# Abstract Algebra: Rings

Last updated: Wed, 14 Nov 2018

The ring $$\mathbb{Z}[\alpha] = a + b \alpha$$ is defined for a complex number $$\alpha$$. $$\alpha$$ is algebraic if it can be expressed as the roots of a polynomial with integer coefficients; otherwise, it is transcendental.

Elements with multiplicative inverses are terms units of a ring. In $$\mathbb{Z}$$, the only units are 1 and -1. In $$\mathbb{R}[x]$$, they are nonzero constant polynomials.

The kernel of a homomorphism $$\varphi : R \to R'$$ is defined as: $ker(\varphi) = \{a \in R \mid \varphi(a) = 0\}$

Closure under multiplication is formalized by a subset $$I \subset R$$, the ideal $$I$$ of $$R$$, such that:

• $$I$$ is a subgroup of $$R^{+}$$
• $$a \in I$$ and $$r \in I$$ implies $$ra \in I$$

Any ring with $$\text{ker } \varphi = 0$$ has characteristic $$0$$. Otherwise characteristic is $$n$$ where $$1 + 1 + ... \text{ (n times)} = 0$$. $$\mathbb{R}$$, $$\mathbb{C}$$, and $$\mathbb{Z}$$ have characteristic $$0$$, while $$\mathbb{F}_P$$ has characteristic $$P$$.

## Homomorphisms

Let $$\pi : R \to \overline{R}$$ be a homomorphism, such that $$\overline{R} = R + I$$ are the set of cosets. Then $$\text{ker } \pi = I$$.

For $$\varphi: R \to R'$$, $$\varphi / \text{ker } \varphi \cong \text{im } \varphi$$. Moreover, there is a bijective correspondence between ideals of $$R$$, and those of $$R'$$.

Notation: $$\mathbb{R}[\alpha]$$ = ring obtained by adjoining $$\alpha$$ to the ring $$\mathbb{R}$$.

Let $$\mathbb{R}[x]$$ denote the polynomial ring generated by $$\mathbb{R}$$ and $$x$$. By adjoining the solution of $$x^{2} + 1 = 0$$ to the ring $$\mathbb{R}$$, we get the isomorphism $$\mathbb{C} \cong \mathbb{R}[x]/(x^{2} + 1)$$.

$$\mathbb{R}[x, y]$$ can be viewed as a polynomial in $$y$$ with coefficients in $$x$$: $$\mathbb{R}[x, y] \cong \mathbb{R}[x][y]$$ is an isomorphism. For example:

$y^{2} x^{2} + 2yx + 2 = (x^{2}) y^{2} + (2x)y + (2)$

## Fields and integral domains

Integral domains are fields without zerodivisors, and fields have the additional property that every element has a multiplicative inverse.

No zerodivisors: satisfies the cancellation law that if $$ab = ac$$, then $$b = c$$.

Fields are characterized by having exactly two ideals: $$(0)$$ and $$(1)$$. In integral domains, on the other hand, every ideal is a principal ideal (these ideals generate the whole ring): for instance, in $$\mathbb{Z}$$, every prime number is a principal ideal.

## Maximal ideals

$$I$$ is termed as a maximal ideal if there are no ideals between $$I$$ and the whole ring $$A$$: in other words, if $$I$$ is the maximal ideal, then there is no $$J$$ such that $$I \subset J \subset A$$. Every maximal ideal $$M$$ has the property that $$R / M$$ is a field, having exactly two ideals.

## Nullstellansatz

Given three functions in two variables:

$f_{1} = x^{2} + y^{2} - 1, f_{2} = x^{2} - y + 1, f_{3} = xy - 1$

We can write $$1 = \sum g_i f_i$$, a linear combination with polynomial coefficients.

## Algebraic Geometry

The set of solutions of $$ax + by + c = 0$$ is a variety, as is a set of points.

Two polynomials of degree $$m$$ and $$n$$ have finitely many points of intersection, and this is known as Bezout bound $$mn$$, as long as they have no common factor.

## Unique Factorization Domain and Principal Ideal Domain

$$\mathbb{Z}[\sqrt{-5}]$$ is not a UFD, because there exist two different factorizations of the element $$6 = 2 . 3 = (1 + \sqrt{5}) (1 - \sqrt{5})$$.

However, a UFD allows for associate factorizations: since the units of $$\mathbb{Z}$$ are $$\pm 1$$, $$2$$ and $$-2$$ are associates: $$2 . -2 = -2 . 2 = -4$$.

Irreducible elements in a UFD are prime, but irreducible and prime are not equivalent in the general case: in $$\mathbb{Z[\sqrt{-5}]}$$, $$2$$ is irreducible because it has no proper factors, but it is not prime because it divides $$6 = (1 + \sqrt{-5}) (1 - \sqrt{-5})$$.

An integral domain in which all ideals are principal (generated by a single element) is termed a principal ideal domain.

## Algebraic numbers

A number $$\alpha$$ is algebraic if and only if it is the root of a polynomial with integer coefficients. Moreover, it's an algebraic integer if and only if the polynomial is monic.