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0

Each member $v\in V$ can be written in a unique way as a binary number with $n$ digits (bits), by prepending zeroes on the left if needed. Choose as your "addition" operator the bitwise exclusive "or" operation. This is just bitwise addition without carry. You can verify that this gives an abelian group where zero is the neutral element. Each element is ...


2

This looks similar enough to the pair-based definition of the complex numbers that it looks promising to explore how far that analogy will take us. First, it is easy to see that multiplying any pair by $(a,0)$ simply scales both of its elements by $a$. If we identify $(a,0)$ with the usual real numbers (that is sort of handwavy but bear with me for a ...


0

The addition and multiplication formulas somewhat remind of the formulas for complex number. We suspect that a map of the form $$ \phi\colon (x,y)\mapsto \alpha x+\beta y$$ for suitable $\alpha,\beta\in\mathbb C$ turns out to be a field isomorphism. As we want $(1,0)\mapsto 1$, we conclude $\alpha=1$. Then $$ ...


1

Note the following theorem of Jacobson (see, e.g., T.Y. Lam, A First Course in Noncommutative Rings, Theorem 12.10). Theorem Let $A$ be a ring such that, for any $x \in A$ there exists an integer $n(x) > 1$ such that $x^{n(x)}=x$. Then $A$ is commutative. Now let $A$ be a ring satisfying your property. Since $x^5=x$ for all $x \in A$, the ring is ...


0

Note that $A$ is a finite field and from $x^4=1$ we get $x=\pm1$ or $x^2+1=0$. If $\operatorname{char}A=2$, then $A\simeq\mathbb F_2$. Otherwise, $A$ can have three elements ($0$ and $\pm1$) and then $A\simeq\mathbb F_3$, or $|A|>3$ and $x^2+1=0$ for $x\ne\pm1$. In the later case $|A|=5$, so $A\simeq\mathbb F_5$.


4

Eoin (+1) already showed that $h$ is invariant under $\phi$. By induction on $n$ we see that $\phi^n(r(x))=r(x+na)$ for all $n\in\Bbb{N}$ and all $r(x)\in k(x)$. From this it follows that $\phi$ is of order $p$. So $\phi$ generates a subgroup $G\le Aut(k(x))$ of order $p$. By basic facts of Galois theory the fixed field $K=\operatorname{Inv}(G)$ satisfies ...


1

By unique factorization, the fixed field of the automorphism is the field of fractions of the fixed ring, since assuming the denominator and numerator are coprime your automorphism permutes the irreducible factors appearing in them. To compute the fixed ring use induction on the degree and polynomial division.


2

We can show that $h$ is fixed by this mapping by direct computation: $$\varphi(h)=h(x+a)=(x+a)^p-a^{p-1}(x+a)=x^p+a^p-a^{p-1}x-a^p=h$$ Since $k\subset k(x)$ is fixed (the map $f(x)=c=f(x+a)$ for every $a$), we now have a fixed field $k(h)\subset k(x)$ by the above computation and noting that all elements of $k(h)$ are rational functions in the "variable" ...


2

Because subfield and extension feild always have the same characteristic? After all the characteristic is obtained from the prime field, i.e., the smallest subfield (which is the same for subfield and extension field, as it it is loosely speaking just generated from $0$ and $1$), so here $\mathbb Q$. Another definition: Consider the kernel of the additive ...


0

As mentioned in the comments, for a field extension $F \subset E$ which contains separable as well as inseparable elements, simply adjoin such elements. For an explicit example, let $p$ be a prime and $F=\mathbb F_p(t)$ the field of rational functions over the finite field $\mathbb F_p$. (You need an infinite field of positive characteristic to find ...


2

You can get a complete answer by checking out the related answer that @JyrkiLahtonen mentions in his comment. But I see this as not so much an algebraic question, as one in complex variable theory and geometry. Do you know the relationship between $2\times2$ complex matrices and fractional-linear transformations of the extended complex plane, also known as ...


0

In addition to the axioms given by kryomaxim, a field also must satisfy (5).... Addition and multiplication are commutative , and (6).... $0 \ne 1$. If all of the requirements are met except commutative multiplication, it is called a Division Ring With Unit. (The "unit" is 1). There is no real number x that satisfies $x^2+1=0$ but there certainly is a ...


1

A field is a set of numbers which satisfy some calculation rules. For the field the following rules hold: Addition rules Addition has the neutral element 0 Addition has an inverse (adding the negative part to a number) Addition is associative Multiplication rules Multiplication has the neutral element 1 Multiplication is always invertible (by ...


