Let $\mathbb{F}$ be a field, $\lambda \in \mathbb{F}$ and $A,B \in M_n(\mathbb{F})$ such that $m_A(x)=m_B(x)=(x-\lambda)^k$ and such that the geometric multiplicity of $\lambda$ in $A$ equals to the geometric multiplicity of $\lambda$ in $B$. a. Prove that for $n=5$ or $n=6$- $A$ and $B$ are similar b. Find an example for $n=7$ for which $A$ and $B$ are NOT similar    My main question here is whether I should try proving it for $n=5$ and $n=6$ seperately? Or should I try proving it for any $n$ and then observe that it holds only for $n=5$ or $n=6$? Plus, any hints or suggestion regarding my main question or the question above will be welcome
123181
I'm trying with matrices over $\mathbb F_2$ and trying to have a look at the Jordan canonical forms of these matrices. If the size of the biggest Jordan block is the same with 1's in all diagonal entries, we do get non-similar invertible matrices with same minimal and characteristic polynomial. But what do I do for satisfying the last condition on eigenspace? Please provide examples if possible and also please explain why it works.
50208
Let $\mathbb{F}_3$ be the field with three elements. Let $n\geq 1$. How many elements do the following groups have?   $\text{GL}_n(\mathbb{F}_3)$ $\text{SL}_n(\mathbb{F}_3)$   Here GL is the , the group of invertible n×n matrices, and SL is the , the group of n×n matrices with determinant 1.
72311
Let $F$ be a field and $u\in F$. Further, let $F\subset G$ be a finite extension and $v\in G$ such that $v^n=u$ for $n&gt;0$.     How would I show that there exists a $\tilde{u}\in F$ such that $u^{[G:F]}=(\tilde{u})^n$?   My first thought was to use the norm map $N_{G/F}(x)=\Pi_{\sigma\in X(G/F)}\sigma(x)$, where $X(G/F)$ is the fundamental set for $F\subset G$. However, I wouldn't know how to continue on this path.
4220
I'm trying to prove that $(ma, mb) = $|$m$|$(a, b)$ , where $(ma, mb)$ is the greatest common divisor between $ma$ and $mb$. My thoughts: If $(ma, mb) = d$ , then $d$|$ma$ and $d$|$mb$ → $d$|$max + mby$ → $d$|$m(ax+by)$. This implies that $d$|$m$ or $d$|$(ax+by)$. This is the same as  $d$|$m$ or $d$|$a$ and $d$|$b$, so $d$|$m$ or $d$|$(a,b)$. This is the same as $d$|$m|$ or $d|(a,b)$, so $d$|$|m|(a,b)$. I don't know what to do.
64659
I am trying to prove that these two definitions of Carmichael function are equivalent.  I am using this definition of Carmichael function:     $\lambda(n)$ is the smallest integer  such that $k^{\lambda(n)}\equiv1 \pmod n\,$ $\forall  k$ such that $\gcd(k,n)=1$.   to prove this one:     $\lambda(n) = \begin{cases}\phi(p^a) &amp; (p=2\text{ and }a\le 2)\text{ or }(p\ge 3)\\\dfrac 1 2 \phi{(2^a)}  &amp; a\ge 3\\ \text{lcm}\left((\lambda(p_1^{a_1}),\cdots,\lambda(p_k^{a_k})\right)&amp; \text{for }n=\prod_{i=1}^k p_i^{a_i} \end{cases}$   Since $a^{\lambda(n)}\equiv 1 \pmod n$,  and $a^{\lambda(m)}\equiv1 \pmod m$ with $\gcd(m,n)=1 \implies a^{\text{lcm}({\lambda(n)},\lambda(m)}\equiv1 \pmod {nm} $   Hence $\lambda (mn)=\text{lcm}(\lambda(n), \lambda(n))$ for $m$ and $n$ coprime to each other.  How can I prove the other relations?
