# Tag Info

2

$A$ is quasi-compact, since it is a closed subset of the quasi-compact set $K^n$. Any quasi-compact discrete set is finite, this is a very easy exercise in basic topology.

1

Just want to mention that there is some new work done in the $k$ not being algebraically closed case. The paper is published in Journal of Algebra this year.

3

$F$ is a finite-dimensional vector space over $K$. Therefore there exists some $n \geq 0$ such that $|K|^n = |F| = 8$. Thus $|K| \in \{2,8\}$. The result follows immediately from this.

7

Furthermore, $|K^\times|$ divides $|F^\times|$, so $|K|-1$ divides $8-1=7$. Since $7$ is prime, there are only two subfields, $F$ and $\mathbb F_2$.

1

Let's write $G = G(\overline{K}\mid K)$. We want to show that $G^{ab}$ is the Galois group of $K^{ab}$ over $K$. Of course, since $G^{ab}$ is a quotient of $G$, the usual Galois theory for fields tells us that $G^{ab}$ is the Galois group of a well-defined field $L$ with $K \subseteq L \subseteq \overline K$, namely the fixed field in $\overline K$ of the ...

0

In your first paragraph it is a bit unclear what direct sums you are talking about. You are right that a non-trivial finite product of fields is not a field. But the direct sum decomposition that you give as an example is one of $K$-vector spaces, not rings (resp., fields). So, more explicitly, the isomorphism preserves addition and multiplication with ...

0

The direct sum (or rather the product) of fields is not even an integral domain. In your context, $K(\alpha)$ is the direct sum of the $K\alpha^i$s as a $K$-vector space.

4

The Galois group of a field extension $L/K$ is profinite, which $\mathbb{Z}$ is not.

2

You know by linear algebra that, if $\mathbb K$ is a field, a matrix $A\in \mathcal{M}_{n\times n}(\Bbb K)$ is invertible if and only if its columns are linearly independent. Now: if $\#\mathbb K=q$, how many $n$-tuples $(v_1,\cdots,v_n)$ of linearly independent vectors are there? $v_1$ can be anything except $0$. So it can be chosen in $q^n-1$ possible ...

2

You can check https://en.wikipedia.org/wiki/General_linear_group where there is a formula and a little explanation for the number of matrices in GL(n,q), which is $$(q^n-1)(q^n-q)(q^n-q^2)\cdots(q^n-q^{n-1})$$

1

About building a quaternion extension of Q (or more generally of any number field) : the problem can be solved completely and explicitly starting from a biquadratic field, using techniques of embedding problem. Actually, given a field K containing a primitive p-th root of unity (where p is any prime), the more general problem of embedding a Galois extension ...

2

You should see the paper "On On_p", by Joseph DiMuro. In it, he gives a characteristic $p$ analogue of the nimbers. The addition is given by adding base $p$ without carries (including for ordinals). The definition of multiplication, unfortunately, is not so simple. But I think it's a good generalization. The definition for the natural numbers is almost ...

1

Suppose $\text{ker}(\phi)$ is a maximal ideal. Let us demonstrate that $A/\text{ker}(\phi)$ is a field directly. It suffices to take an arbitrary coset $x + \text{ker}(\phi) \neq \text{ker}(\phi)$ and show there exists a coset $b + \text{ker}(\phi)$ such that: $(x + \text{ker}(\phi))(b + \text{ker}(\phi)) = 1_A + \text{ker}(\phi) = 1_{A/\text{ker}(\phi)}$ ...

3

By the first fundamental isomorphism theorem $B$ is isomorphic to $A/\ker(\phi)$. Now let $a\in A$ where $a\not\in\ker(\phi)$. Then $(a)+\ker(\phi)$ is an ideal properly containing $\ker(\phi)$. Thus $(a)+\ker(\phi)=A$. So $\exists$ $b\in A$ and $m\in\ker(\phi)$ such that $1=m+ab$. Then $\overline b$ is the multiplicative inverse of $\overline a$. Thus ...

