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1

Note that $X^4 - 12 = (X^2 - 2 \sqrt{3})(X^2 + 2\sqrt{3})$ so that the minimal polynomial of $w$ over $\mathbb{Q}(\sqrt{3})$ is of degree 2. This tells you that $[\mathbb{Q}(\sqrt{3}, w) : \mathbb{Q}] = 4$. A basis can be given by $\{ 1, \sqrt{3}, w, w\sqrt{3} \}$. Alternatively, you can note that $\mathbb{Q}(w)$ already contains $\sqrt{3}$, ...

1

Let $a = \sqrt{4 - \mathrm{i} \sqrt{20}}$, $b = \sqrt{4 + \mathrm{i} \sqrt{20}}$, and $F = \Bbb{Q}(b)$. You have shown that some polynomial in $a$ is in $F$. This is not the same as showing $a \in F$. Too see this, consider that $\mathrm{i}^2 \in \Bbb{Q}$ but $\mathrm{i} \not \in \Bbb{Q}$. There are two generic ways to attack this. The first is to find ...

0

Assume that there was some linear dependence relation of the form $$\sum_{k=1}^n c_k \sqrt{p_k} + c_0 = 0$$ where $c_k \in \mathbb{Q}$ and the $p_k$ are distinct prime numbers. Let $L$ be the smallest extension of $\mathbb{Q}$ containing all of the $\sqrt{p_k}$. We argue using the field trace $T = T_{L/\mathbb{Q}}$. First, note that if $d ... 3 If you multiply the two elements you get $$\sqrt{4+i\sqrt{20}}\cdot\sqrt{4-i\sqrt{20}} = \sqrt{16-20i^2}=\sqrt{16+20}=\pm 6. \label{A} \tag{A}$$ So you have that $$\sqrt{4-i\sqrt{20}}=\frac{\pm 6}{\sqrt{4+i\sqrt{20}}}\in \mathbb{Q}\left(\sqrt{4-i\sqrt{20}}\right).$$ Note that the first equality in my equation \eqref{A} is not really valid since we are ... -1 An important idea here is that if your minimal polynomial is irredicible and seperable, the Galois group must be transitive on the roots of$f$. So suppose your Galois group has the four elements$\{\mathrm{id},\tau,\pi,\sigma\}$and each of these automorphisms acts on$\sqrt{4 + i\sqrt{20}}$by sending it to a distinct root of$f$. These automorphisms ... 1 My “easiest” way of finding the$\Bbb Q$-irreducible polynomial for$\alpha=\sqrt5+i$is this: Certainly the$\Bbb Q(\sqrt5\,)$-irreducible polynomial for$\alpha$is$(X-\sqrt5\,)^2+1=X^2-2\sqrt5X+6$. And so$(X^2-2\sqrt5X+6)(X^2+2\sqrt5X+6)=X^4-8X^2+36$has the following three properties. It is the only factorization of the polynomial you found into ... 0 The implication doesn't hold. You have that $$[F(a,b):F]=\underbrace{[F(a,b),F(a)]}_{=[F(b):F]}[F(a):F]=[F(a):F][F(b):F].$$ Now$[F(a)\cap F(b):F]$divide$[F(a):F]$and$[F(b):F]$. Therefore,$[F(a)\cap F(b):F]=1$if$\gcd([F(a):F],[F(b):F])=1$. Then, if$F(a,b)=F(a)F(b)$, it will hold since $$[F(a)F(b):F]=[F(a):F][F(b):F]\iff \gcd([F(a),F],[F(b);F])=1.$$ ... 2 Note that$ \mathbb{Q}(\sqrt{5}, i)/\mathbb{Q} $is Galois of degree 4, and the stabilizer of$ \sqrt{5} + i $in the Galois group is the trivial group, which means that$ \mathbb{Q}(\sqrt{5} + i) = \mathbb{Q}(\sqrt{5}, i) $so that$ \sqrt{5} + i $has degree$ 4 $over$ \mathbb{Q} $. This means that the polynomial you've found is irreducible, and it is ... 2 Suppose$\sqrt3\in Q(\sqrt2)$, then write$\sqrt3=a+b\sqrt2, a,b\in Q$, Remark that$a,b$are not zero, if$a=0$, then$\sqrt2\sqrt3=b(\sqrt2)^2=2b=\sqrt6$, this is not true since$\sqrt6$is not in$Q$. Suppose$b=0$, then$\sqrt3=a\in Q$this is not true. you obtain$3=(a+b\sqrt2)^2=a^2+2ab\sqrt2+2b^2$. This implies that ... 4 If you know that$\sqrt{3}$is not in$\mathbb{Q}(\sqrt{2})$, then you know the degree is greater than$1$. But$\sqrt{3}$is a root of the equation$x^2-3=0$, which has coefficients in$\mathbb{Q}(\sqrt{2})$, so the degree is exactly$2$. 2 Any finite extension of a field of characteristic$0$is separable, so in particular$\Bbb{Q}(2^{⅓})\supset\Bbb{Q}$is. It is however not normal, as otherwise it would be the splitting field of its minimal polynomial$X^3-2$, which you already know it isn't. To identify all intermediate fields$\Bbb{Q}\subset K\subset E$using Galois theory, in stead of ... 