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6

Hint: $$\frac{1}{a+b\sqrt3}=\frac{a-b\sqrt3}{(a+b\sqrt3)(a-b\sqrt3)}=A+B\sqrt3.$$


2

The vector space is actually a $4$-uple $(V, \Bbb F, +, \cdot)$, where $V$ is a set, $\Bbb F$ is the underlying field, and $+$ and $\cdot$ are operations $+: V\times V \to V$, $\cdot : \Bbb F \times V \to V $ satisfying some conditions. We abuse notation and call $(V,\Bbb F, +, \cdot)$ only $V$, the rest being understood from context. If you have a complex ...


0

This answer was already pointed out in the comments. Suppose $K$ is a field and $L$ is a finite $K$-algebra, meaning $\dim_K L < \infty$. Then for any $a\in L$, we have a $K$-linear map map $L_a(x): L \rightarrow L, x \mapsto ax$. By definition ${\rm{Tr}}_{L/K}(a)$ and ${\rm{Nm}}_{L/K}(a)$ are the trace and determinant of $L_a$ as a linear transformation ...


1

I do not think so. If you take $\phi_i:\mathbb{P}^1\to\mathbb{P}^1$ given by $\phi_i(x)=x^{i+1}$, then the fields are linearly disjoint, but the fiber product is singular.


0

P Vanchinathan's answer is absolutely correct (I just upvoted it) but his curve is reducible. Here is an irreducible example: Consider the curve $C\subset \mathbb A^n_k$ given parametrically by $(t^n,t^{n+1},\cdots,t^{2n-1})$. It is irreducible since it is an image of $\mathbb A^1_k$. However its tangent space at $O=(0,\cdots,0)$ is $T_O(C)=k^n$ : Indeed ...


1

The claim is rather that for all $n\ge1$ and $a\in k$ with $a\ne 0$ the polynomial $X^n-a$ has a root. As $k$ is assumed algebraically closed, such root certainly exists.


-1

Let $\sqrt3 = a + b\theta + c\theta^2 + d\theta^3$ with $a,b,c,d \in \Bbb Q$. Since the trace function($T$) is linear and $T(x) = x$ for $x \in \Bbb Q$: $$ T(\sqrt3) = a + bTr(\theta) + cTr(\theta^2) + d(\theta^3)$$ Everything except $a$ is $0$ here. Therefore, $a = 0$ as well. Divide both sides by $\sqrt 2$ and take the trace again, repeat till ...


1

$\operatorname {Frac}(\mathbb Z[[x]])\subsetneq \mathbb Q((x))$ because $\sum\frac {1}{n!}x^n\in \mathbb Q((x))\setminus \operatorname{Frac}(\mathbb Z[[x]])$


0

I'm going to say no (even though the answer is really yes) and here's why: The process you are describing above is an algorithm -- given a finite set of numbers define recursively the next number and then solve. For algebraic objects, of course there are constructions that you could imagine mimic these operations but they are very different. For example, a ...


2

We use that $k\subset k'$ is separable, a result of Grothendieck which says: If $k'$ and $K$ are extension fields of $k$ such that either $k'$ or $K$ is finitely generated over $k$ and if $k'$ is separable over $k$, then $k'\otimes_kK$ is regular. and the Theorem 33.2(i) from Matsumura, CRT, which asserts: If $A\to B$ is a faithfully flat ...


0

Certainly you can make a field from two elements. Every field of a prime number, $p$, of elements is isomorphic to $\mathbb{Z}_p$. Here $p=2$. If $a+a = a$ and $b+b=b$ then $a=0$ and $b=0$ by the additive inverse. So your operations would not be valid for a field where $a\neq b$. Another way to see it is $b+a = a+a \implies b=a$. This is how the operations ...


1

The cardinal of a finite field must be the power of a prime. See this.


2

As Bernard points out, your statement is false the way you've written it. For example, the fields $\overline{\mathbb{F}_p}$ and $\mathbb{F}_p(x,y)$ are not simple extensions of $\mathbb{F}_p$. However, what you probably meant to write is that if $F$ is a finite field, then any field extension $K/F$ of finite degree is simple. This is true. Since the degree ...


1

It is false: if it were, any algebraic extension of a finite field $F$ would be a finite extension of $F$, hence its algebraic closure would be a finite field. It's easy to show an algebraically closed field can't be finite.


0

Hint: Since $C$ is the companion matrix of $m$, you know that $m$ is the minimal polynomial of $C$. Furthermore, you (should) know that there is only one (up to isomorphism) extension of $\text{GF}(p^k)$ of degree $d$.


1

As has been pointed out, this exercise is as easy as you thought. In particular, your argument is complete. The only mistake is when you said "zero divides everything"; I believe you meant everything (other than the zero) divides zero.


