# Tag Info

7

Yes you can ! When you have a cube, you can divide that cube into $n^3$ smaller cubes by cutting it into $n$ equal slices along each side. Hence this operation adds $n^3-1$ cubes. $$5772=1+(17^3-1)+(9^3-1)+(5^3-1)+(2^3-1)$$ So take a cube, and in any order : Choose one cube and cut it into $17^3$ cubes Choose one cube and cut it into $9^3$ cubes Choose ...

5

Note that $\langle \zeta\rangle$ is a finite, cyclic group of order $7$, so every non-trivial element has order $7$. In particular $\zeta^3$ is another primitive $7^{th}$ root of $1$, hence is another roots of the irreducible polynomial $\Phi_7(x)={x^7-1\over x-1}$. But then as this is irreducible by Eisenstein's criterion applied to $\Phi(x+1)$, we get that ...

5

The canonical projection $G \to G/K$ composed with your isomorphism does the job.

4

This problem you are intereted in is call the inverse Galois problem. Every finite group is a Galois group, every profinite group is a Galois group. In general, it is quite hard, for a general group $G$, to say if $G$ is a Galois group or not. I would recommand JP Serre's "Topics in Galois theory" for an authorative reference and state of the art (a bit ...

3

Let $r$ and $s$ two lines in the plane through the origin, and denote with the same letters the reflections through them. Then ${\rm ord}(r)={\rm ord}(s)=2$. But the isomorphism type of the group $G=<r,s>$ depends strongly on the angle $\theta$ between $r$ and $s$. If $\theta=\pi/4$ then $G=\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$. If $\theta\notin\Bbb ... 2 That's totally fine. I might think it is more commonly written as $$\sum_{\substack{ 1 \leq i \leq n \\ 1 \leq j \leq m}} \phi_{v_i}\,\phi_{l_j},$$ or if the ranges of summation are clear, then $$\sum_{i,\,j} \phi_{v_i}\,\phi_{l_j}.$$ 2 Suppose$a\in\operatorname{Rad}(\operatorname{Rad}(I))$. Then$a^m\in\operatorname{Rad}(I)$, for some$m$. Then$(a^m)^n\in I$for some$n$. 2 Normally, one does not define algebraic sets associated to radical ideals only, but rather defines them for any ideal or even for any set, like$V(S)= \{p \ \colon p(s) = 0 \ \forall s \in S\}$. Then, one shows that every algebraic set is in fact already associated to the algebraic set given by some radical ideal. Since$V(S) = V(\langle S \rangle)$and ... 2 One should point out the obviousness of the first statement:$\alpha$is the first map of an injective composition, hence injective.$\gamma$is the second map of a surjective composition, hence surjective. The second statement can be also seen this way without using the snake lemma: If$\alpha$is an isomorphism, consider the diagram $$\require{AMScd} ... 2 If G is a group of prime order, then every nontrivial element generates G. 2 An idea: let \;X:=\{h_i\;:\;\;i\in I\}\; be a generator set of \;H\;, and let \;\{g_1+H,..,g_k+H\}\;,\;\;g_i\in G\; be a free generator set of \; G/H\; == Show that \;x\in\langle g_1,...,g_k\}\cap\langle X\rangle\implies x=0\; == Show that \;\langle\{ g_1,...,g_k\}\cup\{ X\}\rangle=G\; 2 What you want to show is that$$ hA = Ah \iff \forall a \in A,\ h a h^{-1} \in A $$Suppose hA = Ah, and let a \in A. Since hA = Ah, and ha \in hA, we have ha = a'h for some a'\in A. Then$$ ha = a'h \implies hah^{-1} =a', $$evidently in A. On the other hand, suppose h a h^{-1} \in A for all a \in A. Let x \in hA. Then x = ha ... 2 Any permutation representation in V=K^n has an invariant subspace W given by the equation x_1+x_2+\cdots+x_n=0 (the vectors with coordinates summing to~0; here of course n=3). Since \dim W=n-1, any complementary subspace to W is spanned by a since vector v\notin W. In order for that complementary subspace to be invariant, v must be an ... 