2

Hint: By Isomorphism theorems we have $A/\ker \phi \cong \text{Im } \phi = B$. Then $B$ is a field if and only if the quotient is a field if and only if $\ker \phi$ is a maximal ideal of $A$.

1

Suppose $\phi: A \rightarrow B$ is a surjective ring homomorphism. If $\ker(\phi)$ is a maximal ideal, then $A/\ker(\phi)$ is a field, and by the first isomorphism theorem, $A/\ker(\phi)$ is isomorphic to $\phi(A)$, its image. But $\phi(A) = B$ since $\phi$ is surjective. So since $A / \ker(\phi)$ is a field, $B$ is a field, too.

0

Your map IS indeed induced by the inclusion of K in K' in the following sense. Denote G = Gal(L/K), H = Gal(L/K'), and let $N_G$, $N_H$ etc. be the corresponding norm maps in the group algebras Z[G], Z[H], etc. So, for instance, $N_G$ = $i_{L/K}$ . $N_{L/K}$ (where $i_{L/K}$ is the obvious inclusion) is an endomorphism of L* . The classical ...

2

Unless I missed something, $R[x]/\mathscr{P}$ is generated as an algebra over $R/\mathfrak{m}$ by the class $\bar x$ of $x$ (because any element of $R[x]$ is a polynomial in $x$ with coefficients in $R$, and its class mod $\mathscr{P}$ is therefore a polynomial in $\bar x$ with coefficients in $R/\mathfrak{m}$). But a field extension which is finitely ...

2

Any subfield of $\mathbb R$ is automatically of characteristic $0$ and thus contains $\mathbb Q$, in particular it is an infinite field. Nevertheless, it is possible that a finite field is a subfield of an infinitie field, for example we have $\mathbb Z/3\mathbb Z \subset \mathbb Z/3\mathbb Z(X)$, where the latter is the function field over $\mathbb ... 0 It is not, because$\mathbb Z_3$is not really a subset of$\mathbb R$, since$\mathbb Z_3=\mathbb Z/(3\mathbb Z)$Of course, you can say that$\mathbb Z_3$can be defined as the set$\{0,1,2\}$, but in that case, the operation$+$defined on the set is not the same as the operation we typically denote as$+$, since$2+2=1$in$\mathbb Z_3$, but$2+2=4$in ... 2 Here is a proof that point 3 is false that doesn't use any difficult theorems. As$\gamma$ranges through all transcendental numbers, the number$2^\gamma$takes on uncountably many values. Since only countably many of these values can be algebraic, there must exist a transcendental$\gamma$for which$2^{\gamma}$is also transcendental. Now let$\alpha = ...

