New answers tagged

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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