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0

For addition, it's obvious. As for multiplication, think of the four-entry polynomial $a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$ as the two-entry polynomial $A+B\sqrt{2}$ with $A,B\in\mathbb{Q}(\sqrt{3})$. Thus $\theta_2(A+B\sqrt{2})=A-B\sqrt{2}$ and that should make things easier.


5

Suppose $S$ is a field, $\mathbb R\subsetneq S\subseteq\mathbb C$. Choose $z\in S\setminus\mathbb R$. Then $z=a+bi$ where $a,b\in\mathbb R,\,b\ne0$, and $i=\frac{z-a}b\in S$, whence $S=\mathbb C$.


8

If $S$ is a field such that $\mathbb R\subseteq S\subseteq \mathbb C$, then $S$ is a vector space over $\mathbb R$ which is a subspace of a vector space of dimension $2$, $\mathbb C$. This means that $S=\mathbb R$ or $S=\mathbb C$.


0

The only algebraic extensions of $L$ are $L$ itself (which is a real closed field) and $ L(i) = \overline{\mathbf{Q}}$ (which is an algebraically closed field).


2

If you have some previous notion of orientation for the plane -- some notion of "clockwise" and "counter-clockwise" -- then you can specify which solution of $z^2+1=0$ is which. And vice-versa: given a choice of $i$ for $\mathbb C$, you get a corresponding orientation for the plane $\mathbb R^2$ using the usual bijection.


-1

The extension $\mathbb{Q}(\pi)$ over $\mathbb{Q}$ is a counterexample to your statement. When such $\alpha_1,...,\alpha_n$ exist, the extension is said to be finitely generated. A very known result is the following: A field extension $E|F$ is finite if and only if it is finitely generated and algebraic. Maybe this is what you meant to ask?


0

Hint: The matrix of the transformation with respect to the basis $(1,0,1,0)^T,(1,1,0,0)^T,(0,1,0,1)^T,(1,1,0,1)^T$ is $$ \pmatrix{ 0&1&0&0\\ 1&0&0&0\\ 0&0&0&1\\ 0&0&1&0 } $$


0

Proposition: Suppose $L/K$ is a field extension and $L$ is algebraically closed. Let $E = \{a\in L\mid a \text{ is algebraic over K} \}$. Then $E$ is an algebraic closure of $K$ in $L$. Proof: We have to show that $E$ is a field and that $E/K$ is algebraic. It is hence enough to show that $E$ is algebraically closed. Suppose $f(x) \in E[x]$ is a nonconstant ...


0

You are given values of $f$ on two vectors; the fact that $f\circ f = id_V$ should allow you to get values of $f$ on two additional vectors. Now you know what $f$ does to four vectors in a four-dimensional vector space; that is (typically) enough information to specify $f$ completely. In particular, if you can write a vector $\vec x$ as a linear combination ...


1

Given any $\delta \in Gal(L/K)$, notice that $h(\delta)$ is exactly the restriction of $\delta$ to the set $M$. The reason that $\delta \rvert_M$ maps $M$ to itself is that if we apply $\delta$ to both sides of the equation $f(l_i) = 0$, we see that $\delta$ will fix all the coefficients of $f$, as well as fixing the right hand side $0$, and the resulting ...


0

The author probably means "an element of an (extension field of degree $n$)" and not "an element of an (extension field) of degree $n$". Thus, $\alpha$ is not required to generate the extension field $L$ of $F$ and $[L:F]=n$. At least $F(\alpha)$ is contained in $L$, so that $\mathrm{deg}(\alpha)=[F(\alpha):F]$ divides $[L:F]$.


0

I am pretty sure Galois includes separability as a criteria. I will assume you mean by Galois "normal and separable". This is an equivalent condition to your question: If $E/K$ is not galois, then there is some element $a \in E, a \notin K$ such that one of its conjugate roots over $K$ is not in $E$. Then let $x_i = g_i(a)$ where $g_i()$ are the ...


