[ra] Rings and Algebras


Paris, Chennai

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:

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

Any ring with $\text{ker } \varphi = 0$ has characteristic $0$. Otherwise characteristic is $n$ where $1 + 1 + \ldots \text{ (n times)} = 0$. $\mathbb{R}$, $\mathbb{C}$, 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 $R'$.

Adjunction of elements

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, although it divides $6 = (1 + \sqrt{-5}) (1 - \sqrt{-5})$, it does not divide either factor.

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

Gauss Primes

The ring of Gauss Integers given by $\mathbb{Z}[i]$ has units $\{\pm 1, \pm i\}$: $3$ is an example of a Gauss prime, while $5 = (2 + i)(2 - i)$ is not.

  1. The prime number $p$ is either a Gauss prime, or the product of complex conjugate Gaussian primes.
  2. Let $\pi$ be a Guass prime. Then, $\pi \bar{\pi}$ is either a prime integer or the square of a prime integer.

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. The cube root of unity $\zeta$ is an example of an algebraic number since it is the root of the monic $x^2 + x + 1$.

Algebraic integers in the quadratic field $\mathbb{Q}[\sqrt{d}], \alpha = a + b \sqrt{d}$ are of two types:

  1. If $d = 2 \text{ or } 3 \text{ mod } 4$, then $a, b \in \mathbb{Z}$
  2. If $d = 1 \text{ mod } 4$, then $a, b \in \mathbb{Z} + \frac{1}{2}$

$\zeta$ is an integer of the second type.