Abstract Algebra Assignment: Prime Ideals and Noetherian Rings

This document presents a comprehensive solution to an Abstract Algebra assignment focusing on ring theory. It addresses key concepts such as prime ideals, both in commutative and non-commutative rings, and explores the relationship between completely prime ideals and prime ideals. The solution provides a proof for the equivalence of the two concepts in commutative rings and offers examples to illustrate the differences. The assignment further delves into Noetherian rings, providing examples of left Noetherian rings that are not right Noetherian, and bi-Noetherian rings that are not Noetherian. The document includes detailed explanations, proofs, and examples to clarify these abstract algebraic concepts, along with a bibliography of relevant sources.

Running head: Abstract Algebra
Abstract Algebra
Name of the Student
Name of the University
Author Note

Abstract Algebra 1
Table of Contents
Question 1:.................................................................................................................................2
Question 2:.................................................................................................................................2
Question 3:.................................................................................................................................4
Question 4:.................................................................................................................................4
Bibliography:..............................................................................................................................6

2Abstract Algebra
Question 1:
If, R is a noncommutative ring then there are 2 various notions of the prime ideals, both of
that are equivalent to definition that is usual, if R is commutative.
The proper ideal P of R is basically a prime ideal. However there are some conditions that are
as follows:
1. For the ideals I, J of R, if IJ ⊂ P then I⊂P or J ⊂ P.
2. For the left ideals I, J of R, if IJ ⊂ P then I ⊂ P or J ⊂ P.
3. For the right ideals I, J of R, if IJ ⊂ P then I⊂P or J⊂P.
4. For all a, b ∈ R if aRb ∈ P, then a ∈ P or b ∈ P.
A proper ideal (P of R) is completely prime ideal. If for all a and b ∈ R, ab ∈ P implies that a
∈ P or b ∈ P.
So, it can be said the R is prime and
If (a2) =R, then b∈R is existed exists there, 1=a2b as 1∈R=(a2).
so, 1=a(ab) and a is invertible.
Hence, b∈R is existed there such that a = a2b.
So, the converse is true and R is commutative (proved).
Question 2:
The major facts are:
An ideal I of R is prime, for that there is a condition. The condition is if and only id, R / I is a
domain that is integral. An Ideal I of R is maximal, id and only if R / I is the field.

3Abstract Algebra
The ring of polynomials in 2 variables, R = Q [x, y] over Q as well as the ideal principle I =
(x) that is generated by x. Q is the quotient ring, where Q [x, y] / x is isomorphic to Q [y]. Q
[y] is an integral domain.
Proof of R [x, y] / (x) ≅ R[ y].
The map: ψ : R [x, y] → R [y]. where x = .
For any polynomial,
g (y) ∈ R [y], let G (x, y) = g (y) ∈ R [x, y]
Then, ψ (G (x, y)) = G (0, y) = g (y)
So, ψ is surjective. (proved)
f(x, y) ∈ ker (ψ) f(x, y) ∈ ker(ψ).
it can be written than
f(x, y) = f0 (y) + f1 (y) x +⋯+ fn (y) xn,
where fi ∈ R [y] fi ∈ R [y] for I = 1,…,n
Since, f(x,y) ∈ ker (ψ) f(x,y) ∈ ker(ψ
Such that,
0 = ψ (f(x,y)) = f(0,y) = f0(y) [ as f(x,y) ∈ ker(ψ)]
Hence
ψ(f(x,y)) = ψ(xg(x,y)) = 0g(0,y) = 0
it implies that
f(x,y) ∈ ker (ψ) f(x,y) ∈ ker(ψ), hence ker(ψ)⊂(x)ker(ψ)⊂(x). (by simplifying)
(x) = ker(ψ)(x) = ker(ψ). (Putting two inclusions together)

4Abstract Algebra
The map ψ:R[x,y]→R[y]is a surjective ring and it is homomorphism with kernel (x).
R[x,y]/(x)≅R[y]. (proved)
Question 3:
The quotient of the free algebra Z (x, y) by the ideal (y^2, yx). The twisted polynomials k [x:
s] for k which is the division ring with s which is the endomorphism that is not actually an
auto morphism. The twist is given from x * a = x * s(a).
The triangular ring [ R R
0 Q ]where [R: Q] is basically an extension of infinite degree. The
triangular ring [ Q Q
0 Z ] where a commutative noetherian domain Z’s value is not identical to
the field of fractions of it which is Q. Let, a field endomorphism ‘s’ is there from K to K like
the image (L) is having an infinite index in K. Defining multiplication on the R = K * K by
(x, y) * (x’, y’) = (xx’, s (x) y’ + yx’). The ring is R.
Question 4:
In math, specially in the area of abstract algebra (ring theory), a bi noetherian ring is basically
a ring which satisfies the condition of ascending chain of the right and left ideal, which is
given an increasing sequence of right or left ideals:
I1 ⊆ . . . ⊆ Ik-1 ⊆ Ik ⊆ Ik+1 . . .
A natural number (n) is existed there such that:
In = In + 1 = . . .
Example:
The ring of rational functions can be generated by the x and y / xn , over the k field is actually
a sub ring of the k (x,y) field in two of the variables only. There are rings which are right

5Abstract Algebra
noetheiran however not the noetherian that is on the left. If L is a sub group of Q^2 which is
the isomorphic of Z. Let a ring is there (R) of the homo morphisms f from the Q^2 for self
satisfying f (L) ⊂ L.
The similar ring R can be described as follows:
R = { [α β
0 γ ] | α ∈ Z. β ∈ Q, γ∈ Q}
The ring is right Noetherian however it is not left noetherian. The subset of the I ⊂ R which
is consisting of the elements with γ=0 and a=0 is actually a left ideal which is not generated
finitely as the left R module. Thus, is can be concluded that it is a bi-Noetherian ring, which
is not Noetherian.

6Abstract Algebra
Bibliography:
Reyes, M. L. (2018, February). Progress with the Prime Ideal Principle. In Conference on
Rings and Factorizations—University of Graz.
Rump, W. (2018). Categories of lattices, and their global structure in terms of almost split
sequences. Algebra and Discrete Mathematics, 3(1).