2

Note that $Gal(E|\mathbb{Q}(a))$ is a subgroup of $Gal(E|\mathbb{Q})$. As $Gal(E|\mathbb{Q})$ is Abelian, $Gal(E|\mathbb{Q}(a))$ is a normal subgroup. So, by the fundamental theorem of Galois theory, $\mathbb{Q}(a)/\mathbb{Q}$ is a normal extension. Thus, all roots of the minimal polynomial of $a$, which happens to be $f(x)$ lie in $\mathbb{Q}(a)$. Thus, the ...


1

As was already said, solution is the limit of infinite sequence of field extensions of rational numbers with square roots of all primes. But I’d point out there is more explicit description. The smallest ring (and ${\mathbb Q}$-algebra) which contains all square roots of positive rational numbers is, obviously, an infinite(countable)-dimensional vector ...


1

Extended hint: The field $\Bbb{Q}(\omega_7+\omega_7^5)$ is contained in the 7th cyclotomic field $\Bbb{Q}(\omega_7)$. That field is an abelian extension of $\Bbb{Q}$, so all intermediate fields are themselves Galois extensions of $\Bbb{Q}$. This follows from Galois correspondence as all subgroups of an abelian group are normal. To make much progress here ...


2

Let $K=\mathbb{Q}(\frac{\pi^3}{1+\pi})$. Clearly $\pi$ satisfies the relation $$\frac{x^3}{x+1}-\frac{\pi^3}{\pi+1}=0$$ Therefore $\pi$ is a root of the following irreducible polynomial of degree $3$: $$x^3+\left(\frac{\pi^3}{1+\pi}\right)x+\left(\frac{\pi^3}{1+\pi}\right)\in K[x]$$ And to answer the question in the title of your post, yes, $\mathbb{Q}(\pi)$ ...


1

If you've studied the cyclotomic polynomial the answer becomes quite simple! Since $\omega_7=e^{2\pi i/7}$ is a root of the seventh cyclotomic polynomial $\Phi_7$, and $\Phi_7$ can be shown to be irreducible, the degree of the extension is $\deg\Phi_7 = \varphi(7) = 6$, because $7$ is prime ($\varphi$ denotes the Euler totient function). In this proof we ...


3

Hint: We have $\Bbb Q \leq \Bbb Q(\omega_7 + \omega_7^5) \leq \Bbb Q(\omega_7)$.


1

not sure how much help you want, but you are probably just over thinking this a little, or have slightly misunderstood the question (unless I have!). I hope the following will help. In particular, try and stick with working directly with the fields. You have by definition that $L(\mathcal{M})$ is the smallest subfield of $M$ containing the elements of $L$ ...


4

No, it is not true. Simplest counterexample? It’s $p=2$, $n=3$, where the cube roots of unity are quadratic over the base, just as they are over $\mathbb Q$. So the extension field is $\mathbb F_4((T))$,


5

First we prove that the ideal $P = (p,a+\delta)$ has $\{p,a+\delta\}$ as a lattice basis. We just need to check that $\delta p$ and $\delta (a+\delta) = a\delta + d$ are integer linear combinations of $p,a+\delta$: indeed, $$\delta p = p(a+\delta) - a(p),$$and $$a\delta + d = a(a+\delta) - \frac{a^2-d}{p}(p).$$ Note that $\overline{\delta} = -\delta$. Then ...


2

Finding an irreducible polynomial (try random polynomials, and check whether they are irreducible) actually is not hard. (See How many irreducible polynomials of degree $n$ exist over $\mathbb{F}_p$? for a count.) If your $k$ is so large that testing polynomials for efficiency becomes an issue, most likely field arithmetic itself would be an issue as well. ...


1

Fix an algebraic closure $\overline{\mathbb{F}}_3$ and take $F = \varinjlim_{n = 3^k} \mathbb{F}_{3^n}\subseteq\overline{\mathbb{F}}_3$ (i.e., $F = \bigcup_{n = 3^k}\mathbb{F}_{3^n}$). Now, the polynomial $x^2 + 1$ is irreducible over $\mathbb{F}_3$ (it's quadratic, so just test to see that it has no roots). Let $i\in\mathbb{F}_9 = \mathbb{F}[x]/(x^2+1)$ be ...


1

Sure. Given any "supernatural number" $\pi=\prod p^{e(p)}$ (i.e. a formal product of primes, with no restriction on how many exponents can be nonzero, and with the value $e(p)=\infty$ available) one can speak of the union (within the algebraic closure $\overline{F}$) of all finite fields of order $p^n$ with $n$ dividing $\pi$. If not all of the exponents are ...


