# Abstract algebra questions and answers

Recent questions in Abstract algebra
Albert Byrd 2022-04-10 Answered

### Use the isomorphism theorem to determine the group $G{L}_{2}\frac{\mathbb{R}}{S}{L}_{2}\left(\mathbb{R}\right)$. Here $G{L}_{2}\left(\mathbb{R}\right)$ is the group of $2×2$ matrices with determinant not equal to 0, and $S{L}_{2}\left(\mathbb{R}\right)$ is the group of $2×2$ matrices with determinant 1. In the first part of the problem, I proved that $S{L}_{2}\left(\mathbb{R}\right)$ is a normal subgroup of $G{L}_{2}\left(\mathbb{R}\right)$. Now it wants me to use the isomorphism theorem. I tried using$|G{L}_{2}\frac{\mathbb{R}}{S}{L}_{2}\left(\mathbb{R}\right)|=\frac{|G{L}_{2}\left(\mathbb{R}\right)|}{|S{L}_{2}\left(\mathbb{R}\right)|},$but since both groups have infinite order, I don't think I can use this here.

Ramiro Grant 2022-04-09 Answered

### Two sided ideal in ${M}_{2×2}\left(\mathbb{C}\right)$

Deangelo Hardy 2022-04-08 Answered

### I was doing some practice abstract algebra questions off the internet since I have a quiz coming up soon. However, I am not very skilled at abstract algebra. In fact, I did very average in my group theory class, so I am struggling in my ring theory one. Can someone please help explain what is happening in this proof? I'm very sorry if it's extremely straightforward, I just think I need some time to get used to the way of thinking that's required to solve these questions.Let be commutative rings with identities and let $R={R}_{1}×{R}_{2}$. The question asks to show that every ideal I of R is of the form $I={I}_{1}×{I}_{2}$ with ${I}_{1}$ an ideal of ${R}_{1}$ and ${I}_{2}$ an ideal of ${R}_{2}$.

Breanna Fisher 2022-04-08 Answered

### Understanding a proof of: In a finite group, the number of elements of ' order p is divisible by $p-1$.

Amya Horn 2022-04-05 Answered

### Fields with involution whose fixed field is ordered?Let K be a field with an involution $\ast$, meaning $\cdot :K\to K$ is an automorphism and $\left(x\ast \right)\ast =x$ for all $\xi nK$. Suppose further that the fixed field of $\ast$ is ordered (i.e., it can be given an ordering that is compatible with the field structure).

Makayla Stevens 2022-04-04 Answered

### Euler's remarkable '-producing polynomial and quadratic UFDsExample of a polynomial which produces a finite number of 's is:${x}^{2}+x+41$which produces 's for every integer $0\le x\le 39$.

Janiyah Hays 2022-04-03 Answered

### Residue of x in polynomial ring $\mathbb{Z}\frac{x}{f}$When we say that α is the residue of x in $\mathbb{Z}\frac{x}{f}$, and $f={x}^{4}+{x}^{3}+{x}^{2}+x$, wouldn't $\alpha$ just be x? Because if we divide with the remainder, we would get ?

Miley Caldwell 2022-04-01 Answered

### How to solve a cyclic quintic in radicals?Galois theory tells us that$\frac{{z}^{11}-1}{z-1}={z}^{10}+{z}^{9}+{z}^{8}+{z}^{7}+{z}^{6}+{z}^{5}+{z}^{4}+{z}^{3}+{z}^{2}+z+1$ can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:Let the roots be ${\zeta }^{1},{\zeta }^{2},\dots ,{\zeta }^{10}$, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group [2,4,8,5,10,9,7,3,6,1] we can partition into the five subgroups of conjugate pairs [2,9],[4,7],[8,3],[5,6],[10,1].$\begin{array}{rl}{A}_{0}& ={x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}\\ {A}_{1}& ={x}_{1}+\zeta {x}_{2}+{\zeta }^{2}{x}_{3}+{\zeta }^{3}{x}_{4}+{\zeta }^{4}{x}_{5}\\ {A}_{2}& ={x}_{1}+{\zeta }^{2}{x}_{2}+{\zeta }^{4}{x}_{3}+\zeta {x}_{4}+{\zeta }^{3}{x}_{5}\\ {A}_{3}& ={x}_{1}+{\zeta }^{3}{x}_{2}+\zeta {x}_{3}+{\zeta }^{4}{x}_{4}+{\zeta }^{2}{x}_{5}\\ {A}_{4}& ={x}_{1}+{\zeta }^{4}{x}_{2}+{\zeta }^{3}{x}_{3}+{\zeta }^{2}{x}_{4}+\zeta {x}_{5}\end{array}$Once one has ${A}_{0},\dots ,{A}_{4}$ one easily gets ${x}_{1},\dots ,{x}_{5}$. It's easy to find ${A}_{0}$. The point is that $\tau$ takes ${A}_{j}$ to ${\zeta }^{-j}{A}_{j}$ and so takes ${A}_{j}^{5}$ to ${A}_{j}^{5}$. Thus ${A}_{j}^{5}$ can be written down in terms of rationals (if that's your starting field) and powers of $\zeta$. Alas, here is where the algebra becomes difficult. The coefficients of powers of $\zeta$ in ${A}_{1}^{5}$ are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have ${A}_{1}$ as a fifth root of a certain explicit complex number. Then one can express the other ${A}_{j}$ in terms of ${A}_{1}$. The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.

Octavio Chen 2022-03-31 Answered

### Prove if $F\left(\sqrt[n]{a}\right)$ is unramified or totally ramified in certain conditions

Rex Maxwell 2022-03-30 Answered

### $\frac{{\mathbb{F}}_{2}\left[X,Y\right]}{\left({Y}^{2}+Y+1,{X}^{2}+X+Y\right)}$ and $\frac{\left({\mathbb{F}}_{2}\left[Y\right]\right\}\left\{\left({Y}^{2}+Y+1\right)\right\}\frac{\left\{\right)}{X}}{\left({X}^{2}+X+\stackrel{―}{Y}\right)}$ are isomorphic

Aleah Choi 2022-03-29 Answered

### I have to prove that if P is a R-module , P is projective $⇔$ there is a family $\left\{{x}_{i}\right\}$ in P and morphisms ${f}_{i}:P\to R$ such that for all $x\in P$$x=\sum _{i\in I}{f}_{i}\left(x\right){x}_{i}$where for each for almost all $i\in I$.

Jasper Dougherty 2022-03-29 Answered

### Proving the generator of$A=\left\{154a+210b:a,b\in \mathbb{Z}\right\}$ is

Hugh Soto 2022-03-28 Answered