algebra notes

3.1

rings

a ring is a nonempty set $R$ with two operations (addition & multiplication) that satisfy:

• closure of addition
• associativity of addition
• commutativity of addition
• existence of additive identity
• existence of additive inverses
• closure of multiplication
• associativity of multiplication
• distributivity

a ring is different from a field because it doesn’t guarantee:

• commutativity of multiplication
• existence of multiplicative identity
• existence of multiplicative inverse

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we can extend the definition of a ring to define a commutative ring which is everything a ring is, with commutative multiplication.

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we can extend the definition of a ring to define a ring with identity which is everything a ring is, with the existence of a multiplicative identity.

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ring examples galore

• the set of even numbers, $E$ is an example of a commutative ring with the usual addition and multiplication as defined on the integers. we have an additive identity (0), but no multiplicative inverse or identity.
• the set of odd number, $O$ fails the first axiom. the sum of two odd numbers is not odd!
• the set $\text{Mat}_{2 \times 2}(\mathbb{R})$ is a ring with identity. in general, if $R$ is a commutative ring with identity, then the set of all $n \times n$ matrices with entries in $R$ form a non-commutative ring with identity.

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integral domain

an integral domain is a commutative ring with identity $1 \neq 0$ such that if the product of two elements is 0, then one of the elements has to be $0$.

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integral domain examples galore

$\mathbb{Z}$ is an integral domain. $\mathbb{Z}_p$ when $p$ is prime is also an integral domain.

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fields

a field is a commutative ring with identity that also has multiplicative inverses.

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theorem 3.1

given two rings $R$ and $S$ the cartesian product defined as follows ensures that $R \times S$ is a ring as well.

$$(r_1, s_1) + (r_2, s_2) = (r_1 + r_2, s_1 + s_2) \\ (r_1, s_1) \cdot (r_2, s_2) = (r_1 \cdot r_2, s_1 \cdot s_2)$$

if $R$ and $S$ are commutative rings, then $R \times S$ is commutative as well. similarly, if $R$ and $S$ are rings with identities, so is $R \times S$.

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subrings

when a subset $S$ of a ring $R$ also satsfies the properties of a ring, we call $S$ a subring of $R$.

example:

let $T$ be the ring of all functions from $\mathbb{R}$ to $\mathbb{R}$. then, the subset $S$ consisting of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$ is a subring of $T$.

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theorem 3.2

suppose that $R$ is a ring and $S$ is a subset of $R$ such that:

• $S$ is closed under addition
• $S$ is closed under multiplication
• $0_R \in S$
• $a \in S \implies -a \in S$

then, $S$ is a subring of $R$.

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subrings example galore

• the set $\{0, 3\}$ in $\mathbb{Z}_6$ forms a subring.
• the set $\mathbb{Z}[\sqrt{2}] = \{a + b\sqrt{2} : a,b \in \mathbb{Z}\}$ is a subring of $\mathbb{R}$.

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3.2

theorem 3.3

additive inverses are unique.

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theorem 3.4

$a + b = a+ c \implies b=c$

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theorem 3.5

given $a, b \in R$, we have the following:

• $a \cdot 0 = 0 \cdot a = 0$
• $a(-b) = -ab$, $(-a)b = -ab$
• $-(-a) = a$
• $-(a+b) = -a-b$
• $-(a-b) = -a+b$
• $(-a)(-b) = ab$

furthermore, if $R$ has an identity:

$$(-1)a = -a$$

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cuidado!

in a non commutative ring:

$$(a + b)^2 = a^2 + ab + ba + b^2$$

but in a commutative ring, we have:

$$(a + b)^2 = a^2 + 2ab + b^2$$

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theorem 3.6

let $S$ be a subset of a ring $R$. then $S$ is a subring of $R$ if:

• $S$ is closed under subtraction
• $S$ is closed under multiplication

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units

an element $a$ in a ring is called a unit if it has a multiplicative inverse $a^{-1}$

$a \cdot a^{-1} = a^{-1} \cdot a = 1$

the only units in $\mathbb{Z}$ are $1$ and $-1$.

the units in $\mathbb{Z}_{15}$ are $1, 2, 4, 7, 8, 11, 13, 14$

every non zero element in a field is a unit

units in sets of matrices usually satisfy $\det(A) \neq 0$.

