# Tag Info

3

The number of such fields is likely finite (up to isomorphy) but it is not know unconditionally. The complete list can be described as: the fields with cardinality a Fermat prime, i.e., a prime of the form $2^k+1$, or $9$. (The former is likely finite yet noone knows.) As you observed correctly the condition for a finite field is that the order of $|F^{\... 0 There is no such field. Take any uncountable field$F$that is not algebraically closed. Let$K$be an algebraic closure of$F$. Then$[K:F] > 1$and hence$K \smallsetminus F$is uncountable. If there are countably many monic irreducible polynomials over$F$of degree more than$1$, then there are countably many elements in$K \smallsetminus F$, ... 3 An algebraic closure must be a splitting field of all polynomials over itself (coefficients from the algebraic closure). So existence and uniqueness of a splitting field of all polynomials over a field$K$does not trivially imply existence and uniqueness of an algebraic closure of$K$. However, it is not hard to prove that an algebraic extension of an ... 2 When you take the splitting field of all irreducible polynomials in a given field$K$then all those polynomials split in the bigger field, but there is no guarantee that all polynomials in the bigger field split since there are many more new polynomials now. The difficulty with proving an algebraic closure exists is precisely this: it is easy to add the ... 1 If you talk about the existence of field contain strictly F as sub field , the answer is yes, as indicated in the comments, with for example$K = F (t)$rational field of polinomial ring$F[t]$. If you talk about finite extension$K$over$F$,then$K$exist if and only if$F$is not algebraically closed. if you talk about a proper sub extension$K$of$...

0

the signification of $L$ is the splitting field of $f$ over $K$ is: in an algebraic closure $\Omega$ of $K$, the field $L$ is the smallest field containing K (as minimum degree of $L$ over $K$) and included in $\Omega$, in which $f$ split completely (that is has all roots in $L$)

0

let $\alpha$ be a root of $f(x)=x^2+1$ in Extended field of $\mathbb{Q}\,$, therefore $$f(\alpha)=f(-\alpha)=\alpha^2+1=0$$ on the other hand $$-\alpha=-(\alpha)$$ as a result $$\mathbb{Q}(\alpha)=\{a+b\alpha|\,a,b\in\mathbb{Q}\}\cong \frac{{{\mathbb{Q}}}[x]}{({{x}^{2}}+1)}=a+bx+\left\langle {{x}^{2}}+1 \right\rangle$$

1

Your $f(x)$ is not a polynomial, which must have a fixed, but arbitrary degree $n\ge 0$. It is a theorem then, that every polynomial of degree $n$ over a field $F$ has at most $n$ zeros. This is no longer true over, say, a skew-field. Consider the polynomial $x^2+1$ over the quaternion algebra, as an example - see here. It has infinitely many roots.

1

The degree must be finite, otherwise $\sum \limits _{n = 0} ^\infty \frac {x^n} {n!}$ would be a polynomial - which clearly it isn't, being $\Bbb e^x$. In fact, a polynomial is just a function from $\Bbb N$ to $F$ with finite support.

1

The reason why we care about real closed fields is that these have very tame geometry. This does include quantifier elimination, which in this case tells us that a projection of a semialgebraic set is semialgebraic. But there are many other nice properties stemming from the elementary equivalence. For example, you can apply more or less any theorem true ...