2

Let $\alpha$ be any transcendental. Then $\beta := \frac{1}{\alpha}$ is transcendental and $\alpha\cdot \beta =1$. Thus 1. is false. Consider $f \colon \mathbb Q(\alpha) \to \mathbb Q(\beta), \frac{x_0 + x_1 \alpha + \ldots + x_n \alpha^n}{y_0 + y_1 \alpha + \ldots + y_m \alpha^m} \mapsto \frac{x_0 + x_1 \beta + \ldots + x_n \beta^n}{y_0 + y_1 \beta + ... 0 Though it is possible to find fields of order$p^2$for a fixed$p$by what you have done but to prove it for an arbitrary field you need to use a concept popularly known as Splitting field. Once you get that(I don't know whether you are accustomed to it or not) you need to consider the splitting field of the polynomial$x^{p^2}-x $over$\Bbb Z_p$and ... 2 Let$p=2$. Note that$x^2+x+1$is irreducible over the two-element field, for it has no roots in that field. Now let$p$be odd. Let$F_p$be the$p$-element field. We have$(-a)^2=a^2$, and if$a\ne 0$then$-a\ne a$. Thus there are at most$\frac{p-1}{2}$non-zero elements of$F_p$that are squares of elements of$F_p$. (Actually, there are exactly ... 4 How many polynomials are there of degree 2? How many reducible polynomials are there of degree 2? Let the leading coefficient in both cases be 1. 2 First, you can argue that any abelian extension of$F$is a composite of cyclic extensions. Next, it suffices to show that any such cyclic extension$E/F$is of the form$F(\sqrt[m]{a})$for some$m$and some$a \in F$. Let$m = [E : F]$, so the Galois group of$E/F$is isomorphic to$\mathbb{Z}/m\mathbb{Z}$. Let$\sigma$generate this Galois group, and ... 0 I can just make you an example. For the zeromello's lemma in every set is possibile to impose a good order (such that every subsets has minimal element) which is very different from the standard order on R. So i belive that the answer to point a is yes order can have really different propreties. 1 A simple example,$F = \mathbb Q(X)$. For any transcendental real number$\alpha$, we can order$F$be letting$X=\alpha$. We can also order$F$by letting$X$exceed all elements of$\mathbb Q$. Or with$0 < X < r$for all positive rationals$r$. There are others, too. 3 Here’s another method. You know that your transformation$\phi$is of order two, and that the “conjugate” of$X$is$1-X$. The minimal polynomial for$X$over the fixed field is accordingly$f(T)=T^2-T+(X(1-X))$. Here I’ve used the sum of the conjugates for the linear coefficient (with the necessary change of sign) and the product of the conjugates for the ... 1 If$\operatorname{char}(K)\neq 2$, then$X$satisfies a polynomial of degree$2$over$K(Y)$(namely,$p(t)=(2t-1)^2-Y$), so$[K(X):K(Y)]\leq 2$. Since$K(Y)\subseteq L\subset K(X)$, it follows that$K(Y)=L$. If$\operatorname{char}(K)=2$, on the other hand, this doesn't work (the polynomial$p(t)$used above is identically$0$). And in fact it is clear ... 0 You can't really find minimal polynomials in a constructive way, unless you know more about the fields themselves and the elements$a,b$. Most of abstract field theory is very nonconstructive stuff, unfortunately. However, you can argue as follows: Let$E \subseteq E'$be fields, and assume that$b$is algebraic over$E$. Let$f(X) \in E[X]$be the ... 2 Since$\alpha \in K$is a root of the irreducible polynomial$f \in L[X]$, then$f$is the minimal polynomial of$\alpha$over$L$. The degree$d$of$\alpha$over$L$is$≤2$, because$[K : L]=2$. If$d=1$, what can you conclude? If$d=2$, write$f(X)=X^2+aX+b=(X-\alpha)(X-\beta)$. What are the relations between$\alpha$and$\beta$? 1 There is a bit of abuse of notation involved here, since$f,g$are being reused to identify their compositions$f\circ h$and$g \circ h$as if they were the same rational expressions in$k(t)$but "restricted" to$k(h(t))$. But the notation needs to be taken with a grain of salt. It really means the compositions$f(h(t)) = g(h(t)$are equal as rational ... 1 You don't need to find the minimal polynomial. That's because whatever the polynomial is for$b$over$K$, call it$p$, it is still a polynomial for$b$over$K(a)$. Therefore, the minimal polynomial for$b$over$K(a)$divides$p$and therefore has degree no greater than$\deg(p)=[K(b):K]$. 1 clearly the extension is of degree 2 and as it is finite extension of finite field so is galois (normal,seprable,finite). then galois group consist of two elements G=(1,a) where a is Automorphism of Field with 9 elements keeping base field fixed. orbit(x)={x,a(x)} for any x in F= field with 9 elements now a(x)=x for each element of the base field by ... 