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zero divisors

an element $a$ in a ring is called a zero divisor if:

• $a \neq 0$
• there exists a nonzero element $b$ such that $ab = 0$ or $ba = 0$.

$b$ doesn’t necessarily need to be unique.

examples:

• $2, 3$ are zero divisors in $\mathbb{Z}_6$.

an integral domain contains no zero divisors by definition duhhhhh

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theorem 3.7

cancelation is valid in any integral domain. if $a \neq 0$ and $ab = ac$, then $b=c$.

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theorem 3.8

every field is an integral domain.

the converse of this if false in general. for example, $\mathbb{Z}$ is an integral domain that is not a field, but this is true in the finite case as we see in the next theorem. patience is key my friend.

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theorem 3.9

every finite integral domain is a field.

proof

let $R$ be our finite integral domain. we know that $R$ is already a commutative ring with identity. so it suffices to prove the existence of multiplicative inverses. since $R$ has a finite number of elements, let’s enumerate them as follows: $\{a_1, a_2, \dots, a_n\}$. given some arbitrary element, $a_i$, we want to show that it’s multiplicative inverse exists in $R$.

now, consider all the pairwise products as follows: $\{a_ia_1, a_ia_2, \dots, a_ia_n\}$. we claim these are $n$ distinct elements. assume for the sake of contradiction there exists some distinct $j, k \in [0, n]$ such that $a_i a_j = a_i a_k$, but by cancleation law (theorem 3.7) we would get that $a_j = a_k$ which is a contradiction.

so we have listed all our elements of $R$ as follows $\{a_ia_1, a_ia_2, \dots, a_ia_n\}$ but we don’t know what order this is. but we know that $1$ exists somewhere in this list. and thus, we’ve found a multiplicative inverse of an arbitrary element $a_i$.

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3.3

isomorphism

a ring $R$ is isomorphic to ring $S$ if we can conveniently relabel elements of one as the other and obtain the same addition and multiplication tables. rigorously,

$R$ is isomorphic to $S$ if there exists a bijective function $f: R \rightarrow S$ such that

$f(a+b) = f(a) + f(b)$

$f(ab) = f(a) f(b)$ for all $a,b \in R$ .

further, $f$ is called the isomorphism.

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the set of $2 \times 2$ matrices that take real entries of the form

$$\begin{pmatrix} a & b \\ -b & a \end{pmatrix}$$

is isomorphic to $\mathbb{C}$ wtfffff

the complex conjugation map is another example of an isomorphism

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homomorphism

a function $f: R \rightarrow S$ is called a homomorphism if

• $f(a+b) = f(a) + f(b)$
• $f(ab) = f(a) f(b)$

a homomorphism may fail to be injective or surjective.

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the zero map is a homomorphism!

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the function $f:\mathbb{Z} \rightarrow \mathbb{Z}_6$ given be $f(a) = [a]$ because of how addition and multiplication are defined in $\mathbb{Z}_6$.

$f$ is surjective, but not injective because $[1]$’s pre image could be $1, 7, 13$ or anything of the form $1 + 6k$ for some integer $k$.

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theorem 3.10

let $f:R \rightarrow S$ be a homomorphism of rings. then,

• $f(0_R) = 0_S$
• $f(-a) = -f(a), \forall a \in R$
• $f(a-b) = f(a) - f(b), \forall a, b \in R$

further, if $R$ is a ring with identity and $f$ is surjective, then:

• $S$ is a ring with identity $f(1_R)$
• if $u$ is a unit in $R$, then $f(u)$ is a unit in $S$ and $f(u)^{-1} = f(u^{-1})$

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theorem 3.11

if $f:R \rightarrow S$ is a ring homomorphism, then the image of $f$ is a subring of $S$.