1

For odd primes $p$, attempt to solve $X^2 - X - 1 = 0$ using the quadratic formula. Your polynomial is irreducible iff it has no root iff $5$ is not a square mod $p$. I'll leave the $p=2$ case to you.


3

You can always do the following. It requires a tiny bit of algebraic number theory. Assume that the polynomial has integer coefficients, so $$ p(x)=(x-r_1)\cdots(x-r_n)\in\Bbb{Z}[x],\qquad(*) $$ where $r_i\in\Bbb{C}$ are the zeros. Let $K$ be the splitting field of $p(x)$ over $\Bbb{Q}$. Then $[K:\Bbb{Q}]<\infty$. Let $\mathcal{O}_K$ be the ring of ...


0

Let's tweak your example above slightly and take $f(x)=x^2-2$. Then over $\mathbb{C}$ we have $f=(x-\sqrt{2})(x+\sqrt{2})$, each factor having minimal polynomial $f$. Over $\mathbb{F}_5$, $f$ is still irreducible, however from what you're saying is that we can take any root in each case so we shall choose the same root $\alpha$ twice. Then ...


1

Let $x=a+b\sqrt{5}$, $y=c+d\sqrt{5}$, where $a,b,c,d\in\mathbb{Q}$. Then $$(a+b\sqrt{5})(c+d\sqrt{5})=19,$$ then $$ac+5bd=19,$$ $$ad+bc=0.$$ There are $a,b,c,d\in\mathbb{Z}$ that are solutions: $(a,b,c,d)=(1,2,-1,2)$, $(a,b,c,d)=(8,3,8,-3)$. So, examples: $$x=1+2\sqrt{5}, y=-1+2\sqrt{5};$$ $$x=8+3\sqrt{5}, y=8-3\sqrt{5}.$$ And other kind of ...


0

It is a field because one has an isomorphism: \begin{align*}\mathbf R[T]/(T^2+2)&\to \mathbf R^2\\ x+y\times T\bmod(T^2+2)&\mapsto (x,y) \end{align*} and $\mathbf R[T]/(T^2+2)$ is a field, since $(T^2+2)$ is a maximal ideal in $\mathbf R[T]$.


0

Let us fix a non-zero element $x \in GF(q)$ and assume $q>2$. Now consider two sets: $A=\{u^2 \, | \, u \in GF(q)\}$ and $B=\{x-v^2 \, | \, v \in GF(q)\}$. Observe that the first set has cardinality $\frac{q+1}{2}$. Can you show that the second set also has the same cardinality. Note both $A, B \subset GF(q)$. If they were disjoint then they exceed the ...


0

This can be proven combinatorially. First, show that if $G$ is a finite group and $A,B$ are nonempty subsets of $G$ then $G=AB$ if $|A|+|B|>|G|$. With this in mind, take $F$ to be a finite field of size $p$ a prime and consider $F$ under addition. Take $A=B$ to be $F^2$. We know this set has cardinality $\dfrac{p+1}2$ (there are $\dfrac{p-1}2$ quadratic ...


16

No. The reason is that the splitting field is not unique up to unique isomorphism, and any terminal object has to be unique in this stronger sense.


3

Clearly, when $b=0$, $v=a$, so it is not a uniform random variable. When $b \neq 0$, we have that $b$ has an inverse in $\mathbb{Z}_p$, so we have for $c \in \mathbb{Z}_p$ $$\mathbb{P} ( v = c) = \mathbb{P} ( a + b r =c) = \mathbb{P} ( b r = c - a) = \mathbb{P}(r =b^{-1}(c-a)) .$$ Since $r$ is uniform, we have $\mathbb{P}( r = \hat{c} ) = \frac{1}{P}$ for ...


1

I know from experience that this can get incredibly confusing. You have correctly identified the fixed field of $\varphi:x\mapsto x+1$ as $k(x^p-x)$, and called your generating element $u$. If you determine that $\bigl[k(x)\colon k(u)\bigr]=p$, you will have identified the complete fixed field as $k(u)$ because the degrees are right. So all you need to do ...


2

Did you notice that $t^p-t$ is invariant under $G$ ?


0

As $C_m$ is a cyclic group of order $m$, so we have $m$th root of unity in $F$ like $a$. If $p=Char(F)|m$, then there exists natural number $r$ such that $rp=m$ we have : $0=a^m-1=(a^r)^p-1=a^r-1$, so the order of $a$ can not be $m$ and it is contradiction. Therefore $p\nmid m$.


1

Edit: I've been unclear about the difference between being fixed pointwise by conjugation (i.e. for all $k \in K$, $\overline{k} = k$) and being closed under conjugation (i.e. $\overline{K} = K$). Hopefully it is now clarified. I'm assuming that you're considering $\mathbb{Q} \subseteq K \subseteq \mathbb{C}$, since otherwise complex conjugation is ...