1 Since$\;K/F\;$is Galois it is then separable, so$\;a\;$has as many conjugates as its minimal polynomial's degree. 0 The first structure is not a field because the distributive law fails to hold. If you use more familiar notation (where T is a 0 and F is a 1) then we get that$0 \cdot 1 = 1$, which seems to imply that perhaps the distributive law is violated. We can show this by computing$(T+T) \vee T$in two ways and we get two different answers:$F \vee T = T$, and if ... 1 It is known that if$F$is a finite field, then for each integer$n$there exists an irreducible polynomial of degree$n$over$F$. So we cannot have any connection between the characteristic of$F$and$n$. 3 You are right about $$R=\left\{\begin{bmatrix}\alpha & \beta \\ \bar\beta & \bar\alpha\end{bmatrix}\mid \alpha,\beta\in\mathbb{C}\right\}$$, especially since$e=\frac12\begin{bmatrix}1&1\\ 1&1\end{bmatrix}$would create zero divisors :$e(1-e)=0$. But presumably what you're reading is about $$R=\left\{\begin{bmatrix}\alpha & \beta ... 1 EDIT : I was wrong, the generators are actually not of order 2 as I thought. The Galois group is actually cyclic. I believe you have the wrong generators. \sigma_2, for example, cannot be an homomorphism as you should have$$2+\sqrt{2}=\left(\sqrt{2+\sqrt{2}}\right)^2=\left(\sigma_2\left(\sqrt{2+\sqrt{2}}\right)\right)^2 = ... 0 It is true that any automorphism$\sigma: F(\alpha) \rightarrow F(\alpha)$can be extended to an automorphism$\tilde{\sigma} :F(\alpha, \beta) \rightarrow F(\alpha, \beta)$. Briefly this is because$\sigma$induces an automorphism from$F(\alpha)[X]/(f)$to itself, where$f$is the minimal polynomial of$\beta$over$F(\alpha)$. As for your exercise, I do ... 1 Whenever$F$is a perfect field, its algebraic closure$\overline{F}$is a Galois extension of$F$. Since$\mathbb{Q}$and$\mathbb{F}_p$for primes$p$are perfect fields, we arrive at easy examples of infinite Galois extensions by considering their algebraic closures. 1 The extension$\mathbb{Q}(\sqrt{2})$is easily seen to be Galois since it is the splitting field of the polynomial$x^2-2$. Similarly one gets all other square roots of positive reals. Also (as you will probably see later on), the Galois groups of the examples mentioned here (finite extensions of finite fields and cyclotomic extensions of the rationals) ... 2 The answer is no. For example, let$y$be an element of$F$and let$g(x) = (x-y)^2f(x)$, where$f(x)$is in$F[x]$and has all simple roots. Then the splitting field of$g$over$F$is the same as that of$f$. 3 I will use the notation$K'/K$for$K\subseteq K'$. First show that$K'L/K'$is a normal extension: we know that$L$is the splitting field of some polynomials$f_i$with coefficients in$K$. I claim that$K'L$is the splitting field over$K'$of the same polynomials. Let$E=K'L$. The field$E$is the splitting field of the$f_i$if the following hold: ... 1 The minimal polynomial of$\zeta_3$is in fact$x^2+x+1$, since$x^3-1$has$1$as a root. Therefore$\mathbb{Q}(\zeta_3)$is a quadratic extension of$\mathbb{Q}$, so there are no intermediate subfields. 1 If your field extension is Galois, then there is a one-to-one correspondence between subgroups of the Galois group and the intermediate fields. The correspondence associates a subgroup to its fixed field. This is called "The Fundamental Theorem of Galois Theory". So different subgroups will have different fixed fields. However if your extension is not ... 2 $$x^n+ax^m+b=\prod_{k=1}^{n}(x-\zeta_k) \tag{1}$$ $$n x^{n-1}+am x^{m-1} = \sum_{h=1}^{k}\prod_{\substack{1\leq k\leq n \\ k\neq h}}(x-\zeta_k)\tag{2}$$ and by evaluating$(2)$at$x=\zeta_i$we get: $$n\zeta_i^{n}+am \zeta_i^{m} = \zeta_i\prod_{\substack{1\leq k\leq n \\ k\neq i}}(\zeta_i-\zeta_k)\tag{3}$$ or: $$(am-an) \zeta_i^{m}-b = ... 