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4.1

theorem 4.1

if $R$ is a ring, then there exists a ring $T$ containing an element $x$ such that $x \notin R$ and the following hold:

• $R$ is a subring of $T$
• $xa = ax, \forall a \in R$. this does not necessarily imply that $T$ is commutative!
• the set $R[x]$ of all elements of $T$ of the form:

$$a_0 + a_1x+a_2x^2 + \dots + a_nx^n , n \geq 0, a_i \in R$$

is a subring of $T$ that contains $R$

• the representation of elements of $R[x]$ is unique
• $a_0 + a_1x+a_2x^2 + \dots + a_nx^n = 0$ iff $a_i = 0_R$

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the elements of $R[x]$ are called polynomials with coefficients in $R$ and the elements, $a_i$ are called coefficients.

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theorem 4.2

if $R$ is an integral domain and $f(x)$ and $g(x)$ are nonzero polynomials in $R[x]$ then

$$\deg[f(x)g(x)] = \deg (f(x)) + \deg (g(x))$$

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corollary 4.3

if $R$ is an integral domain, so is $R[x]$.

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corollary 4.4

let $R$ be a ring. if $f(x)$, $g(x)$, and $f(x)g(x)$ are nonzero in $R[x]$, then we have

$$\deg[f(x)g(x)] \leq \deg (f(x)) + \deg(g(x))$$

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corollary 4.5

let $R$ be an integral domain and $f(x) \in R[x]$. then, $f(x)$ is a unit in $R[x]$ iff $f(x)$ is a constant polynomial that is a unit in $R$. in particular, if $F$ is a field, the units in $F[x]$ are the nonzero constants in $F$.

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theorem 4.6

division algo equivalent

let $f(x), g(x) \in F[x]$. then there exists unique $q(x), r(x) \in F[x]$ such that

$$f(x) = q(x)g(x) + r(x)$$

either $r(x) = 0_F$ or $\deg(r(x)) < \deg(g(x))$

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4.2

theorem 4.7

let $a(x), b(x) \in F[x]$ with $b(x)$ nonzero.

• if $b(x)$ divides $a(x)$ then $c \cdot b(x)$ divides $a(x)$ for each nonzero $c \in F$
• every divisor of $a(x)$ has degree less than or equal to $\deg(a(x))$

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polynomial gcd definition

let $a(x), b(x) \in F[x]$ not both zero. the gcd of $a(x)$ and $b(x)$ is defined as the monic polynomial of highest degree that divides both $a(x)$ and $b(x)$

let $d(x) = \gcd(a(x), b(x))$. then:

• $d(x)|a(x)$ and $d(x)| b(x)$
• if $c(x)|a(x)$ and $c(x)|b(x)$ then $\deg(c(x)) \leq \deg (d(x))$

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theorem 4.8 (bezout)

let $a(x), b(x) \in F[x]$. then there is a unique gcd $d(x)$. furthermore, there are polynomials $u(x), v(x)$ such that $d(x) = a(x)u(x) + b(x)v(x)$.

similarly, we can derive a division algorithm for polynomials. let $r_0 = a(x), r_1 = b(x) \in F[x]$, not both zero. perform division as follows:

$$r_0(x) = q_1(x)r_1(x) + r_2(x) \\ r_1(x) = q_2(x)r_2(x) + r_3(x) \\ \vdots \\ r_{n-2}(x) = q_{n-1}(x)r_{n-1}(x) + r_n(x) \\ r_{n-1}(x) = q_x(x)r_n(x) + r_{n+1}(x)$$

where $r_{n+1}(x) = 0$. then $\gcd(a(x), b(x))$ is the monic associate of the last nonzero remainder, $r_n(x)$

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corollary 4.9

let $a(x), b(x) \in F[x]$ and let $d(x) = \gcd(a(x), b(x))$. then the following are true:

• $d(x) | a(x)$ and $d(x) | b(x)$
• if $c(x)|a(x)$ and $c(x) | b(x)$ then $c(x) | d(x)$

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theorem 4.10

let $a(x), b(x), c(x) \in F[x]$ and assume $a(x) | b(x)c(x)$ and assume $a(x)$ and $b(x)$ are relatively prime. then, $a(x) | c(x)$.