0

Hint $\ $ Complete the square $\ 4(x^2\!+\!x\!+\!1) = (2x\!+\!1)^2\!+\!3 = X^2\!+\!3,\ $ for $\,X = 2x\!+\!1$


1

As per Jyrki Lahtonen's hint, I will instead try $\phi(x^2+x+1) = c(x^2+3)$. So then since $x \mapsto ax+b$, we have $a^2x^2 + (2ab+a)x + b^2+b+1 = cx^2 + 3c$, so $a^2 = c$, $2ab+a = 0$, and $b^2+b+1 = 3c$. Since $a^2 = c$ and the only squares mod 5 are 1 and 4, we have that $c = 1,4$. Having already gleaned that 1 will not suffice, let us try $c = 4$. Then ...


1

HINT: You want $(\mathbb{Z}/5)[x]/(x^2+x+1) \to (\mathbb{Z}/5)[x]/(x^2+3)$, $x \mapsto a x + b$, so $x^2+x+1 \mapsto (ax+b)^2+ (ax + b) +1$, and you want the latter a multiple of $x^2+3$. Divide $(ax+b)^2+ (ax + b) +1$ by $x^2+3$ and you get the remainder $$a\,(2 b+1)\, x + (b^2 + b+1 - 3 a^2)$$


4

No. Take $g(x)=x^4$, and consider the polynomial $x^4+q$, where $q$ is a rational number. If $q$ is negative, then the roots of $x^4+q$ in $\mathbb{C}$ are $\sqrt[4]{-q}, i\sqrt[4]{-q}, -\sqrt[4]{-q}$ and $-i\sqrt[4]{-q}$, where $\sqrt[4]{-q}$ is the positive real fourth root of $-q$. Then the decomposition field of $x^4+q$ is $$ \mathbb{Q}( \sqrt[4]{-q}, ...


1

You are right $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$ are not isomorphic. If $\varphi$ was an isomorphism between $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$ you would have $\varphi(\sqrt 2) = a + b \sqrt 3$ with $a,b \in \mathbb Q$ and therefore $(a + b \sqrt 3)^2 - 2=0$ which cannot happen as $\sqrt 2$ and $\sqrt 3$ are irrational. The $M_i$ are ...


0

I think what you needed as a hypothesis was that the extension is Galois. By this, I mean that the polynomial $f(x)$ splits into distinct linear factors in an extension, so I am just assuming the negation of the problem about having repeated roots that other folks were suggesting. In this case, the transitive action on the roots of $f(x)$ should give ...


0

Both (1) and (3) look sound. As for (2), consider that if $E = \{ e_{k} : k \in \mathbb{N} \}$ is a basis for $\mathbb{R}$, then we can write every real number $\mathbb{R}$ as a finite sum $\sum_{i = 1}^{N} e_{k_{i}} q_{1}$, where $q \in \mathbb{Q}$. Let $S_{N} = \{ \sum_{i = 1}^{N} e_{k_{i}} q_{i} : q_{i} \in \mathbb{Q} \}$. Then if we consider the map ...


0

(1) If for example $\;K=F(\alpha)\;,\;\;\alpha\;$ algebraic over $\;F\;$ , then $\;[K:F]=\deg(\alpha)\;$ , and $\;K\;$ is then clearly finite . But as any algebraically closed field is infinite, if $\;K=\overline F:=\;$ the algebraic closure of $\;F\;$ , then $\;K\;$ is (countable) infinite. 2) If $\;F=\Bbb R\;,\;\;K=\Bbb C\;$ , then both fields are ...


0

First of all, the first question: If $|F|$ is finite, what about $|K|$? Well, I will show that if $K$ is the algebraic closure of $F$, then $|K|$ have to be infinite. Suppose $|F|=p^k$, then an algebraic simple extension $F(\alpha)$ has degree $n$ on $F$, where $n$ is the degree of the minimal polynomial of $\alpha$, so $|F(\alpha)|=p^{nk}$. There ...


1

To do 2, you can use an argument similar to your first one. Do you know how big $\bigoplus_{n \in \Bbb N} \mathbb{Q}$ is? If $[\Bbb R : \Bbb Q]$ were countable, this is what $R$ would look like as a vector space over $\Bbb Q$. Edit: your answers to 1 and 3 look good to me.


0

$\mathbb Q(a,b)$ has degree at most $6$ and must be a multiple of both $2$ and $3$, and so is $6$. $a+b \in\mathbb Q(a,b)$ and so has degree a divisor of $6$. If $a+b$ has degree $2$, then $b=(a+b)-a$ is in an extension of even degree at most $4$ and $b$ cannot have degree $3$. One is left with the case when $a+b$ has degree $3$. More on this later.


