In ex. 1.2.3 on p.21 of Diamond's A first course in modular forms, I was trying to show that $$|SL_2(\mathbb{Z}/N \mathbb{Z})|=N^3 \prod_{p|N}(1-1/p^2).$$  First of all, I knew that If $F$ is a field with $q$ elements, then $|GL_2(F)|=(q^2-1)(q^2-q)$. If we consider the surjective homomorphism $\det : GL_2(F) \rightarrow F^*$ with kernel $SL_2(F)$, then we have $|SL_2(F)|=(q^2-1)(q^2-q)/(q-1)=q(q^2-1)=q^3-q$. Therefore if we replace $F$ by $\mathbb{Z}/N \mathbb{Z}$, we get $|SL_2(\mathbb{Z}/N \mathbb{Z})|= N^3-N$, but thats' not agree with the book, so what's wrong with me...? In fact, in the part (a) of the problem, it asks us show $|SL_2(\mathbb{Z}/p^e \mathbb{Z})|=p^{3e}(1-1/p^2)$ for $p$ prime (that's already doesn't agree with my conclusion...), and uses chinese remainder theorem to conclude.
120663
the answer should be $$|\operatorname {SL}_2(\mathbb Z/N\mathbb Z)|=N^{3}\prod_{p|N}(1-{1 /p^2})$$ But first how to prove $$|\operatorname {SL}_2(\mathbb Z/p^e\mathbb Z)|=p^{3e}(1-{1 /p^2})$$
119971
Prove that $\mathbb Z_{m}\times\mathbb Z_{n} \cong \mathbb Z_{mn}$ implies $\gcd(m,n)=1$.  This is the converse of the Chinese remainder theorem in abstract algebra. Any help would be appreciated.  Thanks!
15141
I need to find $|GL_2(F_p)|$, I am very very uncomfortable in counting, so please help here I proceed,  $\begin{pmatrix}a&amp;b\\c&amp;d\end{pmatrix}$ be such an invertible matrix, so we need and sufficient that $ad\neq bc$, so if i chose all non zero $a,b,c$ arbitrarily and chose $d\neq a^{-1}bc$ and form matrices then the cardinality   will be $(q-1)^3\times (q-2)$, well now if one of $a,b,c,d$ be zero then other three can be anything nonzero so in this case cardinality will be $4(q-1)^3$, so finally if two entries are $0$ then they must be opposite to each other i.e diagonally, but other two must be nonzero, so cardinality in that case $2(q-1)^2$, so total number is just sum of three cases.
90126
There is something I do not understand in my lecture notes, mainly because I do not understand quotients very well. We consider a number field $K = \mathbb{Q}(\sqrt{d})$ where $d$ is squarefree.  If $\mathcal{O}_K$ denotes the ring of integers of $K$, it has a basis $\langle 1, \omega \rangle$ where $\omega$ is either $\sqrt d$ or $\frac{1+\sqrt{d}}{2}$ depending on $d \bmod 4$. We let $f(x) \in \mathbb{Z}[x]$ be the minimal polynomial of $\omega$ (so $f$ is monic of degree 2).  What I do not understand is first of all that we can write  $ \mathcal{O}_K = \mathbb{Z}[x] / f(x) $. Also, if $p \in \mathbb{Z}$ is prime, the notes then say $$ \frac{\mathcal{O}_K}{(p)} = \frac{\mathbb{Z}[x]}{(p,f(x))} = \frac{\mathbb{F}_p[x]}{f(x) \bmod p} $$  I do understand how to go from the first expression to the second(trivial when you know the above) but not how to go from the second to the third.  Also, the lecturer then gives the expression $\frac{\mathcal{O}_K}{(p)}$ when $p$ is ramified, split or inert. I do not really see why, in the sense how does it help to know that when $p$ is inert then  $$ \frac{\mathcal{O}_K}{(p)} = \frac{\mathbf{F}_p}{f(x)} = \mathbf{F}_{p^2} $$ for example ?  Thank you very much for your help !
86847
I've just left an exam of algebraic structures, and one of the questions asked to compute the order of $SL_2(\mathbb{Z}/ 5\mathbb{Z})$ (the 2x2 matrices with determinant $1$ over the field $\mathbb{Z}/ 5\mathbb{Z}$). The answer should be $120$, and a friend of mine could achieve that, but by methods of raw counting. My question is: Is there a way to compute the order of $SL_2(\mathbb{Z}/ n\mathbb{Z})$ for all $n$?  OBS: Although $120=5!$, I believe this is purely a coincidence.
