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1

Hint : Since you have shown that the Jordan decomposition exists when $f$ splits in $\mathbb{F}$, the same proof you found should work when $f$ splits over a field $K$. So you have a Jordan decomposition of $M$ on $K$, where $K$ is the splitting field of $f$. Then use the fact that since $\mathbb{F}$ is perfect, you can be sure that $K$ is a Galois extension ...

2

The polynomial $x^{16}-x\in\mathbb{F}_2[x]$ has roots precisely equal to the elements of $\mathbb{F}_{16}$, and the subfields of $\mathbb{F}_{16}$ are $\mathbb{F}_{16}$, $\mathbb{F}_{4}$, and $\mathbb{F}_{2}$, which have degrees $$[\mathbb{F}_{16}:\mathbb{F}_2]=4\qquad [\mathbb{F}_{4}:\mathbb{F}_2]=2\qquad [\mathbb{F}_{2}:\mathbb{F}_2]=1$$ Therefore ...

1

The only place that you use the fact that $F$ is algebraic over $\mathbb{Q}$ is in the primitive element theorem. If for some reason your irreducible polynomial will have only one parameter (instead of three), then your solution will still work. Suppose that your polynomial is $f(x)=x^3+Ax^2+Bx+C$. First consider the polynomial ...

10

In fact, they can be isomorphic. For example $\mathbb{R}(X^2) \subset \mathbb{R}(X)$ is a field extension of degree $2$ but the two fields are isomorphic (as fields) by $X^2 \mapsto X$. This isomorphism (as fields) does however not induced an isomorphism of $F$-vectorspaces between $F$ and $K$, which is why $[K:F]\neq [F:F]$ is not a problem even when ...

2

The conjugates of an algebraic number are (by definition) the roots of its minimal polynomial. The number of (distinct) roots of an irreducible polynomial over the rationals is equal to its degree, that is four. Thus once you know the minimal polynomial "it is clear." There is some wiggling room as one might or might not count the number itself among its ...

6

In general we would guess that $\sqrt a+\sqrt b+\sqrt c$ has eight conjugates, obtainable by toggling signs individually for the surds. However, in this special case we see that $\sqrt a\sqrt b\sqrt c=30$, which cannot change its sign. Hence once we picked the sign of two of the surds, the sign of the third is determined.

4

$$\sqrt{6}+\sqrt{10}+\sqrt{15}=\sqrt{2\cdot 3}+\sqrt{2\cdot 5}+\sqrt{3\cdot 5} = \frac{1}{2}\left(\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2-(2+3+5)\right)$$ where $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is an algebraic number of degree $8$ over $\mathbb{Q}$, having conjugates $\pm\sqrt{2}\pm\sqrt{3}\pm\sqrt{5}$, whose minimal polynomial is an even function. It follows ...

1

No. A counter-example is given by the algebraic closure $\overline{\mathbb{F}}_p$ of a finite field $\mathbb{F}_p$ of characteristic $p>2$. Indeed, since any automorphism $\sigma$ of $\overline{\mathbb{F}}_p$ fixes all elements of the prime field $\mathbb{F}_p$, then this $\sigma$ is an element of the absolute Galois group of $\mathbb{F}_p$, which is ...

1

Yes, $\mathbb{C}_p\cong\mathbb{C}_q\cong\mathbb{C}$. However, the isomorphism is only an isomorphism of fields. The natural topology on the three fields are different, so they are nonisomorphic as topological fields. However, the isomorphism is impossible to fully specify using a finite number of symbols/words. Yes, for $x\in\mathbb{C}_p$, it does make ...

0

Here is a link to a short note establishing that the roots of a polynomial are $C^\infty$ functions of the coefficients using the implicit function theorem.

4

The argument given in this answer on MO shows that if $A$ is any set equipped with a collection of finitary operations which satisfy some equational axioms such that $2\leq|A|\leq\aleph_0$, then there is an infinite-dimensional contractible CW-complex $X$ which can be equipped with corresponding finitary operations which are continuous and which satisfy the ...