2

Yes, the degrees of sums in separable field extensions multiply, if they are relatively prime, see here. Since $gcd(2,3)=1$ we obtain, that the degree of $a+b$ is equal to $2\cdot 3=6$, and $\mathbb{Q}(a,b)=\mathbb{Q}(a+b)$.


2

The argument of orangeskid can be generalized to all cases where the base field is infinite. Unfortunately, for the general case my impression is that no such basic proof is known. Typically, the statement is given as a corollary of the rational canonical form, which is a normal form for square matrices under the equivalence notion of similarity. In ...


1

If you have an order in your field, you will want your order to be compatible with the field operations. This basically means that the sum and product of positives should be positive; and it has many consequences. One of them is that if $a <b $, then $b-a>0$. So if $2i <3i $, $$ 0 <3i-2i=i. $$ Now $i $ is positive, and by squaring we get ...


1

Let $\lbrace P_1,\dotsc,P_r\rbrace$ be the places in $S$ where $v_{P_i}(x^{-1})>0$ (there are only finitely many.) Then the inequalities you wrote are satisfied by any $z$ with $$v_{P_0}(z)=\max\lbrace 0,v_{P_0}(x^{-1})\rbrace$$ and $$v_{P_i}(z)=v_{P_i}(x^{-1}),\,\,i=1,\dotsc,r$$ and $$v_P(z)\geq 0\,\,\textrm{if}\,\,P\in S\backslash\lbrace ...


1

Let's write $K$ for the field "generated by $1$" and $L$ for the intersection of all subfields of $F$. It should be clear that $L\subseteq K$. This is because $K$ is a subfield of $F$, and is therefore one of the fields in the intersection that created $L$. What does it mean to be "generated by $1$", and why is $K\subseteq L$? Saying that $K$ is generated ...


0

I can show you an example of an element of $Q[[x]]$ that is not a fraction of two elements of $F[[x]]$, but I’m afraid it’s ridiculously advanced. I’m sure that others will give better examples. I’m going to take $F=\Bbb Z_p$ and $Q=\Bbb Q_p$, the $p$-adic integers and numbers respectively. A basic tool for handling $\Bbb Z_p[[x]]$, or indeed power series ...


3

$\mathbb{Q}(i)\otimes_{\mathbb Q}\mathbb{Q}(\sqrt{2})\simeq\mathbb Q(i)\otimes_{\mathbb Q} \mathbb Q[X]/(X^2-2)\simeq\mathbb Q(i)[X]/(X^2-2)$ which is a field since $X^2-2$ is irreducible over $\mathbb Q(i)$. (This field is in fact $\mathbb Q(i,\sqrt 2)$.) However, if you consider $\mathbb{Q}(i)\otimes_{\mathbb Q}\mathbb{Q}(i)\simeq\mathbb ...


3

Recall that the tensor product is bilinear (over $\mathbb{Q}$ in your example). Therefore, $$(a + b\sqrt{2})\otimes(c + di) = ac(1\otimes 1) + ad(1\otimes i) + bc(\sqrt{2}\otimes 1) + bd(\sqrt{2}\otimes i).$$ Note that $\{1\otimes 1, 1\otimes i, \sqrt{2}\otimes 1, \sqrt{2}\otimes i\}$ is a basis for $\mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(i)$ ...


2

No, there isn't. Here's a case where this is particularly clear: if $F$ is a normal extension of its prime field, then every field automorphism of $E$ will send $F$ to itself. For example, consider $E=\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $F=\mathbb{Q}(\sqrt{2})$. Then $\mathrm{Aut}(E|F)$, according to your definition, consists of all four automorphisms of ...


3

Brute-Force Method: Let $\alpha:=\sqrt{5}+\sqrt[3]{2}$. Then, $\alpha^3-3\sqrt{5}\alpha^2+15\alpha-5\sqrt{5}=(\alpha-\sqrt{5})^3=2$, whence $\left(\alpha^3+15\alpha-2\right)^3=5\left(3\alpha^2+5\right)^2$, or $\alpha$ is a root of the polynomial $f(x):=x^6-15x^4-4x^3+75x^2-60x-121 \in \mathbb{Q}[x]$. If this polynomial is reducible, then consider it ...


2

Let's show that $s=\sum \sqrt[n_i]{d_i}$ ($d_i >0$ rationals ) generates $\mathbb{Q}(\sqrt[n_i]{d_i})$. Consider all in larger Galois extension $K \supset \mathbb{Q}(\sqrt[n_i]{d_i})\supset \mathbb{Q}$. Now, to show that $s$ generates all the $\sqrt[n_i]{d_i}$ it's enough to show that whenever a Galois transformation $\phi$ of $K$ preserves $s$ it must ...


