# Tag Info

1

If I'm not mistaken in understanding of your question. Let's $G$ be an abelian group of order $p^n$. Does a subgroup of order $p^k$ exist? Yes, it exists: If $G$ is cyclic then subgroup of order $p^k$ is generated by element $g = p^{n-k}$. It is an implication of the fact that $g$ has order $p^k$. Every abelian group of order $p^n$ can be represented as a ...

1

First, let me point out that it doesn't make sense to say that there exists $\sigma \in G(E/F)$ such that $\sigma \alpha = \alpha'$, because the root $\alpha$ lives in $\overline{E}$, not necessarily in $E$. Here is how I suggest you prove this: write $f(x) = \prod_{i=1}^n e_i(x)$ for the factorization of $f$ in $E[x]$. Let $G$ act on the set $\{e_i\}$. ...

1

Do this by definition. $\mathbb{Q}(\sqrt{2})$ is by definition the intersection of all subfields that contain $\mathbb{Q}$ and $\sqrt{2}$. So, to prove that $\mathbb{Q}(\sqrt{2}) \subseteq L$, it is enough to establish two things: (1) $\mathbb{Q} \subseteq L$ and (2) $\sqrt{2} \in L$. Fact (2) is given to you for free. It remains to prove fact (1): ...

2

Hints: Every field of characteristic zero has the rationals as a prime field, meaning: any field of characteristic contains an isomorphic copy of the rationals. This already solves (A) . In a positive characteristic $\;p>0\;,\;\;p\;$ a prime, the prime field is $\;\Bbb F_p:=\Bbb Z/p\Bbb Z\;$

1

A $p$-adic integer can be expressed uniquely in the form $$\sum_{i=0}^{\infty}a_{i}p^{i}$$ where $a_i$ is an integer with $0\le a_i<p$. Moreover, any $p$-adic integer can be written in a unique way as $$up^k$$ with $u$ invertible and $k\ge0$. In particular, $$\frac{r}{s}=\frac{r}{up^k}=\frac{u^{-1}r}{p^k}$$ Write ...

1

This is essentially correct. However, you should give the field $F(\alpha\beta)$ a name; you refer to it in the proof as just $F$, which is not correct. I would rewrite the portion of the proof starting with this sentence as follows: Now, $\alpha\beta = l\alpha^{Bn}\beta^{Bm} = l(\alpha\beta)^{Am}\alpha^{Bn}\alpha^{-Am}$. Let $L = F(\alpha\beta)$; then ...

6

Sadly, the statement doesn't hold in general. For instance for $(n_1, n_2, n_3) = (2, 3, 8)$ we can choose $$x = \frac{1}{2}\left(\sqrt{3} - 1 +\sqrt{2}\sqrt[4]{3}\right)$$ and $$y = \frac{1}{2}\left(\sqrt{3} - 1 -\sqrt{2}\sqrt[4]{3}\right).$$ Now it's easy to see that $\gcd(2,3,8) = 1$, $x^2+y^2 = 2$, $x^3+y^3 = 2$ and $x^8+y^8 = 8$, and both numbers ...

-1

$\mathbb R$ is a $\mathbb Q$-vector space, so every $\mathbb Q$-linear function is in $Aut(\mathbb R)$ and $End_{\mathbb Q}(\mathbb R)$ isn't abelian, so $Aut(\mathbb R)$ isn't abelian. If you mean $Aut(\mathbb R)$ like the group of ring automorphism of $\mathbb R$, the only one is the identity.

1

You are off to a good start. Let me clarify a bit: If $f(X)$ is the minimal polynomial for $a$ over $F$, then $f(x)$ factors into irreducibles in $F(a)[X]$: $f(X) = (X-a)f_1(X)\cdots f_k(X)$, where $k\leq n-1$. If K is an intermediate field, $F \subsetneq K \subsetneq F(a)$, then the minimal polynomial of $a$ over $K$ is a product of some collection of ...

1

The cycle type is determined by the way you identify $\phi_p$ as an element of $S_n$. There is no canonical way of doing this, which makes your question a bit underdetermined, and may also be the root cause of your difficulties. It looks like you are doing it (or, expected to do it) as follows. Let $X=\{\alpha_1,\alpha_2,\ldots,\alpha_n\}$ be the set of ...

3

Like @Martín-Blas Pérez Pinilla: said, $$\underbrace{1+ \cdots + 1}_{p^n}=0$$ and so $$\underbrace{p \times \cdots \times p}_{n}=0$$ in a field, so $p=0$.