1 No. This is false. Here is a counter-example. Let \alpha be a generator of \mathbf F_{2^6} with minimal polynomial$$ p=x^6 + x^5 + x^3 + x^2 + 1, $$and let \beta be a generator of \mathbf F_{2^{10}} with minimal polynomial$$ q=x^{10} + x^9 + x^8 + x^3 + x^2 + x + 1 $$over \mathbf F_2. Then \mathbf F_2(\alpha,\beta)=\mathbf F_{2^{30}} since ... 5 This is indeed not true. Take for example K = \mathbb Q(\sqrt[4]{2}), E = \mathbb Q(\sqrt 2) and F = \mathbb Q. Then [K:E] = [E:F] = 2, so that K/E and E/F are normal. But K/F is not normal, since \sqrt[4]{2} is conjugate to i\sqrt[4]{2} (both are roots of the irreducible polynomial X^4-2\in \mathbb Q[X]) but i\sqrt[4]2 does not lie in ... 1 I think what you are asking is how to prove the following: Let f \in \mathbb{Z}[X] be a monic irreducible polynomial. Let L be the splitting field of f over \mathbb{Q}. If the reduction of f modulo p is squarefree, then L/\mathbb{Q} is unramified at p. If \mathbb{Q} \subseteq K \subseteq L are fields, p a rational prime, then L ... 1 A standard way of doing this would be: First check that the cubic polynomial$$ f(x) = x^3+x^2-2x-1$$is irreducible. If f(x) weren't irreducible, since it is a cubic, there would be a rational root. However, the only possible rational roots are 1 or -1, and f(1)\not=0 and f(-1)\not = 0. So f is irreducible. Therefore the Galois group is ... 1 Let f(x)=x^3+x^2-2x-1. Since the degree of f is 3, we know that f has a real root a. Suppose a\in Q. Write a={p\over q} where gcd(p,q)=1. We have {p^3\over q^3}+{p^2\over q^2}-2{p\over q}-1=0. This implies that p^3+qp^2-2q^2p=p(p^2+qp-2q^2)=q^3. This implies p divides q^3 and p=1 or -1 since gcd(p,q)=1. We can also write ... 2 Take x\in R and consider the ideal (x) generated by x. Then (x)=R and hence 1\in (x), it follows that there exists a y\in R such that xy=1=yx. Thus y=x^{-1}. Since x was chosen arbitrarily, the result follows. 2 With a little linear algebra, that at this stage is usually well known already, I think it can be pretty easy and short: For any \;n\in\Bbb N\;,\;\;\Bbb F_{p^n}\; is a linear space over the prime field \;\Bbb F_p\; and of dimension \;n\; , so:$$\Bbb F_{p^a}\hookrightarrow\Bbb F_{p^b}\iff \Bbb F_{p^a}\;\;\text{is a linear subspace of}\;\;\Bbb ... 1 I think the section around proposition 7.2 is indeed somewhat problematic. After making the definition of a not necessarily finite unramified extension one should first check that for finite extensions it agrees with the original definition. To do this, one already needs the proposition 7.2 and its corollary 7.3 (the compositum of two unramified extensions ... 0 A lot of info was scattered about, so here's a condensed bit of the developments in Gregory Grant's answer and the comments. The final solution was the the existence of a field structure on every infinite set$X$is equivalent to the full axiom of choice. Firstly, to show choice implies the existence, we borrow Gregory Grant's construction. Let$K$be a ... 1 Yes: for a finitely generated$k$-algebra, for instance, you can still take your additive function$\lambda$to be$\dim_k$, and you get that the Hilbert series$\mathcal P(V,t) = \sum_{i \geq 0} \lambda(V_i)t^i$is a rational function. See e.g. http://tartarus.org/gareth/maths/notes/iii/Commutative_Algebra_2013.pdf, p. 31 onwards, for an approach to it. 1 By ordering your basis of$L$over$F$correctly, you can make the matrix of$T_a$a block diagonal matrix. Recall that$k = [L:F(a)]$and$\deg(m_a)=m.$Let$V_i \subseteq Lbe defined by: \begin{align*} V_i &= \textrm{span}_F\{a_1l_i, a_2l_i,\dots,a_ml_i\}\\ &= l_i \textrm{ span}_F\{a_1,a_2,\dots,a_m\}. \end{align*} Then clearlyL = ...