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4.3

associates

we say an element $a$ is an associate of $b$ if $a=bu$ for some unit $u \in R$

for example, in $\mathbb{Z}$ the associates of $n$ are $\pm n$

in $F[x]$ the associates of $p(x)$ are $c \cdot p(x)$ for non zero $c$

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irreducible polynomials

$p(x) \in F[x]$ is irreducible if:

• $p(x)$ is a non constant polynomial
• the only factors of $p(x)$ are the nonzero constant polynomials and the associates of $p(x)$

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theorem 4.11

a non zero polynomial $f(x) \in F[x]$ is reducible iff it can be expressed as a product of two polynomials of lower degree.

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theorem 4.12

for a non constant polynomial in $F[x]$ the following are equivalent:

• $p(x)$ is irreducible
• if $p(x) | b(x)c(x)$ then $p(x)|b(x)$ or $p(x)|c(x)$
  • if an irreducible polynomial divides a product of two things, then the polynomial has to divide one of those two things
• if $p(x) =r(x)s(x)$ then $r(x)$ or $s(x)$ is a nonzero constant polynomial
  • if an irreducible polynomial can be written as a product, one of the two things in the product is a constant

note: $p(x)$ being irreducible is sorta the “prime” equivalent in $\mathbb{Z}$

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corollary 4.13

if $p(x)|a_1(x)a_2(x)\cdots a_n(x)$ then $p(x)|a_i(x)$ for some $i$ in $1\leq i \leq n$

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theorem 4.14

every nonconstant polynomial $f(x)$ is the product of irreducible polynomials. this factorization is unique in this sense:

if $f(x) = p_1(x) \cdots p_n(x)$ and $f(x) = q_1(x) \cdots q_s(x)$ then

• $r=s$
• $p_i(x)$ is an associate of $q_i(x)$ for all $i=1,\dots, r$

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unique factorization domains

integral domains in which every non-unit element factors uniquely into a product of irreducibles are called unique factorization domains.

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hierarchy

so generally speaking, if you start with an integral domain, here is how you get to a field:

• integral domain (no zero divisors)
• unique factorization domain (every element can be uniquely factored into “primes”)
• principal ideal domain (every ideal is principal)
• euclidean domain (division always gets you a quotient and a remainder)
• field (inverses always exist)

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4.4

polynomial induced functions

usually given some polynomial $f(x) \in R[x]$ induces a polynomial function. if $f(x) = a_0 + a_1x+a_2x^2 + \dots + a_nx^n$ then the polynomial induced function is $g:R \rightarrow R$ where $g(r) = a_0 + a_1r + a_2r^2 + \dots + a_nx^n$

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insert normal definition of a root/zero for a polynomial

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theorem 4.15

let $f(x) \in F[x]$ and $a \in F$. the remainder when $f(x)$ is divided by the polynomial $x-a$ is $f(a).$

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theorem 4.16

let $f(x) \in F[x]$ and $a \in F$. then $a$ is a root of the polynomial $f(x)$ iff $x-a$ is a factor of $f(x)$ in $F[x]$

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corollary 4.17

a non zero polynomial $f(x) \in F[x]$ of degree $n$ has at most $n$ roots in $F$

this is not necessarily true for more general rings.

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corollary 4.18

an irreducible polynomial $f(x) \in F[x]$ of degree $\geq 2$ has no roots.

• the converse is generally not true. it is NOT necessarily true that if a polynomial has no roots, then it is irreducible.

