Hot answers tagged

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 ...


6

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 ...


5

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.


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.


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 ...


3

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 ...


2

$\frac{-1+i\sqrt 3}{2}$ 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 ...


2

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 $i \sqrt{3}$ ...


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 $6$.


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

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.


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 ...


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 $f$ ...


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 ...


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.


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 ...


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 ...


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

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

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 ...


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. ...


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 $x^{n+1}> P(x)$.


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

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 ...


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 ...


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

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 ...


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

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 $f(x)=g(x^2)$, 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 ...



Only top voted, non community-wiki answers of a minimum length are eligible