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4

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 ...


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}) : ...


3

Since the extension is Galois, the fundamental theorem of the Galois theory says that there exists a bijection between subfields of $E$ and subgroups of $Gal(E/F)$ by the correspondence which sends $H\subset Gal(E/F)$ to $E^H$. The subgroup $H$ such that $F(\alpha)=E^H$ is the trivial group $\{1\}$ here thus $E=F(\alpha)$.


3

Your extensions are of the form $\mathbf Q(a,b)$ and $\mathbf Q(ab)$. Now the generator $ab$ of the second extension is obviously a member of the first. Hence the elements of the field it generates, which are rational functions in $ab$, are ipso facto rational functions of the two variables $a$ and $b$. i.e. elements of the field they generate.


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

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

For the second function, let's show that $\phi_2$ is an homomorphism. You need to verify $3$ properties in order to prove that. Let $f,g \in \mathbb{C}(x)$ : $$\phi_2((f+g)(x))=(f+g)(1-x)=f(1-x)+g(1-x)=\phi_2(f)+\phi_2(g)$$ $$\phi_2((fg)(x))=(fg)(1-x)=f(1-x)g(1-x)=\phi_2(f)\phi_2(g)$$ ...


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 ...


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

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

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 ...


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

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

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 ...


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

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 ...


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 ...


1

Let $k\mathbb{Q}=\{kx\; :\; x\in \mathbb{Q}\}$. There are infinitely many irrational numbers of the form $m=\sqrt{n}$, where $n$ is positive natural number. Hence, the set $ \{ (\sqrt{n}\mathbb{Q}+\mathbb{Q},+,\cdot) \; : \; n \mbox{ is a positive integer}\} $ contains an infinite sub-fields of $(\mathbb{R},+,\cdot)$. Since every natural number is a finite ...


1

Define $p = \mathrm{char}(F)$, the characteristic of the field $F$, as the integer $p$ for which the map $\mathbb Z \to F$ has kernel $p \mathbb Z$ and thus gives the injection $\mathbb Z/p \mathbb Z \to F$. In characteristic zero this gives the canonical inclusion $\mathbb Q \to F$, and in characteristic $p > 0$ this gives $\mathbb F_p \to F$. This ...


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

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

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 ...


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 $n>0$ ...


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 $V_i \subseteq L$ be 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 clearly $L = ...



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