4

What makes you think that $$a + b \zeta_{12}^5 + c\zeta_{12}^{10} + d\zeta_{12}^{15}= a + b \zeta_{12} + c \zeta_{12}^2 + d \zeta_{12}^3$$ is true? The powers of $\zeta_{12}$ repeat modulo 12. In fact since the $12$th cyclotomic polynomial is $\Phi_{12}=x^4-x^2+1$, we have that \begin{align*} \zeta_{12}^5&=\zeta_{12}^5+0\\\\ ... 2 Right before that he says Let us keep the hypotheses of prop. 9. If \mathfrak{P} is a non-zero prime ideal of B... Proposition 9 says Proposition 9. If A is Dedekind then B is Dedekind. and in a Dedekind domain, every non-zero prime ideal is maximal. Thus \mathfrak{P} and \mathfrak{p}=\mathfrak{P}\cap A are indeed maximal, and thus ... 4 Here's an approach using the fact that the discriminant is a symmetric polynomials in the roots. First, consider the ring R=\mathbb{Z}[T_1, \ldots, T_n], and look at the "universal" monic polynomial P(X) = \prod_{i=1}^n (X-T_i) \in R[X]. Its discriminant is D=\prod_{1\leq i<j\leq n}(T_i-T_j)^2. This is a symmetric polynomial in the T_i, so it ... 2 (1) Let f(x) be the minimal polynomial of \alpha. Let k=\alpha^p (note k\in K). Since \alpha^p-k=0, we must have f(x)\mid (x^p-k), and so \beta is also a root of x^p-k, thus \alpha^p=k=\beta^p~\Rightarrow (\alpha/\beta)^p=1. (2) No. There's no reason for them to in general. The needn't be the same size even. For instance, over \Bbb Q, ... 0 The proof should be available in any abstract algebra textbook, the one I have at hand right now is Fraleigh (where it is Theorem 48.3). Since you are familiar with Zorn's lemma, we apply it to the set S of pairs (L,\lambda), where L is an intermediate field between k(\alpha) and K (inclusive) and \lambda is an injective morphism of L into K ... 1 Every element in a field of characteristic p has up to 1 pth root. This is because X^p is a ring endomorphism (the Frobenius endomorphism). This implies that the polynomial must factorize into (X - b)^p in some extension field. Therefore if the polynomial can be reduced in F[X] into fg (where f is of the lowest possible degree) then f must ... 5 If X^p-a has no roots in F, then consider the splitting field K of X^p-a. Over K, X^p-a=(X-b)^p for some b \in K. Hence, X^p-a is reducible over F if and only if (X-b)^n\in F[X] for some integer n with 1\leq n <p, in which case b^p \in F and b^n\in F, whence b=b^{\gcd(p,n)}\in F as well, which is a contradiction. Hence, ... 2 Let A=K be a field extension of k, and l=k(x) a purely transcendental extension of k. Then A\otimes_kl\simeq S^{-1}K[x], where S=k[x]\setminus\{0\}. This is clearly a normal integral domain which is not necessarily a field. However, if the extension k\subset l is algebraic, and K\otimes_kl is an integral domain, then it is a field since the ... 1 Apply the more general fact below to G \subseteq (F,+) and to G\setminus0 \subseteq (F\setminus0,\cdot). Let M be a monoid with cancellation. If S is a finite submonoid of M, then S is a subgroup of M. We need to prove that every element a\in S has a two-sided inverse in S. Indeed, take a\in S and consider f: S \to S given by ... 2 A finite (and non-empty) subset S of any group (G,\ast) closed under the group operation (\ast) is a subgroup (G need not be finite). The proof is similar to Zev Chonoles': Let a \in S. By closure, all positive powers of a: a, a\ast a = a^2, a\ast a\ast a = a^3, etc. are all in S. Since S is finite, these all cannot be different. Hence ... 2 Your reasoning is great so far. Here is a hint for both additive and multiplicative inverses: If A is a finite group with operation \star, then for any given a\in A, the inverse of a is a power of a; that is, for some n, we havea^{-1}=a^{n}=\underbrace{a\star a\star\cdots\star a}_{n\text{ times}}$$That's because any element of a finite ... 4 Use the fact that: If G is a finite group and H be a subgroup of G such that a,b\in H\implies ab\in H then H is a subgroup of G .(Why?) Only thing that remains to show is that a\in H\implies a^{-1}\in H this is true since the group is finite for any a\exists n\in \mathbb N such that a^n=1\implies a.a^{n-1}=a^{n-1}.a=1 .thus a ... 3 I'll rephrase slightly to try to make things clearer. Let L=\overline{\mathbb{F}_p}(z,w) where z and w are indeterminates. The polynomial \mathbf{X}^p-z\in L[\mathbf{X}] is irreducible over L. Therefore if \alpha\in\overline{L} is a root of it, we have$$[L(\alpha):L]=\deg(\text{minimal polynomial for $\alpha$ over $L$})=p$$Similarly the ... 0 A field is an algebraic structure allowing the four basic operations +, -, \cdot, and :\,, such that the usual rules of algebra hold, e.g., (x+y)\cdot z=(x\cdot z) +(y\cdot z), etcetera, and division by 0 is forbidden. The elements of a given field should be considered as "numbers". The systems {\mathbb Q}, {\mathbb R}, and {\mathbb C} are ... 2 This answer doesn't address directly to the OP's question, but to the exercise in Vakil's textbook. Let A\subset B be a faithfully flat extension of integral domains, and B is integrally closed. Then A is also integrally closed. This follows immediately from A=B\cap K(A). 0 Pick any polynomial P(t) \in \mathbb C[t]. By long division you have$$P(t)=Q(t)(t^2+1)+at+b $$Now, in the factor you have t^2+1=0 thus P(t) becomes "the same" as at+b. Here the same means, is in the same class. Now, identifying each polynomial with its remainder when divided by t^2+1 is picking ONE single polynomial from each equivalence ... 