117564
This is an exercise from Miles Reid, Undergraduate Algebraic Geometry. The proof has two parts, one of which I can do and one of which I can't. I could have some misunderstandings about notations here as well so anything people can point out will be helpful.  a) Let $k$ be an infinite field and $f\in k[X_1,...,X_n]$ be nonconstant. Prove  $$V(f)=\{P\in \mathbb{A}^n_k\mid f(P)=0\}\neq \mathbb{A}^n_k.$$  $A^n_k$ here is the point set $k^n$.   My proof (which follows from a hint of his) is write $f=\sum a_i(X_1,...,X_{n-1})X_n^i$. Then $V(f)$ consists of two kinds of points   $P=(p_1,...,p_{n-1},p_n=0)$ $P=(p_1,...,p_{n-1},p_n\neq 0 )$ with $\sum a_i(p_1,...,p_{n-1})p_{n}^i=0$   In the first case we have $V(f)\neq \mathbb{A}^n_k$. Assume we have $P$ like in the second case and now write $f=\sum b_i(X_1,...,\hat{X}_{n-1},X_n)X_{n-1}^i$. Here again, either $P=(p_1,...,p_{n-1}=0,p_n\neq 0)$ and $V(f)\neq \mathbb{A}^n_k$ or $p_{n-1}\neq 0$, and by induction we should get to $V(f)\neq \mathbb{A}^n_k$.  b) Now let $k$ be algebraically closed. Let $a_m(X_1,...,X_{n-1})X_n^m$ be the leading term of $f$. Show that when $a_m\neq 0$, there is a finite set of points of $V(f)$ corresponding to every value $(X_1,...,X_{n-1})$. Thus, $V(f)$ infinite for $n \geq 2$.  So I have kinda no idea what to do. A given value of $(X_1,...,X_{n-1})$ corresponds to points in $V(f)$ $P=(p_1,...,p_{n-1},p_n)$ where $f(P)=0$, but I don't know anything about the value of $p_n$; if it vanishes then clearly there is a correspondence $(X_1,...,X_n)\to (p_1,...,p_{n-1},0)$ but I guess I need to show there are a finite number of $p_n$ where this is true and I don't know how.
29506
says that the number of different monomials of degree $M$ in $N$ variables is $$\frac{(M+N-1)!}{M!(N-1)!}\; .$$ Can anyone explain why this is true?
113362
I need some help to show that:  If  $| \langle u,v \rangle | = \|u\| \|v\|$ then $u=\lambda v$ for some scalar $\lambda$.  We have to consider this over an arbitrary field $\mathbb{F}$.  I appreciate your suggestions.  Thanks.
2005
Let $f:\mathbb{N} \to \mathbb{Q}^+$ defined as follows : $$\begin{cases} f(0) = 0 \\ f(2n) = \dfrac{1}{f(n)+1} \\ f(2n+1) = f(n)+1 \end{cases}\,.$$ It is asked to prove the injectivity then the surjectivity of $f$.  After examining first values taken by $f$, I proved by induction that $f(2^k)=\dfrac{F_k}{F_{k+1}}$ where $(F_n)$ is the fibonacci sequence. Also one notes that if $m$ and $n$ have different parity then we can't have $f(m) = f(n)$ since $f(2\mathbb{N}) \subset (0,1)$ and $f(2\mathbb{N}+1) \subset [1,+\infty)$. Any ideas on how to finish the proof of injectivity and any thoughts about surjectivity are welcome. Thanks
87995
If $\gcd(m,n) =1, $ then $m\mathbb{Z}/ mn\mathbb{Z} \cong \mathbb{Z}/(n)$.   Here is my proof to the a/m problem. But I did not make use of $\gcd(m,n) =1.$ Could anyone hence point out and explain the mistakes in my proof? Thank you.   $\text{Define}\ \phi: m\mathbb{Z}/ mn\mathbb{Z} \to \mathbb{Z}/(n)$ by $\phi(ms + mn\mathbb{Z})= [s']_n, \text{where}  \ s \in [s']_n \ \text{and} \ s' \in \{0,...,n-1\}$  $\phi$ is well-defined:  $mx + mn\mathbb{Z} = my + mn\mathbb{Z}$  $\implies mx = my + mr,$ for some $r\in \mathbb{Z}$  $\implies n | x-y$  $\implies [x]_n = [x']_n= \phi(mx + mn\mathbb{Z}) = [y]_n = [y']_n= \phi(my+ mn\mathbb{Z})$  $\phi$ is injective:  It suffices to prove that $\ker(\phi) =\{mn\mathbb{Z}\}.$  Since $\phi(0+mn\mathbb{Z}) = [0]_n, \ker(\phi) \neq \emptyset.$ Given $x = ms+mn\mathbb{Z} \in \ker(\phi),\ $ $ \phi(x)= [s]_n = [0]_n$ $\text{hence}\  s = nk,  \text{for some}\  k \in \mathbb{Z}.$  Clearly, $mnk + mn\mathbb{Z}\subseteq mn\mathbb{Z}.$ Given $y \in mn\mathbb{Z},$ let $y = mnk + (mnc -mnk),$ for some $c \in \mathbb{Z}.$ Then $y \in mnk + mn\mathbb{Z},$ i.e. $mnk + mn\mathbb{Z} = mn\mathbb{Z} $ and hence $\ker(\phi) =\{mn\mathbb{Z}\}.$  $\phi$ is homomorphism:  Let $a = ms + mn\mathbb{Z}, a' = ms'+ mn\mathbb{Z} \in m\mathbb{Z}/mn\mathbb{Z}. $ Then, $\phi(a+a')= \phi(m(s+s') + mn\mathbb{Z}) $  $\exists r \in \{1,...,n-1\}$ such that $[r]_n = [s+s']_n$ and so $\phi(m(s+s') + mn\mathbb{Z}) = \phi(mr + mn\mathbb{Z}) = [r]_n = \phi(a) +_n \phi(a').$   $\phi$ is surjective:  $\text{Given} \ [r]_n \in \mathbb{Z}/(n),\  \text{let}  \ x= mr + mn\mathbb{Z}. \ \text{Then}   \ \phi(x) =[r]_n$
116277
Let $\mathbb{F}, \mathbb{F'}$ be fields, $\mathbb{F} \subset \mathbb{F'}$, and $\alpha \in \mathbb{F'}$. Assume $\mathbb{F(\alpha)}$ is finitely generated algebra over $\mathbb{F}$.   Is this true: $(\mathbb{F}(\alpha):\mathbb{F}) &lt; \infty$ ?   First i thought it's a consequence of this theorem:   But we don't know if $\alpha$ is algebraic or not.