2

There is a big Mathoverflow discussion with the same title and 10 answers, some of them are quite informative depending on reader's background. It begins just like this question, I have heard people say that a major goal of number theory is to understand the absolute Galois group of the rational numbers $G=$Gal$(\overline{\mathbb{Q}}/ℚ)$. What do people ...


2

The answer of @BenS. is best, but another way of answering this would be to say that we can understand the full Galois group when we can describe all finite quotients (by normal closed subgroups). For $\Bbb Q$, this would mean describing the Galois groups of all normal extensions of $\Bbb Q$. As I’m sure you know, we’re not yet able to do this.


0

The field is the same as $Q(\sqrt{-2})$ which is the splitting field of $x^2+2$.


5

One possible interpretation is that we would like to be able to write down an explicit name for the group; that is, we would like to identify what group it actually is, and what the properties of that group are: what does its center look like, what are the irreducible representations, etc. Frequently, what is meant is that we would like to understand the ...


1

I assume $a \in \mathbb{E}\setminus \mathbb{F}$ You have to ask yourself a fondamental question: WHERE I'm valuating the polynomial $f$? By your definition you are saying that $I$ is the ideal of the polynomials with coefficients in $\mathbb{F}$ such that have $a$ as rooth as polynomials of $\mathbb{E}[x]$. Now $I$ is an ideal and you can take its ...


2

Before we delve into $\Bbb{F}_2[x]$ let's consider the irreducibility of $f(x)$ in $\Bbb{Z}[x]$. Because $$ f(x)(x^n-1)=x^{3n}-1, $$ the zeros of $f$ are roots of unity of order dividing $3n$.This makes us consider the cyclotomic polynomial $\Phi_{3n}(x)\in \Bbb{Z}[x]$. Its degree is given by the Euler (totient) function $\phi(3n)$. If $n$ is divisible by ...


0

The cases $p=2,3,5,7$ can be handled by inspection. For $p \ge 11$ we can even show that any non-$0$ member of $Z_p$ is equal to $a^2+b^2$ for some non-zero $a,b$, with $a^2 \ne b^2$. Proof: Let $S= \{ {x^2 | 0 \ne x \in Z_p} \}$, let $T=(Z_p - S) - \{ 0 \}$, let $ A= \{ y+z | y,z \in S , y \ne z\ \} - \{ 0 \}$. Observe that if $c \in S \cap A$ then $S=cS ...


1

There is an overkilling solution, using the Chevalley-Warning Theorem. Let $q$ be a prime power. The polynomial $X^2+Y^2+Z^2 \in \mathbb{F}_q[X,Y,Z]$ has one root $(X,Y,Z)=(0,0,0)$. The degree of this polynomial is $2$, which is less than the number of variables (namely, $3$). By the Chevalley-Warning Theorem, it has another root $(X,Y,Z)=(x,y,z)$. ...


4

Write your equation as $X^2 = -1-Y^2 \pmod{p}$ and use the pigeonhole principle. Note that $x \mapsto x^2 \pmod{p}$ assumes $ \frac{p+1}{2}$ many different values. This method even shows that $-1$ can be replaced with an arbitrary value.


7

Let $(x_i)_{i\in I}$ be a transcendence basis of $\mathbb R$ over $\mathbb Q$, and $S=\mathbb Z[x_i:i\in I]$. Then $Q(S)$, the field of fractions of $S$, is $\mathbb Q(x_i:i\in I)$, and $Q(S)\subset\mathbb R$ is an algebraic field extension. Now let $R$ be the integral closure of $S$ in $\mathbb R$. We have $R\subsetneq\mathbb R$ (why?) and $Q(R)=\mathbb R$ ...


1

Let $A$ be a UFD, and $f\in A[Y]$ irreducible with $\deg f\ge 1$. Then the field of fractions of $A[Y]/(f)$ is $K[Y]/(f)$, where $K$ is the field of fractions of $A$. Set $S=A-\{0\}$. Then $K[Y]/(f)=S^{-1}A[Y]/S^{-1}(f)\simeq S^{-1}(A[Y]/(f))$, so $K[Y]/(f)$ is a ring of fractions of the integral domain $A[Y]/(f)$ and moreover it is a field, so ...


2

As it turns out ${\displaystyle \cos\bigg({\pi \over 5}\bigg) = {1 + \sqrt{5} \over 4}}$, which can be seen for example by using the equation $\cos(3x) = \cos(2x)$ for ${\displaystyle x = {\pi \over 5}}$, writing everything in terms of $\cos(x)$ using double and triple angle formulas, and then solving for $\cos(x)$. So we have $$\sin^2\bigg({\pi \over ...