3

Since you know that $[\mathbb Q[\sqrt[3]{2},\sqrt 5]:\mathbb Q] = 6$, you know that each of the six values $$1,\sqrt{5},\\\sqrt[3]{2},\sqrt[3]{2}\sqrt{5},\\\sqrt[3]{4},\sqrt[3]{4}\sqrt 5\tag{1}$$ are linearly independent over $\mathbb Q$. Now $$(\sqrt[3]2+\sqrt 5)^2=5\cdot 1 + 2\cdot \sqrt[3]2\sqrt 5 + 1\cdot\sqrt[3]4$$ Is it possible for ...


2

Clearly $LHS\subseteq RHS$. Now it suffices to write down the minimal polynomial of $\sqrt{5}+\sqrt[3]{2}$ and note it has degree $6$.


0

If you make use of the following characteristic of a field of fractions, the proof is constructive: The field of fractions $Q$ of integral domain $F$ is such that: $Q$ is a field. If any other field $K$ contains $F$, then there exists a unique injective ring homomorphism $\phi: Q \rightarrow K$. Now, we note that $Q(x)$ is a field by ...


1

My comment essentially says it all, so let me use the answer space to give a more philosophical answer. Fields are wonderful little worlds. There are a ton of axioms that go into their definition, but that is compensated by the tons of structure that lie within them. In thinking about a field $K(\beta)$, these structures give rise to several ways of ...


3

This question was previously asked and answered on Math Overflow: Quick proof of the fact that the ring of integers of $\mathbb{Q}[\zeta_n]$ is $\mathbb{Z}[\zeta_n]$? Alternatively, a proof can be found here.


0

You should have seen that there is an equivalent characterization of $k(S)$ where $k$ is a field, and $S$ is a subset (or element) of a larger field $\Omega$. The definition you’ve apparently seen is “the set of all rational expressions in $S$, with coefficients in $k$”. But you can also define $k(S)$ as the intersection of all subfields of $\Omega$ that ...


4

The easiest way here is to show that $\sqrt 2 = (\sqrt 2+2)-2 \in \mathbb Q(\sqrt 2+2)$ and $\sqrt 2+2 \in \mathbb Q(\sqrt 2)$ - which is trivial. From the first we get $\mathbb Q(\sqrt 2) \subseteq \mathbb Q(2+\sqrt 2)$ and from the second we get the reverse inclusion. The two fields are therefore equal and therefore isomorphic.


2

Hint : You can write $S_1=\{c+d(\sqrt{2}+2): c,d \in \mathbb{Q}\}$ as $S_2=\{(c+2d)+d(\sqrt{2}): c,d \in \mathbb{Q}\}$. Logically you don't need to find any isomorphism. All you need to know is that as sets $S_1=S_2$. Then since they are both subsets of $\mathbb{R}$ they obviously have the same inherent field structure.


3

What does the Orbit-Stabilizer theorem tell us? If $\Gamma = \textrm{Gal}(L/\mathbb{Q})$, then because the extension is finite, $$ |\Gamma | = |\Gamma \cdot \alpha| \cdot | \textrm{Stab}_{\Gamma}(\alpha)|. $$ In particular, the number $|\Gamma \cdot \alpha|$ of Galois conjugates of $\alpha$ is less than or equal to the order $|\Gamma|$ of the Galois group. ...


1

Closure$$f(x)=\sum_{j=1}^na_jx^j,\qquad a_1=1,a_r=0\text{ if }r\notin\{1,2,..,n\}\\ \text{coeff of } x^r\text{ in }f(x)^j={1\over r!}{d^r\over dx^r}{f(x)^j}\bigm{|}_{x=0}=\sum_{n_1+\cdots+n_j=r}\prod_{k=1}^j{f^{(n_k)}(x)\over n_k!}\bigm{|}_{x=0}=\sum_{n_1+\cdots+n_j=r}\prod_{k=1}^j{a_{n_k}}\\ \text{coeff of } x\text{ in }f(x)^j=\delta_{1j}a_1=\delta_{1j}\\ ...


3

Yes. One may note that since associativity, closure, and identity are obvious, it would suffice to prove that, for every element $G$, all we need is that there is some $k$ such that $G^k$ is the identity. (In fact this is equivalent to it being a group, as the set is finite) This proves to be reasonably easy, however. Firstly, consider any polynomial of the ...


2

Note that $f(x)=x^3+x$ is not a counterexample, because it's reducible. If $\alpha\in L$ is a root of the irreducible polynomial $f(x)\in K[x]$, then $K[\alpha]$ has dimension $\deg(f)$ over $K$. By the dimension theorem, $$ [L:K]=[L:K[\alpha]]\,[K[\alpha]:K] $$ so $\deg(f)$ divides $[L:K]$.



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