98612
Let $F$ be a field. When happens that the additive group of $F$ is isomorphic to the multiplicative group?  It is easily to work out that $F$ must have characteristic $0$, but then what?
12430
Let $\mathbb{F}_{27}$ denote the finite field of size 27. For each $\alpha\in\mathbb{F}_{27}$ , define $ A_{\alpha}$ $=$ { $ 1, 1+\alpha , 1+\alpha+{\alpha}^2 , 1+\alpha+{\alpha}^2+{\alpha}^3 , .... $} . Then  prove that  the number of  $\alpha\in\mathbb{F}_{27}$ such that $|A_\alpha|$=26 equals 12.  $ 0\in A_\alpha $ if and only if $\alpha\neq 0$ For the first proof , the multiplicative group of the field $\mathbb{F}_{27}$ is cyclic and its order is 26  , so there are 12 generators in that group.   Now  I need  to  show  that  $ 1+\alpha+{\alpha}^2+.. {\alpha}^{25} $ =0  , where order of $\alpha$=26 . Since   $\alpha ^{26}$ =1 ($\alpha$-1)(1+$\alpha$+${\alpha}^2$+...+${\alpha}^{25}$)=0 , since $\alpha\neq 1$  , So $ 1+\alpha+...+{\alpha}^{25} $ =0 My question is , can I write these last two  lines?  I mean are those two lines alright to write? If those two lines are invalid then I need some help to show  $ 1+\alpha+{\alpha}^2+..+{\alpha}^{25} $=0  .Thanks
35983
1) How do I prove the following:   Let A = {6a + 4b ∈ Z : a, b ∈ Z} and B = {2a ∈ Z : a ∈ Z}. Show that A = B.  Thank you all for the help!
122229
I was preparing for an area exam in analysis and came across a problem in the book Real Analysis by Haaser &amp; Sullivan. From p.34 Q 2.4.3, If the field F is isomorphic to the subset S' of F', show that S' is a subfield of F'. I would appreciate any hints on how to solve this problem as I'm stuck, but that's not my actual question.  I understand that for finite fields this implies that two sets of the same cardinality must have the same field structure, if any exists. The classification of finite fields answers the above question in a constructive manner.  What got me curious is the infinite case. Even in the finite case it's surprising to me that the field axioms are so "restrictive", in a sense, that alternate field structures are simply not possible on sets of equal cardinality. I then started looking for examples of fields with characteristic zero while thinking about this problem. I didn't find many. So far, I listed the rationals, algebraic numbers, real numbers, complex numbers and the p-adic fields. What are other examples? Is there an analogous classification for fields of characteristic zero?
55082
How do you construct, for each $n\geq 1$, an ideal in $\mathbb Z[x]$ of the form $(a_1,a_2,\dots,a_n)$ with $a_i\in \mathbb Z[x]$ such that it is impossible to have $(b_1,b_2,\dots,b_m)=(a_1,a_2,\dots,a_n)$ with $m&lt;n$ and $b_j\in\mathbb Z[x]$?   Thank you!
27132
I want to prove that if $d \mid n \in \mathbb{N}$, then $\varphi(d) \mid \varphi(n)$.  It's given that $d \mid n$, so we know that $n = dm$, for some $m \in \mathbb{Z}$. Now, I want to show that $\varphi(d) \mid \varphi(n)$, i.e. that $\varphi(n) = k\varphi(d)$, for some $k \in \mathbb{Z}$. Drawing a blank of where to go from here.