1 There are two ways to see this: 1) The first is to give an isomorphism from$$\frac{\mathbb{R}[t]}{<t^2+1>} \rightarrow \mathbb{C}.$$Define \phi : \frac{\mathbb{R}[t]}{<t^2+1>} \rightarrow \mathbb{C} by \phi(r) = r for r \in \mathbb{R} and \phi(t) = i. We can easily verify that this gives an isomorphism between the two required set. ... 0 The solutions to the equation$$t^2+1=0$$are the complex numbers commonly denoted i and -i. (They are numbers which square to -1.) In the ring$$R[t] / (t^2+1), $$the polynomial t^2+1 is equivalent to 0 (since it is certainly in the coset (t^2+1)R[t]. By my initial comment, another way of thinking about the fact that t^2+1=0 in this ring is ... 1 It is easy to see that this is a 2-dimensional vector space over \mathbb{R}? Now since in the quotient t^2+1\cong 0 then t^2\cong -1. By labeling t as i you are getting the "regular" complex numbers. 2 When taking quotient by (t^2+1), the image of t satisfies t^2 + 1 = 0, or, equivalently, t^2 = -1, so t works as the imaginary unit. Now, any element of the quotient can be uniquely represented by a polinomial with degree \leq 1 (using the fact that applying the identity t^2 = -1 reduces the degree). So, any element can be written as a + tb. ... 2 The complex number a+bi corresponds to the equivalence class of polynomials that contains the polynomial bt+a. For example 2+3i corresponds to the equivalence class$$ \{3t+2, t^2+3t+3, -t^2+3t+1, t^3+4t+2, \pi t^4+\pi t^2+3t+2,\ldots \} $$You should be able to check that addition and multiplication of polynomials correspond to addition and ... 1 Let u = x + \langle f\rangle \in \Bbb Z_p[x]/\langle f\rangle = F. Clearly f(u) = 0, since f(x) \in \langle f\rangle. Since a \mapsto a +\langle f\rangle for a \in \Bbb Z_p is an embedding of \Bbb Z_p in F, we can consider F an extension of \Bbb Z_p. Note that \Bbb Z_p(u) \subseteq F, and since f is irreducible over \Bbb Z_p, we ... 3 Assuming that F \neq \mathbb{F}_3, let x \in F \setminus \mathbb{F}_3. By assumption x is algebraic, so [\mathbb{F}_3(x):\mathbb{F}_3]<+\infty. This means that the unit group \mathbb{F}_3(x)^\times is a finite group. Let n = \vert \mathbb{F}_3(x)^\times \vert be the order of the unit group. Then by Lagrange's theorem x^n = 1, so x is a ... 0 Note that any finite extension F of \mathbb F_3 with [F:\mathbb F_3] = n is isomorphic to the splitting field of the polynomial$$X^{3^n}-X,$$which clearly contains the roots of$$X^{3^{n}-1}-1,$$ie: the 3^{n}-1-th roots of unity. Since any nontrivial algebraic extension of \mathbb F_3 contains such a subfield, we are done. 0 By definition a real-closed field is an ordered field. The required interaction between the arithmetic and the linear order implies character zero. I do not know whether an infinite field F with char(F)> 2 and no square root of (-1) can be algebraically completed by adjoining it to F. 1 You used the assumption of finiteness of the extension when you assumed that exist  \alpha  as you choose. The existence of such  \alpha  is ensured by the Primitive element theorem and not necessary true for infinite extensions. From then on, your solution is absolutely correct. 3 E=K^H, being a subextension of the Galois extension K/F is a finite separable extension of K, hence by the Primitive element theorem, there exists \alpha\in E such that E=K(\alpha). Now, it isresults from the Galois correspondence that:$$H=\{\sigma\in G\mid \sigma(\alpha)=\alpha\}.$$0 You can easily compute a^2+b^2+2ab=7. Thus an invariant under g, is the element a+b=\sqrt{7}. So the fixed field for <g>=Q(\sqrt{7}). You can easily compute a^2+b^2-2ab=-1. Thus an invariant under h=fg, is the element a-b=\sqrt{-1}. So the fixed field for <h>=Q(\sqrt{-1}). Then it follows easily that ... 2 Maybe I'm missing something here, but what about this? Suppose that a \in K_1 is not in the domain of f and let p(X) \in F[X] be the minimal polynomial of a. Then define g(b) = f(b) for every b in the domain of f, while g(a) = \alpha where \alpha is any fixed root of p(X) in K_2 not in the range of f. After extending (pointwise) g ... 3 see https://oeis.org/A003172 and https://en.wikipedia.org/wiki/List_of_number_fields_with_class_number_one#Real_quadratic_fields for some detail. The difference, quite enormous, from imaginary quadratic fields is that those have only a handful of class number one, and it was a huge endeavor to prove the known list complete. ... 1 For an easy approach, there exist sets S\subsetneq \mathbb{R}^2 such that a polynomial f\in \mathbb{R}[x,y] that vanishes at every point in S must be identically zero, for example S=\mathbb{R}^2\setminus\{\overline{s}\} for any point \overline{s}=(s_1,s_2) (because a polynomial function is continuous in the standard topology on \mathbb{R}^2). ... 2 There's nothing wrong, because the splitting field of x^8-1 is the same as the splitting field of x^9-x: we're adding a root that's already in the field. This is known to be the field with nine elements. You can factor the polynomial as$$ x^8-1=(x^4-1)(x^4+1)=(x-1)(x+1)(x^2+1)(x^4+1) $$and note that$$ ...