1

You can prove: If every extension of F splits, then F is perfect. Let F ⊆ K ⊆ E a tower of fields. If E/F is algebraic and splits, then K/F splits. And finally use the definition of algebraically closed.


1

All normal extensions are splitting fields and vice versa. In your particular case you have two algebraic numbers, that is, $i$ and $j\sqrt[3]{2}$. The spliting field of the first is $\mathbb Q(i)$, while of the second is $\mathbb Q(j, \sqrt[3]{2})$. Now take their compositum and this is the normal closure.


2

Yes, there is a best perfect extension field $K^{p^{-\infty}}$ of any given field $K$ of characteristic $p$, called the perfect closure of $K$ (more about the article "the" below). The field $K^{p^{-\infty}}$ consists of all elements $l\in \bar K$ in an algebraic closure $\bar K$ of $K$ with the property that $l^{p^r}\in K $ for some $r\geq 0$. It is ...


1

Observe that $\alpha^2-(\alpha+\beta)\alpha + \alpha \beta = 0$. Hence, $\alpha$ is algebraic over $\mathbb{Q}(\alpha+\beta,\alpha\beta)$. But $\mathbb{Q}(\alpha+\beta,\alpha\beta)$ is algebraic over $\mathbb{Q}$. Hence, $\alpha$ is algebraic over $\mathbb{Q}$.


0

The number of elements in $V$ is $p^n$. An one-dimensional subspace has $p$ elements. The intersection of two different one-dimensional subspaces has only one element ($0$), and every non-zero element generates an one-dimensional subspace. Then, if $k$ is the number of such subspaces: $$p^n=kp-(k-1)=1+k(p-1)$$ Therefore, ...


2

Hints: == Every $\;1$-dimensional subspace of $\;V\;$ is of the form ${}$ Span$\,\{v\}\;,\;\;0\neq v\in V\;$ == There are exactly $\;p-1\;$ vectors in $\;U:=$Span $\,\{v\}\;$ for which their span equals $\;U\;$ itself.


3

Recall that for some field $J$ so that $L \subset J \subset M$ you have that the degree of the extension $L \subset M$ is the product of the degrees of the extensions $L \subset J $ and $ J \subset M$. Use this for example with $J=K(X^p,Y)$, applying the result you know twice.


3

Using the name $x$ for both the indeterminate of the polynomial and the indeterminate of the field extension is just asking for trouble. So let's factor $y^2+2y+2$ in $\Bbb F_3[x]/(x^2+1)$. $y^2+2y+2 = (y+1)^2+1 = (y+1)^2 - x^2 = (y+1-x)(y+1+x)$


2

When constructing an automorphism of $\mathbb{Q}(\zeta)$, the question is where $\zeta$ goes to. But because roots from $x^{15}-1$ should be interchanged, there are only $15$ possibilities. These are the possibilities: $$f(\zeta)\in\{\zeta,\zeta^2,\zeta^4,\zeta^7,\zeta^8,\zeta^{11},\zeta^{13},\zeta^{14}\}$$ There are no other possibilities, because if for ...


0

Hints: For all prime $\;p\;$ and $\;n\in\Bbb N\;$ : $$\begin{align}&\bullet\;\; \Bbb F_{p^n}=\left\{\;\alpha\in\overline{\Bbb F_p}\;:\;\;\alpha^{p^n}-\alpha=0\;\right\}\\{}\\ &\bullet\;\; \Bbb F_{p^n}\le\Bbb F_{p^m}\iff n\mid m\;\;\end{align}$$


1

Often, when dealing with many field extensions, it is useful to draw a diagram depicting all of the fields that are/could be relevant and how they relate. The case of finite extensions of finite fields is rather simple, however, since there is exactly one isomorphism class of field extensions of each degree (where the degree ranges over the positive ...


2

Since $a$ satisfies a quadratic polynomial with coefficients in $K(a)$ we have $[K(a):K(a^2)]=1$ or $2$. In the former case there's nothing to prove. In the latter case we get $$ [K(a):K]=[K(a):K(a^2)][K(a^2):K]=2[K(a^2):K]. $$ But this is impossible because $[K(a):K]$ is odd.