-1

If $E/F$ is an algebraic extension, $K/F$ is an extension. Prove that $EK/K$ is algebraic. Assume $H=\{a\in EK|a \text{ is algebraic over } K\}$. It is easy to see that $H$ is a field and $K<H<EK$. $E/F$ is algebraic, so all the elements in $E$ are algebraic over $F$, also over $K$; so $E\subseteq H$; then we have $EK\subseteq H$, which means $EK=H$. $... 0 What about$\Bbb{F}_{11^3}\approx \Bbb{F}_{11}^3$the isomorphism being an$\Bbb{F}_{11}$-algebra isomorphism. The uniqueness up to (field) isomorphism of a finite field of cardinal$p^q$shows that$\forall P$irreducible in$\Bbb{F}_{11}[X]$of degree$3$$$\Bbb{F}_{11}[X]/(P)\approx \Bbb{F}_{11^3}$$ 1$x^3+a$with$a \in \mathbb F_{11}$is always reducible because$x \mapsto x^3$is a bijection in$\mathbb F_{11}^\times$since$\gcd(3,10)=1$.$x^3+x^2+2$is irreducible over$\mathbb F_{11}$. 1 The difference between real closed and algebraically closed fields isn't really rooted in model theory. Recall that a field$K$is algebraically closed if every nontrivial polynomial over$K$has a root in$K$. Note that if$K$is actually an ordered field,$K$can't be algebraically closed, since$x^2 = -1$can't have a solution (exercise: the square of any ... 3 This is not true:$\Bbb Q(\sqrt 2) = \Bbb Q(1+\sqrt 2)$even if the minimal polynomial$x^2-2$of$\sqrt 2$is different from the minimal polynomial$x^2-2x-1$of$1+\sqrt 2$. The true result is: if$a$and$b$have the same minimal polynomial$f$over$\Bbb Q$, then$\Bbb Q(a) \cong \Bbb Q(b)$, because they are both isomorphic to$\Bbb Q[X]/(f)$. You ... 0 Consider the subspace of$\mathbb{R}^n$consisting of vectors whose coordinates are not all distinct. This is a union of positive codimension subspaces. For example, the subspace of vectors with first two coordinates equal but otherwise pairwise distinct has codimension$1$, just like the line$y=x$in the plane$\mathbb{R}^2$. Since every monic polynomial ... 11 Let we take$x=\exp\left(W(\log 2)\right)$, i.e. a solution of$x^x=2$. Step 1.$x\not\in\mathbb{Q}$. Assuming$x=\frac{p}{q}$with$\gcd(p,q)=1$, we have$p^p=2^q\cdot q^p$, absurd. Step 2.$x$is not an algebraic number. Assuming that$x$is in algebraic number, the Gelfond-Schneider theorem gives that$2$is a transcendental number. It is not, so:$\...

1

Here is a very elementary solution which shows the power given to us by all those nice results the other solutions use. Any morphism of unitary rings $$f\colon \mathbf F_3[x]/(x^2+1)\rightarrow \mathbf F_3[x]/(x^3+2x^2+1)$$ is dominated by a morphism $$g\colon \mathbf F_3[x]\rightarrow \mathbf F_3[x]$$ in the following sense: the diagram $$\begin{... 4 If \Bbb F_9 \subset \Bbb F_{27}, then the groups of invertible elements should be in the same relation, i.e. \Bbb F_9 ^* \subset \Bbb F_{27} ^*. But \Bbb F_9 ^* has 8 elements, while \Bbb F_{27} ^* has 26, and 8 \nmid 26, so \Bbb F_9 \not\subset \Bbb F_{27}. In fact, the argument can be extended: if \Bbb F_{p^m} \subset \Bbb F_{p^n}, then ... 3 There are many standard ways of showing the general result. Jef Laga's is a textbook method for proving one direction. Here's something specifice to your construction. Let use denote by \alpha the coset x+\langle f_1(x)\rangle\in\Bbb{F}_3[x]/\langle f_1(x)\rangle. Because \alpha is a zero of f_1, we see that \alpha^2=\alpha^2-f_1(\alpha)=-1. ... 4 Do this argument satisfy your needs? If \mathbb{F}_{9}\subset\mathbb{F}_{27} then \mathbb{F}_{27} would be a vector space over the field \mathbb{F}_{9} (since it is an abelian group with compatible multiplication of \mathbb{F}_9). If \{a_1,\dots,a_n \} is a basis for this vector space, then every element in \mathbb{F}_{27} can be written ... 6 To answer all three questions it suffices to construct a field of any given infinite cardinality, since a field is a ring which is additively a group. To that effect, let A be an infinite set. Then the field \mathbb Q(A) of rational functions over \mathbb Q using the elements of A as indeterminates is a field with the same cardinality as A. 6 Yes, many results of this form follow from the Löwenheim–Skolem theorem, which asserts that if a first-order theory (a particular way to write down axioms something should satisfy; this includes groups, rings, fields, and more) has an infinite model then it has a model of every infinite cardinality. The Löwenheim–Skolem theorem has much weirder ... 4 The claim is false. For p \neq 2, take K = \mathbb{Q}_p, L = \mathbb{Q}_p(\zeta_p), and x = \zeta_p. For what it is worth, the extension L/K is tamely totally ramified. For the exercise, there is a concrete description of tamely totally ramified extensions L/K with degree n: L = K(\sqrt[n]{\pi}) for some uniformizer \pi of K. The ... 0 The idea is that if  f  is irreducible in  K[x]  where  K  is a field, then the ring homomorphism  \phi : K[x] \to K[\alpha]  ( \alpha  is a root of  f ) defined by  \phi(p(x)) = p(\alpha)  has kernel  (f) , so by the first isomorphism theorem we have  K[x]/(f) \cong K[\alpha] . Since  (f)  is maximal, this quotient is actually a field, ... 0 let \alpha is root of f(x)=x^2+1 in Extended field of \mathbb{Z}_3,therefore$$f(\alpha)=f(-\alpha)=\alpha^2+1=0$$on the other hand$$-\alpha=2\alpha$$as a result$$\mathbb{Z}_3(\alpha)=\{a+b\alpha|\,a,b\in\mathbb{Z}_3\}\cong \frac{{{\mathbb{Z}}_{3}}[x]}{({{x}^{2}}+1)}=a+bx+\left\langle {{x}^{2}}+1 \right\rangle$$0 Yes of course. It's a basic theorem in field theory that for each field F and irreducible polynomial f(x) \in F[X], there exists some field extension E of F such that f has a root (say \alpha) in E. Since your polynomial has degree 2, \mathbb{Z}_3(\alpha) is necessarily the splitting field of x^2+1 over \mathbb{Z}_3 1 Yes, there are proper connected subfields of \mathbb{C} other than \mathbb{R}. A construction can be found in a paper by J.Dieudonné: J.Dieudonné, Sur les corps topologiques connexes, C.R. Acad. Sci. Paris vol. 221 (1945) pp. 396-398. I don't have access to the above paper at the moment. It can in fact be proved that the only proper path-connected ... 0 Hint: A subring will be a vector space over the prime field; if it has to be proper, its dimension must be 1, since the dimension of F_{p^2} over the prime field is 2. Thus a proper subring has p elements; how many choices do you have, taking into account that subrings must contain 1? Having a unity is irrelevant: a subring (not supposing it ... 3 Any proper subfield of \;\Bbb Z_p\; would have to be, in particular, a sugbroup of its additive group. But this is a group of order a prime \;p\; , and as such it has no proper divisors, so by Lagrange's Theorem that's impossible. 2 It’s a proper subset, but it’s not a subfield at all: for example, 2^2=1 in \Bbb F_3, and 2^2=4\ne 1 in \Bbb F_5, so they don’t have the same operations. Similarly, 1+2=0 in \Bbb F_3, but 1+2=3\ne 0 in \Bbb F_5. 2 I see no difficulty. Since 1+x^k+x^{2k}=\frac{x^{3k}-1}{x^k-1} (putting k=3^n), the order of x is at most 3k, so it cannot generate the whole multiplicative group, so the polynomial is not primitive. 0 Lemma: Assume m \mid n. The homomorphism$$(\mathbb{Z}/n\mathbb{Z})^{\ast} \rightarrow (\mathbb{Z}/m\mathbb{Z})^{\ast}$$given by t + n\mathbb{Z} \mapsto t + m\mathbb{Z} is surjective. Proof: Write n = p_1^{e_1} \cdots p_j^{e_j}, and m = p_1^{d_1} \cdots p_j^{d_j} with d_i \leq e_i. We have a commutative diagram$$\begin{matrix} \mathbb{Z}/n\...