2 Well, if you've actually proved the biconditional statements you mentioned, then you're done. Alternatively, show that $$\Bbb Q\subseteq\Bbb Q(i)\cap\Bbb Q\bigl(\sqrt2\bigr),$$ which I leave to you. Then, suppose$z\in\Bbb Q(i)\cap\Bbb Q\bigl(\sqrt2\bigr).$Since$z\in\Bbb Q\bigl(\sqrt2\bigr),$then$z\in\Bbb R.$From there, we can use the fact that ... 1 yes it is true$Q\sqrt2$and$Q(i)$are extension of degree$2$of$Q$so the degree of$Q(i)\cap Q(\sqrt2)$is either 1 or$2$, if it is 2 it implies that$Q(i)=Q(\sqrt2)$thus$\sqrt2=a+bi, a,b\in Q$, by writing$(\sqrt2-a)^2=-b^2$, you obtain$a=0$unless$\sqrt2\in Q$a fact which is not true. If$a=0$,$\sqrt2=bi$thus the$2=-b^2$this not true also. 3 A number$a$is called constructible here if there exists a classic geometric construction (that is: using straightedge and compasses [and for the sake of completeness: picking a generic point]) that can construct a line segment of length$a$times as long as a single given line segment. Since addition, subtraction, multiplication (regarding the given length ... 3 If$x^2 + xy + y^2=0$, then$x^3 - y^3 = (x-y)(x^2+xy+y^2)=0$. But$3$does not divide$31$, so$a\mapsto a^3$is injective, and therefore$x=y$. 0 See the book "Lectures on the Mordell-Weil Theorem" of Jean-Pierre Serre. 3 To see this directly: If there is an element of order$2k$, then there is an element of order$2$. But$x^2-1 = (x-1)^2$in a field of characteristic$2$, so there are no nontrivial square roots of$1$. 3 If$\;k(X)=k\left[ \frac{f_i(X)}{g_i(X)}\;,\;\;1\le i\le n\right]\;$, then there is a finite number of prime elements that can appear as factors of the the denominator of any element generated as a polynomial in the above$\;n\;$elements and coefficients in$\;k\;$. It is enough then to show there is an infinite number of prime elements in$\;k[X]\;$, ... 3 Hint 1 : For your first question, notice that an element in$K$is of the form$P(\theta)$, with$P(X) \in \Bbb Q[X]$and that for any$g \in G$,$g(P(\theta)) = P(g(\theta))$. Answer 1 : 2 In a finite group, the order of an element divides the order of the group. A finite field of characteristic$2$has$2^n$elements for some positive integer$n$, so its multiplicative group has odd order. 1 If$y = 0$, clearly also$x = 0$. So suppose$y \ne 0$, multiply by$y^{-2}$, and set$t = x y^{-1}$to obtain$t^{2} + t + 1 = 0$. Now you should know that the solutions of the latter equations are the two elements different from$0, 1$in the field of order$2^{2} = 4$. Since$32 = 2^{5}$, the field of order$4$is not a subfield of$F$. So no solutions ... 1 Hint $$x^3-y^3=(x-y)(x^2+xy+y^{2})$$ Hint 2 If$x \neq 0$and$y \neq 0$you get $$x^{3}=y^3 \\ x^{31}=y^{31}$$ From here you get immediately$x=y$, which you can plug in the original equation. The case$x=0$OR$y=0$is easy... 4 In a commutative ring$R$we have$\frac{R}{I}$is a field if and only if$I$is a maximal ideal. When we have a field$F$we have that$F[x]$is a PID, so an ideal is maximal if and only if it is generated by an irreducible polynomial. In this case however$\mathbb Z$is not a field, so we cannot conclude$\mathbb Z[x]$is a PID. So we cannot conclude ... 1 Let$T$denote the transposition transformation, and let$I$denote the identity transformation. We note that $$(T - I)^2 = T^2 - 2T + I = T^2 - I = 0$$ It follows that the minimal polynomial of$T$divides$x-1$, which is to say that$1$is its only eigenvalue. If$T$is diagonalizable, it must then be equal to the identity. Since$T$is not the ... 0 To the case of infinite fields: If you search for a field, in which you cannot solve some equation, you should always start with a function field, since they are very far from being algebraically closed. So lets take$F=\mathbb F_2(t)$and the equation $$x^2+fy^2=0.$$ Cleary if one of$x,y$is non-zero, both are non-zero, hence after multiplication with ... 1 I assume that you know how to show that$\bar L$is a field. (Tell me if this is unclear for you). In order to show$\overline{\bar{L}}=\bar{L}$, you have to prove two parts :$\bar{L} \subseteq \overline{\bar{L}}$. This part is easy since for any subfield$M \subseteq K$, one has$M \subseteq \overline M$.$\overline{\bar{L}} \subseteq \bar{L}\$. This ...

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