1

What you’re missing is the observation that $c=\cos(\pi/5)$ is a quadratic irrationality over $\Bbb Q$. Once you know that $c^2+Ac+B=0$ for well-chosen rational numbers $A$ and $B$ with $A$ nonzero, the equation $c=(B-c^2)/A$ becomes $c=(s^2+B-1)/A$, where $s=\sin(\pi/5)$, and you have it.


0

Hint : show that $c=\frac{3}{2}-2s^2$, where $c=\cos(\frac{\pi}{5})$ and $s=\sin(\frac{\pi}{5})$. You can check it using Euler's identities for example.


1

What do you mean we don't get examples for $\Bbb{R}$? Say I want to extend $\Bbb{R}$. You can adjoin $i$, an element such that $i^2=-1$. You can also adjoin $k$, another element such that $k^2=-1$. Then $\Bbb{R}(i)\neq \Bbb{R}(k)$ but they are obviously isomorphic. Really, in algebra, treat "isomorphism" as equality. Isomorphic things are "the same ...


1

What does unequal mean for isomorphic fields after all? Is $\mathbb Q[\sqrt 2]$ really so much different from $\mathbb Q[X]/(X^2-2)$? When considering field extensions of $\mathbb Q$ living inside $\mathbb C$, you might for example consider $\mathbb Q[\sqrt[3]2]$ and $\mathbb Q[\sqrt[3]2\,e^{\frac23\pi i}]$, which are isomorphic by the obvious map, but only ...


2

Not sure what your background is, so some of this you'll understand and some of it you may not. Feel free to ask for more details. To make life easy, let's assume we're in characteristic zero. So you're given a field $F$, and you want to adjoin all the roots of all the polynomials $f_n(X) =X^n - 1$ for $n = 2, 3, ...$. If $\zeta_n$ is a primitive $n$th ...


1

If it is reducible, you can write it as the product of two linear polynomials. For degree reasons it has to be $p(x,y)q(x,z)$. Now $$x^2-yz=(ax+by)(cx+dz)\iff ac=1,\ bd=-1,\ ad=0,\ bc=0.$$ These equations are incompatible. In UFDs, irreducible elements are prime.


1

For a polynomial in several variables one can sometimes apply Eisenstein's criterion. Your particular polynomial is Eisenstein at the prime $(y)$ in $(K[y,z])[x]$, for example, but also at $(z)$ in $(K[y,z])[x]$. But looking at your polynomial in $(K[x,y])[z]$, it doesn't seem to be Eisenstein. After all if $x^2\in\mathfrak{p}$ for some prime ...


8

We have that $f=X^3-9X^2+8X=X(X-1)(X-8)$. So the splitting field of $f$ is $\mathbb{Q}$ itself. So the Galois group of $f$ over $\mathbb{Q}$ is trivial.


6

I hope that the below is not too long to be useful. I just thought that I would expand upon my above comment. But, then I realized that the literal words weren't helpful without some explanation of what they themselves mean geometrically. Things spiraled out of control, and I ended up writing a novella. Some background on étale morphisms: A morphism of ...


1

Let $r\in R$, $r\notin F$; then $F[r]\subseteq R$. Since $r$ is algebraic over $F$, $F[r]$ is a field. So… If you don't know that $F[r]$ is a field, consider a relation $$ a_0+a_1r+\dots+a_{n-1}r^{n-1}+r^n=0 $$ with $a_0,\dots,a_{n-1}\in F$, $a_0\ne 0$ and $n\ge1$. Then $$ -a_0r^{-1}=a_1+\dots+a_{n-1}r^{n-2}+r^{n-1} $$ Multiply both sides by ...


1

Since $\alpha^2\in F[\alpha]$, $F[\alpha^2]\subseteq F[\alpha]$. Thus $[F[\alpha]:F]=[F[\alpha]:F[\alpha^2]]*[F[\alpha^2]:F]$ since $\alpha$ is a root of $x^2-\alpha^2$ in$F[\alpha^2][x]$, the extension $[F[\alpha]:F[\alpha^2]]\leq 2$. Thus it must be 1 since the total extension is of odd degree showing that $F[\alpha]=F[\alpha^2]$.


4

Let $K\subset L$ be a finite extension of fields of degree $n$. Then it is a Galois extension if and only if $L\otimes_K L\cong \prod L$, for $n$ copies of $L$. In geometric terms, if $f:X\to Y$ is a finite map of irreducible curves, then it is Galois if and only if $X\times_Y X$ has $n=\deg f$ irreducible components (all necessarily birational to $X$).



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