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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 $i(ij) = i(a + bi + cj) = -b + ai + c(ij) = -b + ai + c(a + bi + cj) = (-b + ac) + (a + bc)i + c^2 j$ ...


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 $F(\gamma^{m})$


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 ...


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 $L=Q$.


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, ...


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 $F$.


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

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

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 $3$.


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

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.


1

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


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

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

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.


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

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

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 ...


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...



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