2

If $$\mathbb{F}_q \subseteq \mathbb{F}$$ is a subfield, then $\mathbb{F}$ is a finite-dimensional $\mathbb{F}_q$ vectorspace, so it must have $q^m$ elements. But $p^n=q^m$ implies $p=q$

4

The additive group of $\Bbb F_{p^n}$ has order $p^n$, that is...

3

Importantly, the natural numbers with $\leq 2^n$ binary digits already form a field $F_n$ of order $2^{2^n}$ with the nim operations. Let's see how this works inductively. $F_n \subset F_{n+1}$ is a degree $2$ extension. We have the (nim) identity $(2^{2^n})^2 = 2^{2^n} + 2^{2^n-1}$. In other words, $2^{2^n}$ is a root of the irreducible polynomial $T^2 ... 1 If it is ordered, that means there is a nonempty subset$P$such that (1) for each$x$exactly one of$x \in P, x=0, -x\in P$(2) If$x,y \in P$then each of$x+y,x*y$is in$P$. There are not many choices for$P$here.$P$cannot have$0$in it, otherwise (1) above is false. And$P$is nonempty. So far the only possible$P$is$P=\{1\}.$But this$P$does ... 3 An ordered field$F$must have characteristic$0$, because $$\underbrace{1+1+\dots+1}_n > 0$$ for all$n>0$. A finite field can't have characteristic$0$. 0 With the following identity, I would claim that the modular inverse of$(1 + x^2) \mod (1 + 2 x + x^3)$is$1/2 (2 + x + x^2 + x^3)$, provided that$x$is even. $$1/2 (2 + x + x^2 + x^3) (1 + x^2) - 1/2 x (1 + x) (1 + 2 x + x^3) = 1$$ For example, if$x =20$, then$1 + x^2 = 401$,$1 + 2 x + x^3 = 8041$,$1/2 (2 + x + x^2 + x^3) = 4211$, and$(401)^(-1) ...

2

From $w^5=1$, after dividing with $w-1$, we get: $$w^4+w^3+w^2+w+1=0$$ As we know, $2 w^5=2$ and $w^8=w^3$, therefore previous equation is equal to: $$w^2+2 w^5+w^8+w+w^4-1=0,$$ which is equal to $$(w+w^4)^2+(w+w^4)-1=0.$$ Therefore, the minimal polynomial for $w+w^4$ is $x^2+x-1$. For other question, it is probably like this: {I, σ∘τ2} for fixed field has ...

2

First of all, concerning your question about $[\mathbb{Q}(\omega+\omega^4):\mathbb{Q}]$. There is a nice fact that is often useful when dealing with roots of unity. Namely, if $\omega_n$ is an $n$th root of unity ($n$ at least 3) and if $\theta_n=\omega_n+\omega_n^{-1}$, we have $[\mathbb{Q}(\omega_n):\mathbb{Q}] = 2\cdot[\mathbb{Q}(\theta_n):\mathbb{Q}]$. ...

1

I got it. $F = \mathbb{Q}$, $E = \mathbb{Q}(\sqrt2)$ and $L = \mathbb{Q}(2^{1/4})$ constitute a counterexample. Thanks a lot Praphulla and Jyrki for the hints.

3

Without loss of generality, let us set $d = 2^r - 1$ and show that $3^r$ multiplications are required. We have $$f(x) = a_0 + a_1x + \ldots + a_{2^r - 1}x^{2^r - 1},$$ $$g(x) = b_0 + b_1x + \ldots + b_{2^r - 1}x^{2^r - 1}.$$ Now the karatsuba algorithm will split each of these in $2$. For example, $$f(x) = f_1(x) + f_2(x)x^{2^{r - 1}},$$ where ...

1

Hints : Any (well behaving) degree $2$ extension is Normal. Not all degree $4$ extensions are Normal.

1

A sketch answer only: The critical model-theoretic fact is that the theory of real closed fields has quantifier elimination. That means that if you have a real closed field $K$ which you are trying to extend (for an inductive proof) with a new element $b \notin K$ (or indeed if we happen to be looking at some $b \in K$), everything that $K$ can tell you ...

2

Being of degree $3$, it's reducible over $\mathbb{Q}$ if and only if it has a root in $\mathbb{Q}$ ( that is, a factor of degree $1$). $\bf{Fact}$: if a rational number is a root of a monic polynomial with integer coefficients that numbers is an integer. $\bf{Fact}$: An integral root of a monic polynomial with integer coefficients is a divisor of the free ...