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corollary 4.19

let $f(x) \in F[x]$ be a polynomial of degree $2$ or $3$ then $f(x)$ is irreducible in $F[x]$ iff $f(x)$ has no roots in $F$

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corollary 4.20

let $F$ be an infinite field and $f(x), g(x) \in F[x]$ then $f(x)$ and $g(x)$ induce the same function from $F$ to $F$ iff $f(x) = g(x)$ in $F$

this is not necessarily true for finite fields as seen below

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4.5

$\mathbb{Z}[x]$ is a subring of $\mathbb{Q}[x]$

for every $p(x) \in \mathbb{Q}[x]$ there exists some $c \neq 0 \in \mathbb{Z}$ such that $c \cdot p(x) \in \mathbb{Z}[x]$

• basically that you can multiply $p(x)$ with the $\text{lcm}$ of the denominators
• furthermore, if you have a root of $p(x)$, it is also a root of $c \cdot p(x)$

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theorem 4.21

let $f(x) = a_0 + \dots + a_nx^n \in \mathbb{Z}[x]$, if $r \neq 0$ and the rational number $\dfrac{r}{s}$ in lowest terms is a root of $f(x)$, then $r \mid a_0$ and $s \mid a_n$

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lemma 4.22

let $f(x), g(x), h(x) \in \mathbb{Z}[x]$ with $f(x) = g(x)h(x)$. if $p$ is a prime that divides every coefficient of $f(x)$, then $p$ divides every coefficient of $g(x)$ or $p$ divides every coefficient of $h(x)$

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theorem 4.23

let $f(x) \in \mathbb{Z}[x]$. then $f(x)$ factors as a product of polynomials in $\mathbb{Q}[x]$ of degrees $m$ and $n$ iff $f(x)$ factors as a product of polynomials of degrees $m$ and $n$ in $\mathbb{Z}[x]$

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theorem 4.24

let $f(x) \in \mathbb{Z}[x]$ be non constant. if there is a prime $p$ such that

• $p \mid a_0, a_1, \dots, a_{n-1}$
• $p \nmid a_n$
• $p^2 \nmid a_0$

then $f(x)$ is irreducible in $\mathbb{Q}[x]$

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theorem 4.25

let $f(x) \in \mathbb{Z}[x]$. let $p$ be a positive prime such that $p \nmid a_n$. if $\overline{f}(x)$ is irreducible in $\mathbb{Z}_p[x]$, then $f(x)$ is irreducible in $\mathbb{Q}[x]$

• find some prime number such that it doesn’t divide the leading coefficient
• reduce your polynomial coefficients from $\mathbb{Z}$ to $\mathbb{Z}_p$
• check irreducibility there

important if you do this and find out that your polynomial IS reducible in $\mathbb{Z}_p$ this gives you no conclusion for if this polynomial is reducible in $\mathbb{Q}[x]$

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5.1

theorem 5.1

defines the congruence relation for polynomials (same shit as integers)

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theorem 5.2

let $p(x) \in F[x]$ be a non zero polynomial. assume $f_1(x) \equiv g_1(x) \pmod{p(x)}$ and $f_2(x) \equiv g_2(x) \pmod{p(x)}$ we have:

$$f_1(x) + f_2(x) \equiv g_1(x) + g_2(x) \pmod{p(x)} \\ f_1(x) \cdot f_2(x) \equiv g_1(x) \cdot g_2(x) \pmod{p(x)}$$

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define a congruence/modulo class the same way we do in integers

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theorem 5.9

given some non constant polynomial $p(x) \in F[x]$ and if $f(x)$ is relatively prime to $p(x)$ then $[f(x)]$ is a unit in $F[x]/p(x)$

further, if $f(x)$ is not relatively prime to $p(x)$ and $p(x) \nmid f(x)$ then $[f(x)]$ is a zero divisor in $F[x]/p(x)$

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5.3

theorem 5.10

let $p(x) \in F[x]$ be a non constant polynomial. equivalent:

• $p(x) \in F[x]$ is irreducible
• $F[x]/p(x)$ is a field
• $F[x]/p(x)$ is an integral domain

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theorem 5.11

let $p(x)$ be an irreducible polynomial in $F[x]$. then $F[x]/p(x)$ is an extension field of $F$ that contains a root of $p(x)$

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corollary 5.12

given a non constant polynomial $f(x) \in F[x]$, there exists an extension field $K$ of $F$ that contains a root of $f(x)$

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6.1

ideals

let $R$ be a ring. a subring $I \subseteq R$ is called an ideal if $r \in R$ and $a \in I$, then $ar \in I$ and $ra \in I$

• double absorption ($ar, ra$) is only necessary when your ring is non commutative. if it’s commutative, then you just need to check one of them

the set consisting of just $0_R$ is the zero ideal. the whole ring $R$ is also an ideal.