0

$x^3-5$ is irreducible, so the Galois group is a transitive subgroup of $S_3$. The only possibilities are $C_3$ and $S_3$. But your automorphism $\tau$ has order $2$, while $C_3$ has no elements of order $2$. So the Galois group is $S_3$. We can also make a similar argument, using the Galois correspondence, by observing that $\mathbb{Q}[\omega]$ is ...

1

In general, for any $r\in R$ the map $\phi: R[x]\rightarrow R$ defined $f\in R[x]\mapsto f(r)$ is a ring homomorphism. To show this, we need to check that for any $f,g\in R[x]$, $(f+g)(r) = \phi(f+g) = \phi(f)+\phi(g) = f(r)+g(r)$, and $(fg)(r) = \phi(fg) = \phi(f)\phi(g) = f(r)g(r)$. But the order in which we replace $x$ by $r$ doesn't matter. For ...

2

Hint: Evaluation at $0$ associates to every polynomial its constant term. So you have to check the constant term of $f(x)+g(x)$ is the sum of the constant terms of $f(x)$ and $g(x)$, ans similarly for their product.

2

We have the root $\sqrt[3] 5$. Now, factor $x^3-5$ using that root: $$x^3-5=x^3-\sqrt[3] 5 x^2+\sqrt[3] 5 x^2-\sqrt[3] 25 x+\sqrt[3] {25} x-5=(x-\sqrt[3] 5)(x^2+\sqrt[3] 5 x+\sqrt[3] {25})$$ Now, the other two roots must be from this quadratic and using the quadratic formula, we can see they are not in $\Bbb{Q}(\sqrt[3] 5)$. Thus, to split $x^3-5$ in ...

2

First note that $\mathbb{Q}(i,\sqrt 3) = \mathbb{Q}(i\sqrt 3, i)$. It remains hence to calculate the degree of the simple field extension $\mathbb{Q}(i\sqrt 3, i)/\mathbb{Q}(i\sqrt 3)$. We find that $i$ is a zero of the polynomial $X^2+1$, which is still irreducible over $\mathbb{Q}(i\sqrt 3)$. Hence the degree is 2.

1

The trace is somewhat easier than the determinant, but the idea is sort of the same. I will show you how to do the trace. You can use the fact the field trace behaves well with multiple extensions, that is(I will use the symbol $S$ for the field trace) $$S_{K/F} = S_{F(\alpha)/F} \circ S_{K/F(\alpha)}.$$ This makes things a lot easier. Evaluating the above ...

