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1

Following Nishant's suggestion the calculation might go as follows. Modulo $p=2$ our polynomial $f(x)=x^5-2x+7$ factors as $$ f(x)\equiv x^5+1=(x+1)(x^4-x^3+x^2-x+1). $$ This tells us many things. For example the only possible non-trivial factorization over $\Bbb{Q}$ must have a linear factor. But the rational root test quickly tells us that those don't ...


1

Using Dummit & Foote's Abstract Algebra book Page 639 Exercise 21, I can tell you that the Galois group is not solvable. This because they know that the Galois group of a quintic in the form $$ f(x) = x^5 + Ax + B $$ is solvable over $\mathbb Q$ (but do not prove it there, they just give a reference ; note that your polynomial has this form) if and only ...


1

for $f(x)=x^5-2x+7$ what is $f'(x)$ how many real roots do you think this would have??


1

Well, you can factor it modulo a bunch of primes to see what cycle types of elements of $S_5$ are in the Galois group, but of course, you can't use primes that divide the discriminant.


0

$L$ is viewed as an extension $L/K$ but the fact is that $L$ just contains an isomorphic copy $K'$ (as field) of $K$ which we identify with $K$ then we think of $K'$ as $K$ (since they are isomorphic as fields) and say $L/K$ and not $L/K'$. So when you write about the restriction to $K$ it is important to note what this theorem gives you - $L:=K[x]/\langle ...


1

Being a root of a polynomial in $F[x]$ is called being algebraic over $F$. There are extensions $K/F$ in which no element of $K\setminus F$ is algebraic over $F$. Indeed any purely transcendental extension will work (i.e. any adjunction of a collection of algebraically independent indeterminates - no matter what the cardinality of this collection). This is ...


2

You are probably referring to the Frobenius theorem that the only associative division algebras that are finite-dimensional as vector spaces over the reals are up to isomorphism the reals themselves, the complex numbers, and the quaternions. If commutativity is added only $\mathbb{R}$ and $\mathbb{C}$ remain. However, it is essential that we require a ...


1

Note that a polynomial over the field with 2 elements has a binary representation (technically this should be called "$x$-adic notation", since you're writing the polynomials in base $x$). In some contexts, it is understood that when we write an integer, that we mean the polynomial that has the same binary representation: e.g. $2$ means $x$ and $3$ means ...


1

The 4-element field has 4 elements but not $0,1,2,3$. The characterestics is 2, so in the field $3=1$ and $2=0$. The other two elements are the solutions of the equation $x^2+x+1=0$. Their order is $3$.


2

The simple answer to what appears to be your most basic question: "is there a finite field with four elements?" The answer is: "yes" The easiest way to construct it is to take $$\Bbb F_2[x]/(x^2+x+1)$$ where $\Bbb F_2$ is the field with two elements, which you might denote by GF(2). The polynomial $x^2+x+1$ has no roots, and so is irreducible, so that ...


2

"Separable" means the roots are distinct in an algebraic closure. The algebraic closure of $E$ is the same field as the algebraic closure of $F$.


10

It seems to me that we can achieve a contradiction more quickly without assuming this anyway: Since $(1, i, j)$ is a basis for the field, $ij = a + bi + cj$ for some unique $a, b, c \in \mathbb{R}$. Then, on the one hand $i^2 j = -j$, and on the other it is $i(ij) = i(a + bi + cj) = -b + ai + c(ij) = -b + ai + c(a + bi + cj) = (-b + ac) + (a + bc)i + c^2 j$ ...


3

Let $X$ be a total space. If we let symmetric difference take the place of addition: $A \Delta B = (A \cup B) - (A \cap B)$, and let intersection be multiplication, then a sigma algebra of subsets of $X$ becomes a boolean ring with the empty set being $0$ and the total space being $1$. In general a family of sets closed under intersection and symmetric ...


1

Your proof is fine. As the comments have pointed out, the proof can be shortened. If you still want to do it in a single chain of equalities, then you can do something like this: \begin{equation*} \begin{split} 0a &= 0a + 0 && \quad \text{by }\textit{identity element }(+ ) \\ &= 0a + 0a + (-0a) && \quad \text{by ...


1

A few pointers: You don't have to use "Now". You could just say "Let $a\in F$." Don't say "meaningless". Rather, phrase it like so: Non-trivially, associativity implies that any parentheses are redundant. Hence, parenthesis will be suppressed and we will thus not explicitly employ associativity. It's not wrong to say that $(F,+,\cdot)$ is a field, ...