36711
I'm searching for a way to generate the group $\mathrm{GL}(n,\mathbb Z)$. Does anyone have an idea?  The intention of my question is that I am searching for an easy proof of the existence of the epimorphism:  $\Phi:\mathrm{Aut}(F_n )\to \mathrm{Aut}(F_n/[F_n,F_n])=\mathrm{Aut}(\mathbb {Z}^n)=\mathrm{GL}(n,\mathbb {Z})$  I know that $\Phi$ is an canonical homomorphism since the commutator subgroup is characteristic in $F_n$. So I need some nice generators of $\mathrm{GL}(n,\mathbb {Z})$ to find their preimages in $\mathrm {Aut}(F_n)$ to prove the surjectivity of $\Phi$.  Thanks for help!
7115
I have some problems with this exercise. I don't know if it can be done. Consider the polynomial $ x^n - a \in \mathbb{Q} $ Can I compute the Galois group of this over $\mathbb{Q}$? Maybe having a nice "basis.  The splitting field is given by $\mathbb{Q}(\zeta_n,\alpha)$ , where $\zeta_n$ is a primitive root of unity , and $\alpha$ is some number such that $\alpha^n = a $. Well first of all, I want a $\underline{"good basis"}$ for the splitting field. In the sense that the minimal polynomials, of the adjoined elements, are different (in this case the computation of the galois group is very simple).  For example one easy case, it's when $a&gt;0$, then $(a)^{\frac{1}{n}} \in \mathbb{R}$ , so clearly the minimal polynomial of $(a)^{\frac{1}{n}} , \zeta_n$ are distincs, and I'm done. If $n$ is odd then , it's also easy, since one root it's also real, for example $x^3-3 $, the real root is $ \root 3 \of { - 3}  =  - \root 3 \of 3  $ , so I can consider the splitting field as $\mathbb{Q}(-\root 3 \of 3 , \zeta_3 )=\mathbb{Q}(\root 3 \of 3 , \zeta_3 )$. The difficult case is when $n$ is even and $a&lt;0$ , for example  $x^8+20$ or $x^4+20$ in some cases as in the second, there are particular cases since there exist algorithms for the Galois group of quartics, but in general. It can be done?  $\underline{Remark}$ I'm searching a $\underline{"good basis"}$ for the splitting field. In the sense that the minimal polynomials, of the adjoined elements, are different since in this case the computation of the galois group is very simple.
65300
Let $SL_2(\mathbb{Z})$ denote the group (under usual matrix multiplication) of $2\times2$ matrices with integer entries and determinant $1$. Let $H$ be the subgroup of $SL_2(\mathbb{Z})$ consisting of those matrices such that the diagonal entries are all equivalent to $1 \pmod 3$ and the off-diagonal entries are all divisible by $3$.   What is the index of $H$ in $SL_2(\mathbb{Z})$? There are a total of $3^4=81$ different equivalence classes of matrices in $SL_2(\mathbb{Z})$ modulo $3$ (each of the entries can have $0,1,2$ as remainders). Now, the given condition implies only one of the possible $81$ combinations modulo $3$. How do we proceed? any hints? Thanks beforehand.
77135
There's a particular property of the elements of $\mathrm{GL}(2,q)$, the general linear group of $n\times n$ matrices over a finite field of order $q$, that I don't understand.  I know that the order of $\mathrm{GL}(2,q)$ is $(q^2-1)(q^2-q)=q(q+1)(q-1)^2$, since there are $q^2-1$ possible vectors for the first column, excluding the $0$ vector, and $q^2-q$ possible vector for the second column, excluding all multiples of the first.  So the order of any element must divide $q(q+1)(q-1)^2$ by Langrange. However, there is a further detail that      any element of $\mathrm{GL}(2,q)$ must have order dividing $q(q-1)$ or $(q-1)(q+1)$.    Is there a reason why one can narrow down the order to divide one of those smaller factors of $q(q+1)(q-1)^2$? Thanks!