1

$$\cos 16\alpha=2\cos^2 8\alpha-1=2(\cos^2 4\alpha-1)^2-1=2[(8\cos^4\alpha-8\cos^2\alpha+1)^2-1]^2-1$$ let $\alpha=\frac{\pi}{48}$ and $\cos\frac{\pi}{48}=x$ we have $$\frac{1}{2}=2[(8x^4-8x^2+1)^2-1]^2-1$$

2

Let $\omega=e^{i \frac{\pi}{48}}$. Then your cos is an element of $\mathbb Q[\omega]$. Now $\omega$ is a root of the $96$-th cyclotomic polynomial, which is irreducible and has degree $\phi(96)$. If you are not familiar with cyclotomic polynomials, try to use the fact that there are $\phi(96)$ primitive $96$-th roots of unity to find $[Q(\omega):Q]$.

2

If I am not mistaken, both Ian and ihf are right, depending on what the $x$ is in $F(x)$. If $x$ is an indeterminate, then the smallest field containing both $F$ and $x$ is the field of fractions of the polynomial ring $F[x]$. If $\alpha$ is the root of a polynomial in $F[x]$, as is the case with $\sqrt{2}$ in $\mathbb{Q}[x]$ then $F(\alpha)$ will be the ...

3

Since you mention transcendence degree, $F(x)$ is probably the field of rational functions in one variable, that is, the field of fractions of the polynomial ring $F[x]$.

3

In the context of field extensions (as you mentioned in the question), $F(x)$ is the smallest possible field containing $F$ and $x$. For example, one can construct $\mathbb{Q}(\sqrt{2})$ by adjoining the numbers $a+b\sqrt{2}$ to $\mathbb{Q}$ where $a,b$ are arbitrary rational numbers. Note that this does result in a field, because sums, products, and ...