0

Here is what I think for the first question: Suppose $f(x) = x^{3} + x + 3$ is not irreducible over $\mathbb{Q}$. Then we can write $f(x)$ as the product $(x + a)(x^{2} + bx + c)$ for some rational numbers $a, b, c$. But then, multiplying this out gives the expression $x^{3} + bx^{2} + cx + ax^{2} + abx + ac = x^{3} + (b + a)x^{2} + (ab + c)x + ac$ And ...

6

There is a countably infinite field of characteristic $2$ whose elements can be identified with the non-negative integers in a natural way. Addition can be defined by $2^n+2^m=2^n+2^m$ if $m\neq n$ and $= 0$ if $m=n$ and extended by the commutative and associative laws. This is "nim addition" Multiplication can be defined by $2^{2^m}\cdot ... 6 Yes, but not a very natural one. Since$\mathbb{Z}$and$\mathbb{Q}$are both countable, there exists a bijection$\Phi: \mathbb{Z} \to \mathbb{Q}$. With the operations $$m \oplus n := \Phi^{-1}(\Phi(m) + \Phi(n)), \qquad m \otimes n := \Phi^{-1}(\Phi(m)\Phi(n)),$$$(\mathbb{Z}, \oplus, \otimes)$is a field, but NB that$\oplus$and$\otimes$won't have ... 1 Yes, but the answer isn't really satisfying. The set of integers is countably infinite. It's therefore in bijection with$\mathbb{Q}$, the field of rational numbers. By transporting the field operations (addition and multiplication) through the bijection, you get a field structure on the set of integers. 1 There is nothing wrong with your proof. In fact, this is exactly the same argument that Serge Lang gives in his Algebra at page 232 (following Artin's argument). 11 A universal property of some object$A$tells you something about the functor$\hom(A,-)$(or$\hom(-,A)$, but this is just dual). For example,$\hom(R[x],S) \cong |S| \times \hom(R,S)$is the universal property of the polynomial ring (where$|S|$denotes the underlying set of$S$). Conversely, we may consider the functor which takes a commutative ring$S$... 3 Typically, when we adjoin a root to a field, it looks something like$\mathbb C=\mathbb R (i)$, which is the closure of$\mathbb R$when we add a root of$x^2+1=0$. So here, call your root of$x^2+x+1$something, so say$j$satisfies$j^2+j+1=0$mod 2. Now, look at the field$\mathbb Z/2\mathbb Z (j)$, the elements of which are going to be ... 0$a$is a finite sum of primitive tensors$a_i\otimes b_i$, so lies in$F[a_i]\otimes K$. 0 Once you view an object as a field you stop seeing it as a vector space on something smaller or over itself: Any field is a vector space over itself. If$\mathbb{K}$then$\mathbb{K}[x]$is the vector space of all the polynomials with coefficient in$\mathbb{K}$. This set is an algebra but not a field. Let$g$be an irreducible polynomial in ... 0$\mathbb{R}$and$\mathbb{C}$are fields as well as vector spaces over$\mathbb{R}$. More generally any field is a vector space over its subfields. This is simple to prove. 7 It is true that vector spaces and fields both have operations we often call multiplication, but they're fundamentally different, and like you say, we sometimes call the operation on vector spaces scalar multiplication for emphasis. The operations on a field$\mathbb{F}$are$+$:$\mathbb{F} \times \mathbb{F} \to \mathbb{F}\times$:$\mathbb{F} \times ...

2

First, we should be explicit that we are talking about an isomorphism of fields over $k$, rather than just an isomorphism of fields. Otherwise, there are strange examples like $\mathbb{C}(t)[X]/(X^2-t) \cong \mathbb{C}(\sqrt{t}) \cong \mathbb{C}(t) \cong \mathbb{C}(t)[X]/(X)$. With this assumption, $k[X]/(f(X)) \cong k[X]/(g(X))$ implies that $f$ and $g$ ...

0

For both $(1)$ and $(3)$ the only counterexamples are fields of characteristic $2$ (e.g. $\mathbb F_2$, $\mathbb F_{2^m}$, $\mathbb F_2(x)$ with $x$ trascendental, etc.. ). For $(2)$ you can take every field that contains the $4$-th primitive roots of unity. For example $\mathbb C$, $\mathbb Q(i)$, $\mathbb F_{p^{2s}}$ where $p$ is a prime, etc

0

The right (i.e. categorical) way to say this (without the ambiguities of words like "smallest", "containing", etc.) ought to be that the inclusion $\iota: D\to Q(D)$ has the following universal property: If $K$ is a field, and $f: D\to K$ is any morphism of rings, then there is a unique morphism of fields $g : Q(D) \to K$ such that $f = g \circ \iota$. ...

