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I assume you use column vectors. One particular Sylow $p$-subgroup is $$L = \left\{ \begin{bmatrix} 1 & 0 & 0\\ a & 1 & 0\\ c & b & 1\\ \end{bmatrix} : a, b, c \in \Bbb F_p \right\}.$$ You may try and compute its normalizer. The index will give you the number of Sylow $p$-subgroups. Alternatively, you may note that $x \in L$ iff $... 3 What's an element of$R[x]$? It's a finite sum that looks like this: $$a_0+a_1x + \cdots + a_n x^n,$$ where the$a_i \in R$. If we forget about the variable$x$and note that all that really matters in this description is the coefficients$a_i$and the order they appear in, we realize that this corresponds in a bijective fashion to an ordered tuple (with ... 2$\mathbf F_{31}^\times$is a cyclic group. Hence if$a$is one of its generators (there are$\varphi(30)=8$of them),$a^{30/10}=a^3$generates the subgroup of order$10$. 2 Here are the key facts:$U_p$is is cyclic when$p$is prime. A cyclic group of order$n$has elements of order$d$for each divisor$d$of$n$.$31$is prime.$U_{31}$has order$30$.$10$divides$30$. You'll find that$3$is a generator of$U_{31}$and so$27=3^3$has order$10$. All elements of order$10$are$27^1, 27^3, 27^7, 27^9 \bmod 31$, that ... 2 Consider the canonical map$\pi\colon\mathbb{Z}\to G=\mathbb{Z}/n\mathbb{Z}$and let$d$be a divisor of$n$; set$n=dk$. Existence. Consider the subgroup$k\mathbb{Z}$. Then$\pi(k\mathbb{Z})$is a subgroup of$G$and it has$d$elements (prove it). Uniqueness. Suppose$H$is a subgroup of$G$having$d$elements; then$\pi^{-1}(H)$is a subgroup of$\...
$R[x_1,\dots, x_n]$ is by definition the free associative and commutative $R$-algebra, $\;R^{(\mathbf N^n)}$, built on the monoid $\;\mathbf N^n$. In one indeterminate, it's the algebra $\; R^{(\mathbf N)}$ of all eventually $0$ sequences of elements of $R$. Addition is defined componentwise, and multiplication of $(c_n)_{n\ge 0}$ and $(c'_n)_{n\ge0}$ is $(... 2 A module is an abelian group, and so every subgroup (and therefore any submodule) is a normal subgroup. Further, it's easy to show that scalar multiplication is well-defined in the quotient module, and so we don't need any added conditions. For your second question, that condition (that only finitely many$x_i$are nonzero) is basically saying it contains ... 2 It is infact never true that$\mathfrak{m}[x]\subseteq R[x]$! All of these questions become simple when you consider the following fact. An ideal$\mathfrak{p}\subseteq R$is prime if and only if$R/\mathfrak{p}$is an integral domain. Moreover$\mathfrak{p}$is maximal if and only if$R/\mathfrak{p}$is a field. The first statement is simple, the second ... 2 The notes means that, for$n\geq 4$,$n!-1$has remainder$3$modulo$4$. It cannot have any prime factor$p\leq n$. It also must have a prime factor$p\equiv 3\pmod 4$because if$N=p_1p_2\dots p_k$and$p_i\equiv 1\pmod 4$then$N\equiv 1\pmod 4$. So given any$n$, there must be a$p$not in$1,2,\dots,n$such that$p\equiv 3\pmod 4$. 1 If$p(x)=x^d+a_{d-1}x^{d-1}+\dots+a_0$, then the quotient is generated (in fact, freely generated) by$S=\{1,x,\dots,x^{d-1}\}$. Indeed, you can prove by induction that$x^n$is in the submodule generated by$S$for each$n$. For$n<d$this is trivial. For$n\geq d$, you have$x^{n-d}p(x)=0$so$x^n=-a_{d-1}x^{n-1}-\dots-a_0x^{n-d}$, which is generated ... 1 Let$g \in Hx_i \cap Ky_j$. Then there are$h \in H$and$k \in K$such that$g= hx_i = ky_j$. Hence we have $$Hg = H(hx_i) = (Hh)x_i = Hx_i$$ and similarly$Kg = K(ky_j) = Ky_j$. And therefore $$Hg \cap Kg = Hx_i \cap Kx_j.$$ 1 Bernard's answer is very good, but here are two more way of looking at a polynomial ring. It is always helpful to have multiple perspectives. Let$K$be a commutative ring, and$X$some set of variables. A polynomial ring$K[X]$on a set of variables$X$is the free commutative$K$-algebra on$X$, i.e the most general commutative$K$-algebra we can ... 1 You define it recursively. $$R[x_1,x_2, \dots, x_{n+1}] = \left\{ \sum_{j=0}^k a_j x_{n+1}^k \mid a_j \in R[x_1, x_2, \dots, x_n], k \in \Bbb{Z}_{\geq 0} \right\}$$ So a generic element is a polynomial in$x_{n+1}$with coefficients in$R[x_1, x_2, \dots, x_n]$. An example of an element of$R[x,y,z]$is$45 + ((3x) + ((4x)y^{12})z + ((x^{12})y)z^3$, ... 1$G = \mathbb{Z}/n\mathbb{Z}$is cyclic, so let$g$be any generator. Of course$\langle g^{n/d}\rangle$is a subgroup of order$d$. Let$a \in G$be any order$d$element. Then$a^d = 1$; in particular$a = g^t$for some$t$so$g^{td} = 1 \implies td \equiv 0 \pmod n$so$t$is a multiple of$n/d$. In particular,$t \in \langle g^{n/d} \rangle$so$\...