-1

Not quite. It is $\{a\sqrt{2}+ b\sqrt{3}+ c : a,b,c \in Q\}$. $Q(\sqrt{2}, \sqrt{3})$ is "the smallest field containing all rational numbers as well as $\sqrt{2}$, and $\sqrt{3}$" so must contain, for example, 1.

4

$\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the smallest field containing $\mathbb{Q}$, $\sqrt{2}$ and $\sqrt{3}$. One can say that $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the field generated by $\{\sqrt{2},\sqrt{3}\}$ over $\mathbb{Q}$. It an important fact that one has the following tower of extensions: $$\mathbb{Q}\hookrightarrow\mathbb{Q}(\sqrt{2})\hookrightarrow\... 3 This is false since we might have more than one element of norm 2. For example, consider the field \mathbb{Q}(\alpha), where \alpha is a root of x^3-3. Then a_1=1-\alpha and a_2=1+\alpha+\alpha^2 both have norm 2 but neither divides the other. In general, if we assume O_K is a unique factorisation domain (which it rarely is, but then we ... 1 You are on the right track. By letting P to be that least common multiple of the denominators of the f_is you get a polynomial P(X) such that P\Phi\in K[X][T]. It may initially happen that the coefficients P_0(X),P_1(X),\ldots,P_n(X) have a non-constant common divisor d(X). But the P(X)=P_n(X) is also a multiple of d(X), and we can use \... 2 For fields we have A polynomial equation of degree n has at most n solutions. For an algebraically closed field, such as \mathbb C, we have A polynomial equation of degree n has exactly n solutions, if you count them with multiplicity. Perhaps a bit surprisingly, the commutativity of fields is essential for this: The equation x^2+1=... 1 I do not know if I' m answering your question i think this will help you. Suppose you have ab irreducible polynomial f(x) with coefficients over a field F then you can consider the extension of the field F, F\leq K, where all the roots of the polynomial f(x) exists there. It is easy to prove that such a field exists for every irreducible polynomial ... 1 the strict subfields are \Bbb{Q}(w), \Bbb{Q}(^3\sqrt{2}), \Bbb{Q}(^3\sqrt{2}\;w) and \Bbb{Q}(^3\sqrt{2}\;\bar{w}) 2 I think we could argue as follows: suppose you prove the claim for some finite normal subextension \;L'/K\; of \;L/K\; . Now, take an element \;\sigma\in\text{Aut}\,\left(L'/K\right)\; fulfilling \;\sigma q= q'\; . Since \;L/K\; is algebraic (this is a "hidden" assumption, I think, since "normal extension", as far as I know, is only applied to ... 0 An alternative approach. \mathbb{F}^* is a cyclic group generated by some g\in\mathbb{F}^*: let M=\left|\mathbb{F}^*\right|. All the elements of \mathbb{F}^* are listed here:$$\{g,g^2,g^3,\ldots, g^M\}  and the map $\psi:x\to x^2$ sends the element $g^a$ into $g^{2a\pmod{M}}$: it follows that if $M$ is even $\psi$ is not surjective, contradicting ...

6

Since the order of a finite field is always the power of a prime $p,$ we only need to show that the characteristic of $\mathbb F$ is $2.$ Consider the group homomorphism $f:\Bbb F^*\rightarrow\Bbb F^*$ sending $x$ to $x^2.$ If the characteristic of $\Bbb F$ is $p\ne2,$ then the kernel of $f$ contains $1$ and $-1,$ which are different elements in $\Bbb F^*,$ ...

4

The statement is true: the kernel of $f$ must be a proper ideal of $F_1$ since $f(1)=1$, and the only proper ideal of $F_1$ is $(0)$. The issue with your example is as follows: we can construct $\mathbb{F}_4$ by adjoining a root $\alpha$ of the polynomial $X^2+X+1$ to $\mathbb{F}_2$. Then $\alpha^2=\alpha+1\not\in\mathbb{F}_2$, so your homomorphism doesn't ...

2

To see why there are infinitely many intermediate fields between $K=\Bbb{F}_p(x^p,y^p)$ and $L=\Bbb{F}_p(x,y)$ you can do the following. Let $z$ be any element of $K$. Consider $w=x+zy$. We see that $w^p=x^p+z^py^p\in K$, so $K(w)$ is a degree $p$ extension of $K$. If $w'=x+z'y$ for some $z'\in K$, $z'\neq z$, then we easily see that $K(w,w')=L$. Therefore ...

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