1 For G a finite group, and F is a field, the group algebra FG is semisimple if and only if J(FG) = \{0\}, where J(FG) is the Jacobson radical off FG, which has several equivalent definitions: one is \{ j \in FG: 1-jx \} is nilpotent for all x \in FG. We note that if j \in Z(FG) is nilpotent and non-zero, then jx is nilpotent for all x ... 1 I believe that this hint helps you to establish that \mathbb Z[X] is countable. For a polynomial f(X)=a_nX^n+\ldots a_1X+a_0\in\mathbb Z[X], a_n\neq 0, define number h(f) with:$$h(f)= n+|a_n|+\ldots+|a_1|+|a_0|.$$The function h:\mathbb Z[X]\longrightarrow \mathbb N measures some kind of a complexity of a polynomial. It is fairly obvious that ... 1 With ordinary integers: Theorem of Gauss that x^2 + y^2 + z^2 integrally represents all positive integers except those$$ 4^k (8 n+7). $$Meanwhile, if x^2 + y^2 + z^2 \equiv 0 \pmod 4, then x,y,z are all even. Put these together, we say that x^2 + y^2 + z^2 is anisotropic in \mathbb Q_2. It is also anisotropic in the real numbers, as the sum ... 1 Ther are lots of good answer for this questions and they have very good explanations. But I think you can just apply the subgroup test. Note that 1=(a^{-1})a\in H\not= \emptyset Let g_1=xa and g_2=ya be two elements of H. Then$$g_1g_2^{-1}=xaa^{-1}y^{-1}=(xy^{-1}a^{-1})a\in H$$Hence$H\le G.$1 Use free groups are projective. Theres a short exact sequence$0\to H\to G \to \mathbb Z^k\to 0$that splits. 1 Take$Y = \mathbf{F}_{p^{n-2}} = $a field with$p^{n-2}$elements and let$X$be the set of subset of$X$with exactly$k$elements. Make act the group$\mathbf{F}_{p^{n-2}}$on$X$by translation, and show that an orbit is never reduced to a single element. What is then necessarily the cardinal of an orbit ? Conclude. 1 Your definition of the order of an element is missing one important feature. The order of an element$g$is the smallest positive integer$n$such that$g^n = e$. (If no such integer exists, then$g$has infinite order.) So no element can have order$0$. EDIT: In response to your comment, I would instead use the following argument. Let$d = \gcd(n,m)$... 1 If$a_i\in I_i$,$a_i=0+\cdots+a_i+\cdots+0\in I_1+\cdots+I_m$. If$a_i\in I_i$for each$1\le i\le m$, then$a_1\cdot\ldots\cdot a_m\in I_1\cap\cdots\cap I_m$. 1 Note$J$this radical of$I$. If$a\in J$and$b\in A$,$a^n\in I$for some$n>0$so that$(ba)^n = b^n a^n \in I$because$I$is a ideal, so that$ba\in J$. Now, if$a,b\in J$, whose respective powers$a^n$and$b^m$are in$I$, then by commutatativity we have$(a+b)^{n+m} = \sum_{k=0}^{n+m} {n+m \choose k} a^k b^{n+m-k}$by the well-known binomial ... 1 Well, you can note that$(1+i)$is maximal because it has a prime norm (hence it is irreducible) and$\Bbb Z[i]$is a Euclidean ring under the norm (hence PID), so prime ideals are maximal. The image should be a field of some sort. We note that$(1+i)|2$--indeed$(1+i)(1-i)=2$--hence$2\equiv 0\pmod{1+i}$, i.e. the field has characteristic$2$. Indeed, we ... 1 You can also think of this in the following way: Look at the composition$\mathbb Z[i] \to \mathbb Z[i]/(1+i) \cong \mathbb Z[X]/(X^2+1,X+1) = \mathbb Z[X]/(2,X+1) \cong (\mathbb Z/2\mathbb Z[X])/(X+1) \cong \mathbb Z/2\mathbb Z$The first isomorphism is given by$a+bi \mapsto a+bX$and the last isomorphism is given by$X \mapsto -1$. Hence the ... 1 A good way to find the minimal polynomial of an element when knowing the Galois group is to compute all the conjuagtes of the element and compute$ \prod_j (X -a_j)$where$a_j$are the conjugates. The conjugates in the first case are$\zeta + \zeta^{-1}$,$\zeta^2 + \zeta^{-2}$, and$\zeta^{3} + \zeta^{-3}\$. Note the others just repeat, for example, ...

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