MATH 4340: Honors Algebra

Rings Recall that rings are sets along with two operations, $+$ and $*$. Additive group: $(R,+)$ is an abelian group, which we call $R$. Multiplicative group: $(R^*,*)$ is a group. The reason we use $R^*$ is because $(R,*)$ doesn’t necessarily have inverses, so we define $$R^* = \{r \in R \mid \exists r^\prime, rr^\prime = r^\prime r = 1\}.$$ We say $R$ is called an integral domain if $R$ is commutative (meaning multiplication is commutative) and if $ab = 0$ implies $a = 0$ or $b = 0$....

Groups Wrap-Up Given a homomorphism $\varphi : G \to H$, we can define the kernel of $\phi$ as $$\text{Ker } \varphi = {g \in G \mid \varphi(g) = e},$$ where $e$ is the identity of $H$. We often say $\text{Ker } \varphi \Delta G$, which means it is a normal subgroup of $G$. Fact: $\varphi$ is injective iff $\text{Ker } \varphi = {e^\prime}$. Fact: If $\gcd(m,n) = 1$, then $$\mathbb{Z}/{mn\mathbb{Z}} \cong \mathbb{Z} / {m\mathbb{Z}} \times \mathbb{Z} / {n\mathbb{Z}},$$...

An action of $G$ on $X$ is a rule that assigns to each element $g \in G$ and each element $x \in X$ another element $g \cdot x \in X$ so that $e \cdot x = x$, and $(g_1g_2) \cdot x = g_1 \cdot (g_2 \cdot x)$. This is equivalent to giving a group homomorphism $$\phi : G_1 \to G_2$$ from $G_1$ to $G_2$. $\phi$ is an isomorphism if it is a bijection....

One example of a group is $(\mathbb{Z}, +)$. It has identity $e = 0$ and inverse $-a$ for $a$. Another example of a group is $(\mathbb{R} \setminus \{0\}, *)$ which has $0$ excluded because $0$ does not have a multiplicative inverse. Both of these are examples of Abelian groups because the order of the operands does not matter. One example of a finite group is the symmetric group on (finite) $X$, which is the set of all permutations on $X$....

The factorial is defined as $n! = 1 \cdot 2 \cdot 3 \cdots n$ for $n \geq 1$ though we may define $0!=1$. We also have binomials: $$\binom{n}{i} = \frac{n!}{i!(n-i)!}.$$ Fact: If $p$ is prime, then $p$ divides $\binom{p}{i}$ unless $i=0$ or $i=p$. We can also write out the binomial expansion: $$(a+b)^n = \sum_{i=0}^n \binom{n}{i} a^ib^{n-i}$$ where $a,b$ are “numbers.” Importantly, the binomial coefficient is counting something, specifically the number of subsets of $1, \ldots, n$ with exactly $i$ elements....