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22

The binomial coefficient $\binom p i$ is divisible by $p$ for $1 \leq i \leq p-1$ One way of seeing this is Legendre's formula on the power of a prime dividing some factorial, http://www.cut-the-knot.org/blue/LegendresTheorem.shtml and http://en.wikipedia.org/wiki/Factorial#Number_theory From the formula, $p$ divides $p!$ with exponent exactly $1,$ ...


11

Let $F$ be a field of characteristic $p$. Let $f = (1 + x)^p \in F[x]$. We want to show that $f = 1 + x^p$. Take the formal derivative: $f' = p(x+1)^{p-1} = 0$ Now we know that $f$ has degree $p$, and its derivative is $0$, so $f$ must be in the form $A + Bx^p$ with $A$, $B \in F$. $f(0) = 1$ so $A = 1$. A product of monic polynomials is always monic so ...


9

Let $u = \sqrt{2} + \sqrt[3]{5}$, we have $$\begin{align} (u - \sqrt{2})^3 = 5 \iff & u^3 - 3\sqrt{2}u^2 + 6u - 2\sqrt{2} = 5\\ \implies & \begin{cases} \sqrt{2} &= \frac{u^3 + 6u - 5}{3u^2 + 2} \in \mathbb{Q}(u)\\ \sqrt[3]{5} &= u - \sqrt{2} \in \mathbb{Q}(u) \end{cases} \end{align} $$ As a result, $\mathbb{Q}(\sqrt{2},\sqrt[3]{5}) \subset ...


5

You are missing the primitive element theorem. That is the critical argument. To be more precise: a separable extension of degree $n$ would have a primitive element, whose minimal polynomial would be irreducible of degree $n$, contradiction!


4

In this case $F[i]=F(i)$. This is because $F(i)$ is the field of fractions of $F[i]$, which is already a field. You can see this becuase the characteristic polynomial of $i$ is irreducible in any subfield or $\mathbb{R}$, or more explicitly by computing $(a+bi)\frac{1}{a^2+b^2}(a-bi)=1$, so that the inverse of any nonzero element exists.


4

I think the algebraic closure of $\mathbb{F}_p$ does the job for $p$ a prime number. Take $x\in\overline{\mathbb{F}_p}$ such that $x\neq 0$ then $\mathbb{F}_p[x]$ is a finite field and hence $x$ is of $\underline{\text{multiplicative}}$ finite order in $\mathbb{F}_p[x]$ and hence in $\overline{\mathbb{F}_p}$. Of course, the last thing to know is that ...


4

You should note that the Galois group is isomorphic to a subgroup of $S_4$, since any element of the Galois group will be determined by how it permutes the four roots of $x^4-2$. The order 8 subgroups of $S_4$ are isomorphic to the dihedral group with 8 elements. If you want to know exactly what the eight maps are, there are four choices as to where ...


3

You can see that $i^n\sqrt[4]{2},\, 0\le n\le 3$ are all the roots, so that $$K=\Bbb Q(i,\sqrt[4]{2})$$ is the splitting field for the polynomial. Since $L=\Bbb Q(\sqrt[4]{2})$ is real of degree $4$, we see that $K$ is a proper extension of $L$, and since $[\Bbb Q(i):\Bbb Q]=2$ we see the total degree of the extension is $2\cdot 4=8$. But then we have that ...


3

Let $F$ be the field you are talking about, let $p$ be the characteristic of $F$. Then since $F$ is algebraically closed, the equation $$ x^{p^{n}}-x=0 $$ has precisely $p^{n}$ solutions in $F$ because its derivative is $1$. The solutions would now constitute a subfield $F_{n}$ of $F$. Multiplication and inverse is trivial, and addition follows from the ...


3

One way would to say that if $\Bbb Q(i) \subset \Bbb Q(\sqrt[4]{2})$, then it would necessarily be an quadratic extension. This would mean that $x^4-2$ would factor over $\Bbb Q(i)$. Any factorization $x^4-2 = (x^2+ax+b)(x^2+cx+d)$ must have $a = -c$ by consideration of the cubic term. If $a$ is nonzero we must have $b = d$ by consideration of the linear ...


3

The polynomial is irreducible over the rationals, because its possible rational roots are to be found among $\pm1$, $\pm2$, $\pm4$ and $\pm8$. A direct check shows these numbers are not roots. Since the polynomial has degree $3$, reducibility over $\mathbb{Q}$ coincides with having a rational root. So, if $\alpha$ is a root of the polynomial, $f$ is its ...


3

Any example will do, such as $F=\mathbb Q$, $\alpha=\pi$. Since $\overline F$ is algebraic over $F$, $\tau(\alpha)$ would have to be algebraic, say $f(\tau(\alpha))=0$ with $f\in F[X]$. Since $f(\alpha)\ne 0$, $\tau$ fails to be monomorphic.


3

Another way to see the result is as follows: You found the irreducible polynomial fo r $\alpha=\sqrt{1+\sqrt 2}$ essentially by noticing that if you square $\alpha$, then subtract $1$, you obtain a anumber that, when squared, gives $2$: $f(X)=(X^2-1)^2-2=X^4-2X^2-1$. But it is of course the case for all roots of $f$ that squaring, subtracting $1$, and ...


3

The key to this is characterizing the units in $\Bbb Z[[x]]$. If you don't know it already, you need to prove that $a_0 + a_1x + a_2a^2 + \ldots$ is a unit in $\Bbb Z[[x]]$ if and only if $a_0 = \pm 1$.


3

Recall that power series (over a domain) is invertible if an only if its constant term is invertible; in this case this means constant term $\pm 1$. Suppose $x^2 + 3x + 2$ is the product of two power series. Then $2$ is the product of the constant terms of these two power series. This is only possible if one of them is $\pm 1$ and thus invertible. ...