73590
I am currently working to find all ring homomorphisms $\phi : \mathbb{Q}(\sqrt{2})\rightarrow \mathbb{Q}(\sqrt{2})$.   My work so far: Since $\mathbb{Q}(\sqrt{2})$ is a field extension of $\mathbb{Q}$, a field, it is obviously a field as well. Therefore, its only ideals are $\{0\}$ and itself. Since every homomorphism's kernel is an ideal, we have that either $Ker(\phi)=\{0\}$ or $\mathbb{Q}(\sqrt{2})$.   In the latter case, we have the trivial homomorphism. In the former case, we have that the homomorphism is injective (and therefore bijective).   My question: I think it is true that in the former case $\phi(1)=1$, but am not sure why. If this is the case, can I prove inductively that $\phi(x)=x, \forall x\in \mathbb{Q}(\sqrt{2})$?  Thank you in advance for your help.
35804
Let $a = \sqrt{2+i}$ and $K$ is the splitting field of minimal polynomial of $a$ over $\mathbb{Q}$. Prove that $Gal(K/\mathbb{Q})$ is $D_4$.   I find the minimal polynomial of $a$ is $p(x)=x^4-4x^2+5$ and its   roots are $\sqrt{2+i},-\sqrt{2+i},\sqrt{2-i},-\sqrt{2-i}$. Let $b=\sqrt{2-i}$. So the splitting field of $p$ is $K=\mathbb{Q}(a,b)$. Also by rational root theorem, $p$ is irreducible over $\mathbb{Q}$. Thus $[\mathbb{Q}(a):\mathbb{Q}]=4$. Also $b\not\in\mathbb{Q}(a)$, and minimal polynomial of $b$ over $\mathbb{Q}(a)$ is $x^2-2+i$. Thus $[\mathbb{Q}(a,b):\mathbb{Q}]=[\mathbb{Q}(a,b):\mathbb{Q}(a)]\cdot[\mathbb{Q}(a):\mathbb{Q}]=2\cdot 4=8.$ Now since $K$ is the splitting field over $\mathbb{Q}$ of a separable polynomial, $K/\mathbb{Q}$ is Galois. Hence $|Gal(K/\mathbb{Q})|=8$. Hence the 8 automorphisms are $$a\to \begin{cases}a\\-a\\b\\-b\end{cases}\quad\text{and}\quad b \to \begin{cases}b\\-b\end{cases}.$$ From this how to conclude $Gal(K/\mathbb{Q})$ is isomorphic to $D_4$?  Thanks
84383
Problem 4.   Let $\Bbb F_2$ be the field with 2 elements and let $R=\Bbb F_2[X]$. List, up to isomorphism, all $R$-modules with 8 elements.      Solution.   We use the classification theorem of modules over a PID. Since $R$ is a finite module, it is in particular an $\Bbb F_2$ vector space. We can write   $$M\cong R/n_1R\oplus R/n_2 R\oplus\cdots\oplus R/n_r R$$   for polynmials $n_1\mid n_2\mid\cdots\mid n_r$. In our case, we have $\sum_{i=1}^r\deg n_i=3$, so we have three options: $r=1,\deg n_1=3$, $r=2,\deg n_2=2$, and $r=3,\deg n_3=1$. The first case yields 8 options. For the second case, we need $n_2$ to be reducible, so we have $X(X+1),X^2,(X+1)^2$ as choices. The first choice yields 2 decompositions, and the latter choices yield 1 decomposition each, for a total of 4. For the linear case, we need the same linear term repeated thrice, which is 2 choices. Therefore there are 14 in all, listed by invariant factors below:   $$\begin{align} &amp;\{X^3+(0/1)X^2+(0/1)X+(0/1)\}\\ &amp;\{X^2,X\},\{X^2+X,X+(0/1)\},\{X^2+1,X+1\}\\ &amp;\{X,X,X\},\{X+1,X+1,X+1\}. \end{align}$$   My question is the problem above. I can understand the solution, but how can I see that the solutions given are from distinct isomorphism classes? For instance, is the below true? $$R/(X)\oplus R/(X)\oplus R/(X)\cong R/(X+1)\oplus R/(X+1)\oplus R/(X+1)\\ \cong F_2\oplus F_2\oplus F_2$$  Thanks for any help.
