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4

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 $\...


2

$R[x_1,\dots, x_n]$ 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 $\...



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