1

You can assume that $F=K(u,v)$. Let $X$ and $Y$ be independent indeterminates over $K$. Since $v$ is algebraic over $K(u)$, there exist $a_i,b_i\in K[X]$ with $$\frac{a_0(u)}{b_0(u)}+ \frac{a_1(u)}{b_1(u)}v+ \dots \frac{a_{n-1}(u)}{b_{n-1}(u)}v^{n-1}+ v^n=0$$ with $n\ge0$. If $n=0$, then $u$ is algebraic over $K$, hence also over $K(v)$. Assume $n>0$ ...

0

$K$ Inherits conditions of being field (Associativity of addition and multiplication, Commutativity of addition and multiplication, Distributivity of multiplication over addition, and Existence of additive and multiplicative identity elements) unless being additive and multiplicative closeness and having additive inverses and multiplicative inverses in it ...

1

As $a,b\in K$ implies $a-b\in K$ and $K$ is not empty, $K$ is a subgroup (additively) of $F$. Similarly, as $a,b\in K\setminus\{0\}$ implies $ab^{-1}$ and $K\setminus\{0\}$ is not empty(!), $K\setminus\{0\}$ is a (multiplicative) subgroup of $F^\times$.

1

Hint: $v$ algebraic over $K(u)$ $\implies$ $v$ is root of a polynomial with coefficients in $K(u)$ $\implies$ $v$ is root of a polynomial with coefficients in $K[u]$ $\implies$ for some two variable polynomial $P\ne 0$ with coefficients in $K$, $P(u,v) = 0$. Expanding this in powers of $u$: $$0 = P(u,v) = P_0(v) + P_1(v)u + \cdots + P_n(v)u^n,$$ i.e., $u$ is ...

0

As I recall, it depends on a strategy like this, modulo suitable care to the hypotheses: If $k\subset L$ are fields and $\psi\colon L\to\Omega$, where $\psi$ is a field morphism and $\Omega$ is an algebraically closed field, then $\{L:k\}=\{\psi(L):\psi(k)\}$. This needs a proof, but it isn’t hard.

4

Observe that $\sqrt{5} \in \mathbb{Q}(\sqrt{1+\sqrt{5}})$ so $\mathbb{Q}(\sqrt{5})$ is an intermediate field of the extension $\mathbb{Q}(\sqrt{1+\sqrt{5}}) / \mathbb{Q}$. It's easy to see that $\sqrt{1+\sqrt{5}}$ is not in $\mathbb{Q}(\sqrt{5})$ and therefore $[\mathbb{Q}(\sqrt{1+\sqrt{5}}) : \mathbb{Q}(\sqrt{5})] = 2$. Because $[\mathbb{Q}(\sqrt{5}) : ... 0 A result is that if$ f \in K[x] $is irreducible and$ f(\alpha) = 0 $, then we have an isomorphism$ K[x]/(f) \cong K(\alpha) $. Therefore, it suffices to find an irreducible polynomial of which$ \alpha = \sqrt{1 + \sqrt{5}} $is a root. We can see that$ (\alpha^2 - 1)^2 - 5 = 0 $, so$ \alpha $is a root of$ X^4 - 2X^2 - 4 $. This polynomial has no ... 2 Assuming that the extensions (and hence also the Galois groups) are all finite. Let us denote the Galois groups by$G=\operatorname{Gal}(E/F)$and$K=\operatorname{Gal}(E/B)$. So all the automorphisms$\tau\in K$fix$B$elementwise. Show that if$\sigma\in G$, then all the automorphisms in$\sigma K\sigma^{-1}$fix the field$\sigma(B)$elementwise. ... 2 To show these are homomorphisms, we use two standard results: (a) evaluation map theorem for polynomial rings over an arbitrary ring (b) Universal property for field of fractions. We proceed as follows: (a) Given a ring homomorphism$\mu:R\rightarrow S$and an element$s\in S$. There exists a unique ring homomorphism$\bar{\mu}:R[x]\rightarrow S\$ from ...

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