MAT401 Galois Theory Midterm Prep Questions and Answers Abstract Algebra Study Guide Comprehensive Questions Answers Detailed Explanations 2025/ 2026

MAT401 Galois Theory Midterm Prep Questions and Answers MAT401
Abstract Galois
AlgebraTheory
Study Midterm
Guide.pdf
Prep Questions and Answers MAT401
Abstract Galois
AlgebraTheory
Study Midterm
Guide.pdf
Prep Questions and Answers Abstract Algebra Study Guide.pdf




MAT401 Galois Theory
Midterm Prep Questions
and Answers Abstract
Algebra Study Guide




Guidehttps://www.stuvia.com/dashboard!@_)#*)(@$)($@*($@)($@*_
MAT401 Galois Theory Midterm Prep Questions and Answers MAT401
Abstract Galois
AlgebraTheory
Study Midterm
Guide.pdf
Prep Questions and Answers MAT401
Abstract Galois
AlgebraTheory
Study Midterm
Guide.pdf
Prep Questions and Answers Abstract Algebra Study Guide.pdf

,MAT401_ Galois Theory Midterm Prep.pdf MAT401_ Galois Theory Midterm Prep.pdf MAT401_ Galois Theory Midterm Prep.pdf




Chap 1: rings


What is a ring? In this course, we treat commutative rings as rings. A ring is therefore a set with
addition and multiplication that follows the following properties:


1. additively, the set is an abelian group with its identity element denoted by 0


2. multiplication is a commutative and associative binary operation with an identity
element denoted 1 such that 1 != 0


3. multiplication distributes over addition, that is:
x(y + w) = xy + xw


Show that the integers modulo n is a ring.


Consider Z_{n} such that: One choice that works is n = 4 as [3] + [2] = [5] = [1], so we are looking for a number
[3] + [2] = [1] that divides 5 with a remainder of 1.


what could n be? is this choice unique? We notice this choice is not necessarily unique, as n=2 would satisfy this.




MAT401_ Galois Theory Midterm Prep.pdf MAT401_ Galois Theory Midterm Prep.pdf MAT401_ Galois Theory Midterm Prep.pdf

,MAT401_ Galois Theory Midterm Prep.pdf MAT401_ Galois Theory Midterm Prep.pdf MAT401_ Galois Theory Midterm Prep.pdf




Informally, define R[x] R[x] is the set of all polynomials over a ring R. That is, it is the set of all p(x) = anx^n
+ an-1x^n-1 + ... + a0


where ai is in R


Formally, define a polynomial over a ring R. We say a sequence f(x) = (c0, c1, c2, ..., cn, 0,0,...) is a polynomial over a ring R if it
happens that ci in R and ci = 0 when i > n.


In our view of polynomials, what actually is x, or the so- Actually, there is a very determined meaning of x. We say that x is the element of
called indeterminate? the polynomial ring:


(0,1,0,0,...)


this allows us to easily go back and forth between the cumbersome sequence
notation, or the more familiar polynomial notation:


f(x) = cnxn + cn-1xn-1 + ... + c0




MAT401_ Galois Theory Midterm Prep.pdf MAT401_ Galois Theory Midterm Prep.pdf MAT401_ Galois Theory Midterm Prep.pdf

, MAT401_ Galois Theory Midterm Prep.pdf MAT401_ Galois Theory Midterm Prep.pdf MAT401_ Galois Theory Midterm Prep.pdf




Define: leading coeff: cn
leading coefficient monic: cn = 1
monic constant term: c0
constant term degree: n
degree root: some alpha in R such that it brings the whole polynomial to 0
root of polynomial


theorem: 1. suppose there are identity elements 1 and 1', then:
1 is unique (1)(1') = 1 and (1')(1) = 1', but since mult is commuative:
0*r = 0 (1)(1') = (1')(1), hence 1 = 1'.
if -r + r = 0, then (-1)r = -r
(-1)(-r) = r 2. 0r = (0+0)r = 0r + 0r.
thus, 0r = 0


3. (-1 + 1) = 0, hence (-1 + 1)r = 0, hence (-1)r + r = 0. However, since additive inverses
are unique, it must be the case that (-1)r = -r.


4. from 3, (-1)(-r) = -(-r). We appeal to the uniqueness of additive inverses again to
conclude -(-r) = r, hence (-1)(-r) = r


Why do we insist that 1 != 0 for rings? Suppose 1 = 0. Then let r in R. We can see trivially that:
r = 1r, but because 1=0, we get r=0r=0,
hence this is a fairly trivial set that only contains a single element.

MAT401_ Galois Theory Midterm Prep.pdf MAT401_ Galois Theory Midterm Prep.pdf MAT401_ Galois Theory Midterm Prep.pdf