k=1$. So up to now the numbers $\gamma,\gamma^2,\ldots, \gamma^{p-1}$ are all either roots of $p(x)$ or $=1$. Assume $\gamma^k=1$ with $1\le k<p$. Then $\gamma$ is also a root of $x^k-1$, hence also of ...


2

The roots solve the equation $$x^p-1=0$$ Since the derivate of $x^p-1$ is $px^{p-1}$, you can see easily that $x^p-1$ has distinct roots. Therefore the roots of the cyclotomic polynomials are distinct as well.


1

As you noted, $\;[F(\alpha):F]=\deg f=15\;$, so if $\;g(x)\;$ is reducible in $\;F(\alpha)[x]\;$, say $\;g(x)=h(x)k(x)\;,\;\;h,k\in F(\alpha)[x]\;$ and say $\;\beta\;$ is a root of $\;h(x)\;$ , then ...


1

The question of whether $\Bbb Q^q$ has any extensions of degree $4$ is very interesting. It does, for: Let $\rho$ be quartic over $\Bbb Q$ with its splitting field $K$ having $S_4$ as Galois group (over $\Bbb Q$). So $[K\colon \Bbb Q]=24$. This $K$ is the normal closure of $\Bbb Q(\rho)$, that is, it’s the intersection of all normal extensions of $\Bbb Q$ ...


1

The following polynomial with galois group $D_{10}$ was found by PARI/GP : ? f %34 = x^5 - 2*x^4 + x^3 + x^2 - x + 1 ? polgalois(f) %35 = [10, 1, 1, "D(5) = 5:2"] ? So, $$x^5-2x^4+x^3+x^2-x+1$$ is an example. I approved the galois-group with GAP.


2

I agree with Captain Lama, it would be hardly possible to determine the relative Brauer group $Br(L/\mathbf Q)$ just by norm computation. I sketch here a theoretical approach. By definition, $Br(L/\mathbf Q)$ is the kernel of the natural map $Br(\mathbf Q) \to Br(L)$ (which is the restriction map, cohomologically speaking). For any number field $K$, it is ...


0

All you have to do is show by induction that $\sqrt{2+a_i} \not \in \mathbb{Q}(a_i)$. The induction step is then to prove that $\sqrt{2+\sqrt{2+a_i}} \not \in \mathbb{Q}(\sqrt{2+a_i})$ That is that the equation $$2+\sqrt{2+a_i}=(a+b\sqrt{2+a_i})^2$$ is impossible. Squaring gives $$2+\sqrt{2+a_i}=a^2+b^2(2+a_i) + 2ab\sqrt{2+a_i}$$ and the induction ...


0

A plan of attack: Show that $a_n=2\cos\dfrac{\pi}{2^{n+1}}$. Let $\zeta$ be a complex primitive root of unity of order $2^\ell$, $\ell\ge3$. Recall (or do the exercise) that $\operatorname{Gal}(\Bbb{Q}(\zeta)/\Bbb{Q})$ is a direct product of the subgroups generated by $\sigma:\zeta\mapsto \zeta^5$ and $\tau:\zeta\mapsto \zeta^{-1}=\overline{\zeta}$. Here ...


0

You can do it as you suggest. Consider say $(2,0)$ and note that after whatever multiplication the second coordinate will still be $0$. Thus, you can never get $(1,1)$ the identity with respect to multiplication. Note that the subset $\{(q,0) \colon q \in \mathbb{Q}\}$ would be a field, yet with a different identity element namely $(1,0)$. (This however ...


2

You know that in a field $\;ab=0\iff a=0\;\;or\;\;b=0\;$ . Now try with $\;(1,0)\;,\;\;(0,1)\;$ in your case


5

$(1, 0) \ne (0, 0) \ne (0, 1)$, but $(1,0) \cdot (0,1) = \dots$


2

Since $k\to \bar k$ is integral so is $k(p)\to k(p)\otimes_k\bar k$ (Atiyah-Macdonald: Chap.5, Ex.3). But then $\dim k(p)=\dim k(p)\otimes_k \bar k$ (Matsumura: Exer. 9.2), so that indeed $\dim k(p)\otimes_k \bar k=0$. Edit: the fibers are actually finite Since the ring $k(p)\otimes_k \bar k$ is noetherian of dimension zero, it is artinian and has thus ...


2

It suffices to show that if $K$ and $L$ are two fields extending a field $k$ and $L$ is algebraic over $k$, then $\dim K\otimes_k L=0$ (let $K=A_p/m_pA_p$ and $L=\overline{k}$). Let $R=K\otimes_k L$; to show that $R$ has dimension $0$, it suffices to show that for any homomorphism $R\to F$ where $F$ is a field, the image of $R$ in $F$ is a field (if ...


2

No, it is still not sufficient to assume that $\xi\notin F$. For example, if $F=\mathbb{Q}(i)$ and $$\xi=e^{2\pi i / 8}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$$ then $[F(\xi):F]=2$ but $\xi\notin F$. Observe that $\varphi(8)=4$, and $$\underbrace{[F(\xi):\mathbb{Q}]}_{4}=\underbrace{[F(\xi):F]}_{2}\underbrace{[F:\mathbb{Q}]}_{2}$$ The Galois group of the ...


2

For any $n$ the cyclotomic polynomial $\Phi_n(x)$ is irreducible over $\Bbb Q$ (this is a non-trivial fact, and I won't prove it here). This is the minimal polynomial of $e^{2\pi i/n}$. Fortunately for you, there is a way to calculate this if you know all the "smaller" cyclotomic polynomials: $\Phi_n(x) = \dfrac{x^n - 1}{\prod\limits_{d|n, d< ...


0

Splitting fields can be taken with respect to any family of polynomials over a fixed field $K$. Fields $E$ that are splitting fields of families of polynomials over a field $K$ are precisely those fields such that $E/K$ is normal. Galois extensions are defined to be normal and separable. If the family is finite, say $f_1,\ldots,f_s$, then $E$ is the ...



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