1

A very big hint: you've already shown that $0$ isn't a multiple root. Now, any non-zero common factor of $f(x)$ and $f'(x)$ is also a common factor of $f'(x) = nx^{n-1}-1$ and $g(x) = x^{n-1}-1 = \frac{f(x)}{x}$; this means that it's also a factor of $f'(x)-ng(x)$. You should find that this expression has a particularly simple form...


6

You are on the right track. Note that if $f(a)=0$ then $a^{n}=a$, which means $a^{n-1}=1$. Now then $f'(a)=na^{n-1}-1=n-1$. So, $f'(a)=0$ implies $n-1=0$ in your field, which means that the field has characteristic $p$ that divides $n-1$. For all other fields there is no multiple root.


6

$\require{cancel}$ Yes. The basic construction is actually quite simple. You want a Galois extension to be generated by your root, so that--by normality--any separable polynomial with a single root in the field splits completely. Since the degree is $3$, this means we want an extension with Galois group $\Bbb Z/3\Bbb Z$. By general theory we know that all ...


3

$1,\gamma, \ldots , \gamma^{m-1}$ are independent over $F(\gamma^{m})$


1

It follows directly from the infinity of primes and the fact that $\left[\mathbb Q[\sqrt p_1,\dots,\sqrt p_n]:\mathbb Q\right]=2^n$ for $p_1,\dots,p_n$ distinct prime numbers. This last fact is the theme of question 113689.


3

Your computations are correct. In particular, the norm form of an algebra of dimension $r$ over a field $K$ is always of degree $r$. However, as you noted, in the case of quaternions, the norm form is the square of the quadratic form of signature $(4,0)$ which can also be obtained as $$ N(q)=q\overline q $$ where $\overline q$ denotes quaternionic ...


2

If $\alpha$ is a root of $f$, then the minimal polynomial $p_\alpha$ of $\alpha$ divides $f$. Recall that $p_\alpha$ has degree $n=[\mathbb Q(\alpha):\mathbb Q]$, since $\mathbb Q(\alpha)\simeq \mathbb Q[x]/(p_\alpha)$. Since $f$ is monic and a multiple of $p_\alpha$ of the same degree, it follows that $f=p_\alpha$ so that $f$ is irreducible. However, ...


3

Let $K$ be a field and $Q$ be its prime field. Let $L$ be a subfield of $K$. Then $L$ contains $Q$. Every field homomorphism $Q \to L$ must be the inclusion because $Q$ is the prime field of $L$. Thus, the image of a field homomorphism $Q \to L$ is $Q$. Therefore, $L$ is isomorphic to $Q$ iff $L=Q$.


0

Your result for the splitting field itself looks good.$^{*}$ Here's one way to complete the degree computation. Let $L:=\mathbb{Q}(\sqrt[6]{20})$. As you say, $[L:\mathbb Q]=6$ and $$\sqrt{5}=\frac{1}{2}(\sqrt[6]{20})^3 \in L.$$ If $\sqrt{2}\in L$, then $\mathbb Q(\sqrt{2},\sqrt{5})\subset L$. I claim that $$[\mathbb Q(\sqrt{2},\sqrt{5}):\mathbb Q]=4. ...


3

a) Every $\alpha \in E$ is of the form $f(Y,Z)/g(Y,Z)$ for $f, g \in F[Y,Z]$. Then $\alpha^p = f^p/g^p$, and it suffices to show that $f^p \in L$ for all $f \in F[Y,Z]$. Writing $f = \sum_{i,j} a_{ij}Y^iZ^j$ and applying Frobenius gives $f^p = \sum_{i,j} a_{ij}^p (Y^p)^i(Z^p)^j \in F[Y^p,Z^p] \subseteq L$. b) If $E = L(\alpha)$ for some $\alpha \in E$, then ...


0

First, note that $x^{3} - 1 = (x- 1)(x^{2} + x + 1)$ and the quadratic factor is irreducible over $\mathbb{F}_{11}$ so that $\mathcal{K} = \mathbb{F}_{11}/(x^{2} + x + 1) \cong \mathbb{F}_{11^{2}}$. Note that $3^{6} = 3$ in $\mathbb{F}_{11^{2}}$ so that $(x^{2} - 3)(x^{3} - 3) = (x - 27)(x+ 27)(x-9)(x^{2} + 9x + 81)$. As suggested in the comments, $p(x) ...