1

Let $F'$ be a smallest field containing an embedding of $D$ ($f:D\to F'$), and $F$ a field of fraction of $D$. We can extend $f$ to morphism of field $\tilde f:F\to F'$ by $\tilde f(a/b)=f(a)/f(b)$. Now we have that $\tilde f(F)\subseteq F'$ and $\tilde f(F)$ containing an embedding of $D$ , by smallest property we have $\tilde f(F)=F'$. So the tow ...

0

Clearly, $\alpha$ is contained in some quadratic extension of $\mathbb F_p$. But there is only one such extension of $\mathbb F_p$ (in general up to isomorphim, but within a fixed closure it is really unique). So, yes, $\alpha\in\mathbb F_{p^2}\subseteq \mathbb F_q$

0

If $p=2$ Yes: $\alpha^{2^2-1}-1=\alpha^3-1=(\alpha-1)(\alpha^2+\alpha+1)=0$. Hence $\alpha\in \Bbb F_{2^2}$.

4

There are many reasonable notions of "limit of fields" depending on your application. Often, "limits" of sets/spaces/algebras are best thought of in terms of unions. For example, fix a prime $p$, and consider the fields $\{\mathbb F_{p^n}\}_{n\in \mathbb N}$ of order $p^n$. If my memory is correct, there is a map $\mathbb F_{p^n} \to \mathbb F_{p^m}$ if ...

1

Hint: if we take the basis $\{1,a,\dots,a^{n-1}\}$. Suppose furthermore that $a^n = \sum_{k=0}^{n-1} c_k a^k$. Then the matrix of this transformation with respect to the basis is $$T_{\ell} = \pmatrix{&&&&c_0\\1&&&&c_1\\&1&&&\vdots \\&&\ddots\\&&&1&c_{n-1}}$$ Now, what is the ...

0

the basic numerical law for finite extensions is: $$[L(\alpha):K] = [L(\alpha):K(\alpha)][K(\alpha):K] = [L(\alpha):L][L:K]$$ thus because of the data on relative primality: $$[K(\alpha):K]|[L(\alpha):L]$$ and in particular, therefore: $$[K(\alpha):K] \le [L(\alpha):L] \le [L(\alpha):K]$$ which makes the required point, since, because $K \subset L$ we ...

0

Assume $[L:K]$ and $[K(\alpha):K]$ are finite and relatively prime. The minimal polynomial $p$ of $\alpha$ over $L$ divides the minimal polynomial $q$ of $\alpha$ over $K$. The degree of $q$ is $[K(\alpha):K]$, so the degree of $p$ divides the degree of $q$. Furthermore, $\deg p=[L(\alpha):L]$, and since $K(\alpha)\subset L(\alpha),\deg q$ divides ...

3

Yes, if $E$ is an algebraic and normal extension of the fixed subfield of the morphism $E_1\to E_2$, but not in general. Here is an example when you can't extend. Consider $E = \mathbb{Q}( \sqrt{1+ \sqrt{2}})\$, $E_1 = E_2 = \mathbb{Q}(\sqrt{2})$ and $E_1 \to E_2$, $\sqrt{2} \mapsto - \sqrt{2}$. Let's show the existence of an extension. There are two ...

1

Let it be that element $\alpha\in A$ has minimal polynomial $\sum_{i=0}^{n}\beta_{i}X^{i}$ with $\beta_{i}\in B$. Then $C\subset B_{0}:=C\left(\beta_{0},\dots,\beta_{n}\right)$ and $B_{0}\subset B_{0}\left(\alpha\right)$ are finite extensions. Consequently $C\subset B_{0}\left(\alpha\right)$ is a finite (hence algebraic) extension. This tells us that ...

0

If you have an element $\alpha \in A$ which is algebraic over $B$, what can you say about the form of $\alpha$? Similarly, if you have an element $\beta \in B$ which is algebraic over $C$, what can you say about the form of $\beta$? Can you somehow write $\alpha \in A$ as an algebraic element over $C$?

1

Hint: Consider the case $F=\Bbb{Q}$, $a_1=\root3\of2$, $a_2=\omega\root3\of2$, where $\omega=(-1+i\sqrt3)/2$ is a primitive third root of unity.

0

$\Rightarrow$ Assume by contradiction that $n \neq 2^k$. Then it has an odd divisor. Use the formula for $a^k+b^k$... $\Leftarrow$ Use the fact that $(X^n+1)(X^n-1)=X^{2n}-1$. Write this as product of cyclotomic polynomials and use induction by $k$.

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