0

Another easy solution: Since $R$ is finite dimensional over $F$, we have $\{1,r,r^{2},...,r^{n}\}$ is a linear dependent set for some finite $n$ over $F$. In particular, if $r \neq 0$ and $r \in R$, then $a_{n}r^{n}+a_{n-1}r^{n-1}+...+a_{0} =0$ has a nontrivial solution where each $a_{i} \in F$. If $a_{0}=0$ then $a_{n}r^{n}+a_{n-1}r^{n-1}+...+a_{1}r=0 ... 3 Yes, this is true. Given a positive integer$n>1$let$p$be any prime such that$p\equiv 1\pmod n$. In other words we want$p=1+an$for some integer$a>0$. By Dirichlet's theorem on infinity of primes in an arithmetic progression there are infinitely many such primes$p$. Let us fix such a prime$p$. Let$c$be a primitive root modulo$p$. I claim ... 2 The Frobenius automorphism$\varphi:\mathbb{F}_{p^n}\to \mathbb{F}_{p^n}$is a bijection that fixes the prime subfield$\mathbb{F}_p$, so$\varphi(a)\in\mathbb{F}_p$if and only if$a\in\mathbb{F}_p$, and in fact for all$k$, we have$\varphi^k(a)\in\mathbb{F}_p$if and only if$a\in\mathbb{F}_{p}$. Thus, whenever$n$is a power of$p$, we have ... 5 The extension$\Bbb Q(T^{1/4})/\Bbb Q(T)$is not a Galois extension since the minimal polynomial of$T^{1/4}$, which indeed is$X^4-T$, can not be factorized in the extension; it has roots that are not in the extension: $$(X^4-T)=(X^2-T^{1/2})(X^2+T^{1/2})=(X-T^{1/4})(X+T^{1/4})(X^2+T^{1/2})$$ But the last factor has no roots in$\Bbb Q(T^{1/4})$. Note ... 3 Your factorization perfectly works.$\mathbb{Z}[T]$is just a factorial ring, so that you may apply what you know about general factorial rings. Its field of fractions is$\mathbb{Q}(T)$. If$K$is any field, then$i$denotes an element in an algebraic extension of$K$with$i^4=1$and$i^2 \neq 1$. The automorphism group of$\mathbb{Q}(T^{1/4})$over ... 0 All the elements of$k(\alpha)$are separable, because all irreducible polynomials over a finite field are separable. Indeed if$f(x)$is not separable$ f(x),\ f'(x)$have a common root in some extension of$k$, hence$\deg\gcd(f(x),f'(x))\ge 1$, which is impossible if$f$is irreducible, unless$f(x)=g(x^p)$, for some polynomial$g$(in which case ... 2 Getting late here, so a quick sketch (or extended hint) with steps only. Show that the zeros of$f(x)$are the three cube roots of$1+i$and the cube roots of$1-i$. Let$L$be the splitting field of$f(x)$over$\Bbb{Q}$. Given that the ratio$(1+i)/(1-i)$is a primitive fourth root of unity show that$L$contains a twelfth root of unity$\zeta$. If$w\$ ...

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