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 $b = 0$. Thus, if you have any two nonzero elements, their product is nonzero.
$\mathbb{Z}$ is an integral domain. However, $\mathbb{Z}/6\mathbb{Z}$ is not because the class of $2$ and the class of $3$ are such that $[2][3] = [6] = [0]$.
Fields
We call $R$ a field if
- it is commutative,
- non-zero elements have (multiplicative) inverses (technically stronger than ring requirement), and
- $0 \neq 1$.
Subrings
We say $S \subset R$ is a subring if is a ring under the same operations. Importantly, this means $0$ and $1$ are in the subring. Apparently, some “subrings” can have different identities from the large ring, which is why we specify “same operations.”
Example: $\mathbb{Z}$ is a subring of $\mathbb{Q}$.
Ring Homomorphisms
A ring homomorphism is a map which is compatible with the operations. In other words,
$$\phi : R \to R^\prime$$
is such that
- $\phi(0) = 0$,
- $\phi(a + b) = \phi(a) + \phi(b)$,
- $\phi(1) = 1$, and
- $\phi(a*b) = \phi(a)*\phi(b)$.
Some people drop the third condition, but we prefer to keep it. Additionally, condition $4$ implies condition $1$. Remember that the elements and operations on the right side are different from the left.
Fact: $S$ is a subring of $R$ is equivalent to saying $S \xhookrightarrow{} R$ is a ring homomorphism, where $\xhookrightarrow{}$ is the inclusion map.
Fact: If $\phi : R \to R^\prime$ is a homomorphism, then $$\text{im}(\phi) = \{\phi(r) \mid r \in R\} \subseteq R^\prime$$ is a subring or $R^\prime$, and $$\text{ker}(\phi) = {r \in R \mid \phi(r) = 0} \subseteq R$$ is not a subring or $R$.
Group Homomorphisms vs. Ring Homomorphisms
If $G$ is any group, and $S$ is the trivial group with $1$ element, then there is only one homomorphism from $G$ to $S$ because every element in $G$ must map to the identity element in $S$.
Additionally, there is only one homomorphism from $S$ to $G$ because the identity in $S$ must map to the identity in $G$.
For rings, let $S$ be the trivial ring with $1$ element, where $0 = 1$. Then, there is a unique map from $R \to S$.
There is always a unique homomorphism from $\mathbb{Z} \to R$. We can do this by setting $\phi(0) = 0_R$ and $\phi(1) = 1_R$ and setting $\phi(2) = \phi(1) + \phi(1) = 1_R + 1_R$ and so on.
Characteristic of Rings
Let $\phi$ be a homomorphism from $\mathbb{Z} \to R$. Then,
$$\text{char} R = \begin{cases} 0 & \phi \text{ is injective}, \phi(n) \neq 0 \text{ for } n \neq 0 \\ n & \phi(n) = 0, n \text{ is the smallest such } n. \end{cases}$$
For integral domains, the characteristic is either $0$ or a prime number.
Products of Groups vs. Products of Rings
If $G_1$ and $G_2$ are groups, then we are forced to make $G_1\times G_2$ such that
$$(g,h)*(g^\prime,h^\prime) = (gg^\prime,hh^\prime).$$
This also allows us to make projections like $p_1((g,h)) = g$. Importantly, there are also two maps, $\iota_1$ and $\iota_2$ which go in the opposite direction. We are forced to let the other element in the output be the identity of the other group:
$$\begin{aligned} \iota_1(g) &= (g,e_2), \\ \iota_2(h) &= (e_1,h). \end{aligned}$$
In the case of rings, if we let $R_1$ and $R_2$ be rings, $R_1\times R_2$ is a ring, and $0$ in the ring $R_1\times R_2$ is $(0,0)$ and the $1$ is $(1,1)$. This time, we have
$$(g,h) + (g^\prime,h^\prime) = (g + g^\prime,h + h^\prime) \text{ and } (g,h)*(g^\prime,h^\prime) = (gg^\prime,hh^\prime).$$
This is pretty expected though one would still have to verify the ring properties.
Additionally, we have the same projection mappings. However, $\iota_1 : R_1 \to R_1 \times R_2$ is not a ring homomorphism because while the image of the function is closed uner addition and multiplication, it is not a subring.