8219
Let $K=\mathbb{Z}_2 (x,y)$, where $x,y$ are independent, and $L$ be a splitting field extension of $(X^2 - x) (X^2 - y)$, then $[L:K] = 4$ and $L = K(\sqrt{x},\sqrt{y})$ where $\sqrt{x},\sqrt{y}$ are roots of $X^2-x$, $X^2 - y$ respectively. What are the subextensions of $L:K$?  I know all elements in $L$ square to something in $K$, so all the intermediate fields are $K(\sqrt{k})$ for some $k\in K$, but some of them are the same, say $K(\sqrt{x/y}) = K(\sqrt{xy})$...   Note: $\mathbb{Z}_2$ means $\mathbb{Z} / 2\mathbb{Z}$
31296
Evidently, $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ has order $4$, but I think it's infinite.   The four cosets are listed as $(0,0) +  \langle (2,2) \rangle$, $(0,1) +  \langle (2,2) \rangle$, $(1,0)+  \langle (2,2) \rangle$ and $(1,1) +  \langle (2,2) \rangle$. However, $(2,0)$ doesn't appear to be in any of these cosets. Maybe the answer I'm being told is wrong.
82895
If $m|n$. Why the map $f\colon \mathbb{Z}_n^\times \to \mathbb{Z}_m^\times$ given by $a \mod{n}\mapsto a \mod m$ is a surjective homomorphism of groups?  Attempt: I proved it is well a well defined homomorphism because the canonical projection $\overline{f}\colon \mathbb{Z}_n \to \mathbb{Z}_m$ is a surjective ring homomorphism, it extends $f$ and maps units to units.  But I can't prove surjectiveness. Given $a\in\mathbb{Z}$ such that $(a,m)=1$ I'm looking for a $k\in\mathbb{Z}$ such that $(a+k\cdot m,n)=1$. I don't know why one of the $n/m$ candidates for such $k$ makes the number $a+k\cdot m$ an unit modulo $n$.
67241
I'm interested in the following problem from Artin's Algebra text:     Determine the structure of the ring $\mathbb Z[x]/(x^2 + 3,p)$, where (a) p = 3, (b) p = 5.   I know that by the isomorphism theorems for rings we can take the quotients successively, and so  $$\mathbb{Z}[x]/(p) \cong (\mathbb{Z}/p \mathbb{Z})[x] $$  as the map $\mathbb{Z}[x] \to (\mathbb{Z}/p \mathbb{Z})[x]$ defined by $\sum_{n} a_n x^n \mapsto \sum_{n} \overline{a_n} x^n$ is a surjective ring homomorphism with kernel $(p)$. Thus it remains to study the quotients  $$(\mathbb{Z}/p \mathbb{Z})[x]/(x^2+3) $$ for $p \in \{3,5\}$.    If $p=3$, $(x^2+3)=(x^2)$ in $(\mathbb{Z}/3 \mathbb{Z})[x]$, and by using polynomial division all distinct coset representatives can be reduced to the following list of 9 elements  $$\{0,1,2,x,1+x,2+x,2x,1+2x,2+2x\}. $$  Moreover, it can shown that the list above gives 9 distinct cosets, as no difference of two distinct elements of the list is a multiple of $x^2$. Since $1$ and $x$ generate two distinct additive groups of order $3$, the additive group of our quotient ring is not cyclic. Elementary group theory then shows   $$(\mathbb{Z}/3 \mathbb{Z})[x]/(x^2)^+ \cong (\mathbb{Z}/ 3 \mathbb{Z})^2 $$  as additive groups.   I was then about to conclude that the multiplication on the quotient is then compatible with the usual one in $(\mathbb{Z}/3\mathbb{Z})^2$, but this is wrong!  It can be seen that the quotient is not isomorphic to $(\mathbb{Z}/3\mathbb{Z})^2$ as a ring, because the former contains a nonzero element (represented by $x$) whose square is zero, while the latter contains no such elements.    If $p=5$, a full list of coset representatives is of length 25  $$\{0,1,2,3,4,x,1+x,2+x,3+x,4+x,2x,1+2x,2+2x,3+2x,4+2x,3x,1+3x,2+3x,3+3x,4+3x,4x,1+4x,2+4x,3+4x,4+4x \} .$$  And once again, one can see that these represent 25 distinct cosets. Similarly to the $p=5$ case, I've managed to prove that the additive group of this ring is isomorphic to $(\mathbb{Z}/5 \mathbb{Z})^2$.    My questions:   Have I made any mistakes in my argument? What exactly am I supposed to do in this question? determine the number of elements? Write down the tables for addition and multiplication?   Any further information about these quotients will be appreciated, thanks!
44559
Prove that $PSL(2,\mathbb{Z})=C_2 \star C_3$.  Now $C_2 \star C_3=\langle a,b\ |\ a^2, b^3 \rangle$ i.e. the free product.  But how do I show that presentation of $PSL(2,\mathbb{Z})$ is this?