1

As mentioned by Ferra, the field of rational numbers has no prime elements. However, it is true that every nonzero rational number has a unique decomposition into a product of prime integers with positive and negative exponents. In this sense, the "primes" of the field of rational numbers are exactly the primes in the ring of integers. The same holds for ...


5

In the field of rational numbers (and in any other field) there are no primes because all nonzero elements are units!


2

I'll denote the image of $x$ in the quotient as $\bar{x}$. It's clear that $\bar{x}$ doesn't have order 1, and similarly it doesn't have order 2 (because then $\bar x$ would be a root of $x^2 - 1 \in \Bbb{Z}_3[x]$, contradicting that the cubic $f(x)$ (which has $\bar x$ as a root) was irreducible). Thus $\bar x$ has order 13 or 26. Likewise, $2\bar x$ ...


1

$\def\QQ{\mathbb{Q}}$ Yes. Moreover, this is true more generally. Let $f$ be any separable polynomial over $\QQ$. Let $K$ be the splitting field of $f$, let $\theta_1$, ..., $\theta_n$ be the roots of $f$ in $K$ and let $G = \mathrm{Gal}(K/\QQ)$. Map the polynomial ring $\QQ[x_1, \ldots, x_n]$ to $K$ by $x_i \mapsto \theta_i$ and let $M$ be the kernel of ...


2

A local version of the assertion seems to be true: the maximal ideal $M\mathbb{Q}[x_1,\ldots x_n]_M$ of the local ring $\mathbb{Q}[x_1,\ldots x_n]_M$ is generated by the set of Galois-invariant polynomials contained in $M$, or of course even a finite subset of them. Let $H$ be the group of automorphisms of $B:=\mathbb{Q}[x_1,\ldots x_n]$ obtained through ...


1

I believe I recall reading the name "solvable closure" for this construction, but Google doesn't seem to turn up anything. This would make sense, however, as $F(f)$ is the smallest extension of $f$ which is closed under the operation of taking solvable extensions. Alternatively, $\text{Gal}(F(f)/f)$ is the largest pro-solvable quotient of ...


1

What we always have in a ring (or field) is addition, subtraction, multiplication. Division $a/b$, that is the existence and uniqueness of a solution to $bx-a=0$ is different. Even with a field there is not always a soltution (namly if $b=0$ and $a\ne 0$), or it may not be unique (namely if $a=b=0$), so even in a field we only have division if what we divide ...


4

You are correct. The most basic ring is $\mathbb Z$, the ring of integers. Can we divide in the ring of integers, and always get an integer? What if we divide 3 by 2? Some rings have zero divisors. That is, nonzero $a$ and $b$ so that $ab=0$. So $a\cdot b=0=a\cdot 0$. Can we divide $0$ by $a$ and get $b$?


1

The simplest example of a ring is that of the integers $\mathbb Z$, and this is in some sense the motivating example of a ring. Division in $\mathbb Z$ is highly restrictive - we can only divide $m$ by $n$ if $m$ is a multiple of $n$. As such, division is not defined as a binary operation - there is no function that gives a multiplicative inverse. The ...


1

This is true because every irreducible polynomial $f(x)$ in $F[x]$ is separable (provided the characteristic of $F$ is zero, or $F^p=F$ for prime characteristic $p$). Indeed, we have $f'(x)\neq 0$ for the derivative, because $deg(f')=deg(f)-1$. Here we have used that a polynomial $f(x)$ is inseparable if and only if $f'(x)=0$.


1

It turns out to be true. Since I don't think your question is a duplicate, but it is nicely addressed at this solution by Andres Caicedo, I'm giving you a community wiki answer to point you to it. If you take a look at the discussion right before the second to last paragraph, you can learn why exactly the rationals are fixed.


3

Hint $\qquad \begin{array}{ccc} & F(\alpha,\beta)\ &\\ \color{#c00}x\nearrow\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!& &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \nwarrow \color{#0a0}y\\ F(\alpha)\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! & &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! F(\beta)\\ & a\nwarrow\qquad\nearrow b \\ & F & \end{array} ...


2

$[F(a,b):F(b)]=\deg f$ iff $f$ is irreducible over $F(b)$. In this case $[F(a,b):F]=\deg f\deg g$. $[F(a,b):F(a)]=\deg g$ iff $g$ is irreducible over $F(a)$. In this case $[F(a,b):F]=\deg f\deg g$. We also have $[F(a,b):F]=[F(a,b):F(b)][F(b):F]=(\deg g)[F(a,b):F(b)]$ and $[F(a,b):F]=[F(a,b):F(a)][F(a):F]=(\deg f)[F(a,b):F(b)]$.