2

Yes. Recall that for a field $K \subseteq L$ and $a_1, \ldots, a_n \in L$ by $K(a_1, \ldots, a_n)$ we denote the smallest subfield of $L$ such that $K \subseteq K(a_1, \ldots, a_n)$ and $a_1, \ldots, a_n \in K(a_1, \ldots, a_n)$. $(\subseteq)$ Since $\mathbb{R} \subseteq \mathbb{R}(a)$ and $a \in \mathbb{R}(a)$, we have $\mathbb{R} \subseteq \mathbb{R}(a) ...


2

Note: I don't address your attempts in this answer, and just dive toward an alternate solution, so look away if you aren't interested. I did leave some stuff for you to do, though. Well, let's first try to prove the claim: $$\Bbb Q(\sqrt{2} + i) = \Bbb Q(\sqrt{2}, i)$$ If this is true, then both will have the same basis and degree. Notice that $\sqrt{2} ...


2

It is obvious that $F(1+a^{-1})\subset F(a)$. And since $$\frac1{(1+a^{-1})-1}=a$$ we have that $F(a)\subset F(1+a^{-1})$. They have the same index because they are the same extension.


2

I think Adam Hughes gave a great answer to your initial question. Let me try to say a bit more about what the algebraic closure of $\mathbb{Q}$ in $\mathbb{Q}_5$ is using algebraic number theory. Let $p$ be any prime, then we can embed $\mathbb{Q} \subseteq \mathbb{Q}_p$. The algebraic closure $N$ of $\mathbb{Q}$ in $\mathbb{Q}_p$ is then the largest ...


2

As if the excellent answers of Adam Hughes and @benh weren’t enough, I’ll give another argument something like that of Adam, but not using Hensel. For simplicity, let’s take $p\ne2$. Look at the Binomial expansion of $(1+x)^{1/2}=1+\frac12x- \frac18x^2 + +\frac1{16}x^3 - \frac5{128}x^4 +\cdots\>$. You’ll notice that the only denominators are powers of ...


2

The other answer gives an example. However, note that in the text you showed us, $D$ is finite dimensional over $K$. This changes everything, since then $\mathop{Z}(D)/K$ is a finite field extension; thus, if $K$ is algebraically closed, actually $\mathop{Z}(D) = K$. So $D$ is a finite $K$-division algebra. But over an algebraically closed field, no such ...


2

Let $R$ denote a commutative ring with unity. This is easiest when using the best-available definition, namely: Definition. Suppose $S$ is a subset of $R$. Then to say that "$x$ is in the annihilator of $S$" is just to say that "$xS \subseteq 0$." Proposition. $x$ is in the annihilator of $R$ iff $x = 0$. $(\Rightarrow)$ Suppose $xR\subseteq 0$. Then $x1 ...


2

You are mentioning the rank-nullity theorem. It says that, if $f\colon V\to W$ is a linear map and $V$ is finite dimensional, then $$ \dim V=\dim\ker f+\dim\operatorname{im}f $$ In the particular case when $W=V$ and $f$ injective, we get $$ \dim V=\dim\ker f+\dim\operatorname{im}f=\dim\operatorname{im}f $$ so $f$ is surjective.


2

One of the "weird" features of real vs complex algebra is that the ordering is part of what makes $\Bbb R$ what it is. Knowing that $\Bbb Q(\sqrt[4]{2})$ is even a subset of $\Bbb R$ relies on the fact that you can use continuity to prove there is such a real number as $\sqrt[4]{2}$, so it's somewhat cherry-picking when you want to say $i\not\in \Bbb R$ is ...


2

Yes. Any algebraically closed field of characteristic $p$ contains $\mathbb Z/p\mathbb Z$, so it must contain the algebraically closure of $\mathbb Z/p\mathbb Z$, and the subfield of size $p^n$ must be all the roots of $x^{p^n}-x$. So once you know the result for the algebraic closure of $\mathbb Z_p$, you know it for all algebraically closed fields of ...


2

The question is false when $s = 0$ and $t$ is not a square, for obvious reasons. So suppose $s\neq 0$. By definition, $\mathbb{Q}(r+s\sqrt{t})$ is some field (actually, the smallest) containing both $\mathbb{Q}$ and $r+s\sqrt{t}$. But if it contains $\mathbb{Q}$, then in particular it contains $r$ and $s$. Now: as it contains $r+s\sqrt{t}$ and $r$ and ...


2

Taking $s\neq 0$ $r+s\sqrt t \in \mathbb Q(r+s\sqrt t) \implies s\sqrt t\in \mathbb Q(r+s\sqrt t)\implies \sqrt t\in \mathbb Q(r+s\sqrt t)$ since $s\neq 0$ Now $\mathbb Q(\sqrt t)$ is the smallest field containing $\mathbb Q$ and $\sqrt t$ Thus $\mathbb Q(\sqrt t)\subset \mathbb Q(r+s\sqrt t)$ similarly $r,s,\sqrt t\in \mathbb Q(\sqrt t)\implies ...


2

You are looking for the primitive element theorem. A quick Google search will give you several different sets of self-contained notes that include the proof. If your second question is asking whether given a collection of algebraic numbers $\alpha_1, \dots, \alpha_n$, there exists a $\theta$ such that $\mathbf{Z}[\theta] = \mathbf{Z}[\alpha_1, \dots, ...


2

For any two fields $K,L$ and field homomorphisms $\varphi,\psi \colon K \to L$, the set $$I(\varphi,\psi) = \{ x\in K : \varphi(x) = \psi(x)\}$$ is a subfield of $K$. This is a generalisation of the often-used fact that the set of fixed points of a field endomorphism is a subfield of its domain [and this special case is what is used here; nevertheless, the ...



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