3

Great questions! The first thing I'll note is that Eisenstein's criterion really doesn't apply here. Eisenstein allows you to leverage data about the prime ideals the coefficients of your polynomial live in to conclude whether or not the polynomial must be irreducible. Since $\mathbb{F}_{3}$ is a field, all the ideals are trivial, and so you can't really ...


0

A ring is a subring of a field iff it is a domain. Your ring is not a domain: $2\cdot3=0$.


5

If $a=-1$ take $a_1=1$, $a_2=-1$. Otherwise $a+1\neq 0$, so you can take $a_1=\frac{a}{a+1}$, $a_2=\frac{1}{a+1}$, $a_3=-1$, $a_4=a+1$, $a_5=-(a+1)$. We have $\sum a_i=0$ while $\prod a_i=a$.


1

Being isomorphic as fields means that you have two different fields whose sum and product tables are essentially the same (you only have different "names"). You can see that every isomorphism of your fields must fix $\mathbb{Q}$, so 3 has to appear in both of the multiplication tables as a square (which is clearly not possible). On the other hand, when ...


0

For every nonconstant monic polynomial $f(x)$, let $x_f$ denote an indeterminate and consider the polynomial ring $F[\cdots,x_f,\cdots]$ generated over $F$ by the variables $x_f$. This is a polynomial ring in infinitely many variables - it has one variable $x_f$ for each nonconstant monic polynomial $f \in F[x]$. And the proof shows that ...


1

Hints: it must be that $$\;\forall\,x\in K\;,\;\;f(x)=f(x\cdot1)=xf(1)$$ So what do you think $\;a\;$ would (must) be?


0

Try Local Fields by J. W. S. Cassels.


2

Fields have arithmetic properties which have to be preserved under isomorphism. So for example, if $\mathbb Q(\sqrt{3})$ and $\mathbb Q(\sqrt{2})$ are isomorphic, then since $3$ is a square in the first field it has to be a square also in the second, and you can easily check that this is not possible. On the other hand the $\mathbb Q$-vector space structure ...


2

Once an ideal in $\Bbb{Z}[x]$ contains both $2$ and $3$, it contains $\operatorname{gcd}(2, 3) = 1$. Then the ideal is all of $\Bbb{Z}[x]$. It's not that you can't have such an ideal; it's just redundant. You can describe it with fewer generators (and you should).


0

This three books in my opinion are excellent to start: -"A Brief Guide to Algebraic Number Theory" by H. P. F. Swinnerton-Dyer -"Algebraic number theory" by Mollin R.A. -"Problems in algebraic number theory" Ram Murty M., Esmonde J.


0

While this book might be a bit advanced for the reader's purpose, Eisenbud's commutative algebra is excelent, although it may not be the best introduction. If you do look it up, you'll notice that it is 'with a view towards algebraic geometry,' which is very important in algebraic number theory.


1

First, note that since $1 \in F$ is contained in every field, you have $F(a^4+a^2+1) = F(a^4+a^2)$. In order to show this degree is finite you only need to prove that $a$ is algebraic over $F(a^4+a^2+1)$ right? But this is clear because $a$ is the root of the polynomial \begin{equation*} X^4 + X^2 - (a^4+a^2) \in F(a^4+a^2)[X] \end{equation*} Note: If you ...


3

Let $p(x) \in F[x]$ be such that $p(ab) = 0$. Let $$p(x) = \sum_{i=0}^{n}\alpha_{i}x^{i}$$ where $\alpha_{i} \in F$. Now consider $$g(x) = \sum_{i=0}^{n}\alpha_{i}a^{i}x^{i} \in F(a)[x]$$ What is $g(b)$?


2

For the first one: $e^{\frac{2 \pi i}{3}} = \cos(\frac{2 \pi}{3}) + i \sin(\frac{2 \pi}{3}) = - \frac{1}{2} + \frac{1}{2} \sqrt{-3}$, so $\mathbb{Q}(e^{2 \pi i/3}) = \mathbb{Q}(\sqrt{-3})$. Can you do the second one now?



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