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theorem 6.1

$\empty \neq I \subseteq R$ is an ideal iff $I$ is closed under subtraction and $I$ satisfies double absorption

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theorem 6.2

given a commutative ring with identity, $c \in R$ and $I$ which is the set of all multiples of $c$ in $R$, then $I$ is an ideal.

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theorem 6.3

given a commutative ring with identity $R$ and some $n$ elements from $R$. the set $I = \{r_1c_1 + \dots r_nc_n \mid r_1, \dots, r_n \in R\}$ is an ideal in $R$. this is denoted as $(c_1, \dots, c_n)$

• such an ideal is called finitely generated
• principal ideal is when $n=1$
• the generators of a finitely generated ideal don’t need to be unique!

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every euclidean domain is a principal ideal domain. examples of principal ideal domains include $\mathbb{Z}, F[x]$.

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congruence

we say that $a \equiv b \pmod I$ when $a -b \in I$

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congruence relation

same shit as you do with integers and polynomials

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congruence class notation

the set of all elements of $R$ that are congruent to an element $a$ is denoted as $a+I$. this is just notation and does not mean that we are adding an element to an ideal. instead of writing $[a]$, we write $a + I$ 💯

also, a congruence class modulo $I$ is also called a left coset of $I$ in $R$.

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theorem 6.6

let $I$ be an ideal in a ring $R$, and let $a, b \in R$. then $a \equiv b \pmod I$ iff $a+I = b+ I$

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corollary 6.7 67676767676767

let $I$ be an ideal in a ring $R$, then two cosets of $I$ are either disjoint or identical.

• i think this is the same thing as whenever you do something like $\mathbb{Z}_n$ you don’t have two integers map to the same equivalence class unless they’re fr equivalent. i still struggle to see why this is corollary is brought up and two congruence class’s intersection would be non empty

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the set of all cosets of $I$ is denoted $R/I$

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now define addition and multiplication just as we did with integers and polynomials

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theorem 6.9 this is very general

• $R/I$ is a ring with addition and multiplication of cosets defined as we do with equivalence classes
• if $R$ is commutative, so is $R/I$
• if $R$ has an identity, so does $R/I$

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back to homo

the image of a ring homomorphism $f:R \rightarrow S$ is always a subring of $S$, but not necessarily an ideal of $S$

• example: $\mathbb{Z} \rightarrow \mathbb{Q}$, the integers are not an ideal in the rationals

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theorem 6.10

let $f: R \rightarrow S$ be a ring homomorphism, then $\ker(f)$ is an ideal in $R$

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theorem 6.11

let $f: R \rightarrow S$ be a ring homomorphism with kernel $K$. then $K=\{0_R\}$ iff $f$ is injective.

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theorem 6.12

let $I$ be an ideal in a ring $R$. then the map $\pi:R \rightarrow R/I$ given by $\pi(r)=r+I$ is a surjective homomorphism with the kernel $I$

$\pi$ is also called the natural homomorphism whatever tf that means

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theorem 6.13

let $f: R \rightarrow S$ be a surjective ring homomorphism with kernel $K$. then the ring $R/K$ is isomorphic to $S$

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now, we want to determine conditions necessary to see when quotient rings are integral domains or fields. basically, generalize the notion of prime numbers, irreducible polynomials, to _______ 😱

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6.3

prime ideal

let $I \subseteq R$ be an ideal. we say $I$ is prime if $I \neq R$ and whenever $bc \in I$ for some $b,c \in R$ then $b \in I$ or $c \in I$

if $I = R$ then $I$ contains a unit

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theorem 6.14

an ideal $I$ is prime iff $R/I$ is an integral domain

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theorem 6.15

let $M$ be an ideal in a commutative ring with identity $R$, then $M$ is a maximal ideal iff the $R/M$ is a field.

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corollary 6.16

in a commutative ring with identity, every maximal ideal is prime.