2

Hints/comments/whatnot: If $K$ is the field of numbers (in $\Bbb{C}$) that are algebraic over $\Bbb{Q}$, then by your contrapositive assumption the polynomial $$ (x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta $$ has coefficients in $K$. Hence its zeros are algebraic over $K$. Therefore... Your anwser is correct. However, your teacher may want you to ...


2

Those are really separate questions and should be asked separately. I think the general fact you're supposed to use is that algebraic numbers are algebraically closed, i.e. a polynomial with algebraic coefficients has algebraic roots (so the result follows immediately from considering $(x-\alpha)(x-\beta)$). For the second one, $\sqrt 2$ and $i$ are ...


3

The Lemma/Observation that a solution depends on is the following. Fact. Let $f:K\to\Omega$ be a homomorphism of fields were $\Omega$ is algebraically closed. Assume that $\alpha$ is an element of some algebraic extension field $L$ of $K$. Then there exists a homomorphism of fields $\tilde{f}:K[\alpha]\to\Omega$ such that $f(z)=\tilde{f}(z)$ for all $z\in ...


3

Consider $x^3-a^3$, where here $a^3$ is the number which generates $F(a^3)/F$. Then $a$ is a root of this polynomial, and so we will let $m_a(x)$ denote the minimal polynomial for $a$ over $F(a^3)$. We know $$\begin{cases}m_a(x)|(x^3-a^3)\\ \deg m_a(x)\le \deg (x^3-a^3)=3\end{cases}.$$ But since $$F(a)\cong F(a^3)[x]/(m_a(x))$$ we know ...


1

Of course, any nonzero vector of a $1$-dimensional subspace generates that subspace. Therefore in $F(a) = \{yd \mid d \in F\}$ you may as well choose $y=1\in F(a)$ as the generator. Thus $F(a) = \{d \mid d \in F\}=F$. The general observation to be made is $[E:F]=1$ iff $E=F$. See if you can't convince yourself of this. You will move ahead by applying this ...


3

Your answer is on the right track. Consider the algebraic closure $F'$ of $F$. Then $K \subset F \subset F'$, and since $F'$ is algebraic over $F$ and $F$ is algebraic over $K$, $F'$ is also algebraic over $K$. Moreover, $F'$ is algebraically closed. This shows that $F'$ is an algebraic closure for $K$ which contains $F$. Because the algebraic closure is ...


2

Take an algebraic closure $\overline F$ of $F$. Then $\overline F$ is algebraic over $K$ and is algebraically closed, hence an algebraic closure of $K$


4

There are lots of subfields of $\mathbb{C}$. Some typical examples: The algebraic closure $\overline{\mathbb{Q}}$ of $\mathbb{Q}$. $\mathbb{Q}(\sqrt[n]{a})$ for some $n \in \mathbb{N}$ and $a \in \mathbb{Q}$. In particular $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(i)$. The cyclotomic fields $\mathbb{Q}(\zeta_n)$, where $\zeta_n$ is a primitive $n$th root of ...


2

No; for example, $\Bbb Q (\sqrt{2}) = \{a + b\sqrt{2}\;|\;a,b \in \Bbb Q\}$ is a subfield of $\mathbb{C}$. Any field extension of $\Bbb Q$ that is a subset of $\Bbb R$ is necessarily a subfield of $\mathbb{C}$.


2

You have $[F(b):F]\mid [F(a,b):F(a)]$. Now bound it from the other direction, $$[F(a,b):F(a)]\le [F(b):F].$$ Note this is true for arbitrary $F,a,b$. Hint: consider $b$'s minpoly over $F(a)$ and $F$; what's a relationship between them. Once the indices are squeezed, they must be equal. Now what? For your second question, you did Case 2 wrong. The number ...


3

Assume we are in an integral domain. If we subtract the first and third equations, we get $a_0^2 - a_1^2 = b a_0 + c a_0 - b a_1 - c a_1$, which is the same as $(a_0 - a_1)(a_0 + a_1) = (b + c)(a_0 - a_1)$. Since $a_0 \ne a_1$, $a_0 - a_1 \ne 0$, and so we can cancel to get $a_0 + a_1 = b + c$. Subtracting the second and third, we get $a_0 a_0 - a_0 a_1 = ...



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