Definition 1.3.1 Let $I$ be an ideal in a commutative ring $R$. If $a\in R$, then the coset $a+I$ is the subset
$$a+I=\{a+i;i\in I\}.$$
The coset $a+I$ is often called $a\bmod I$. The family of all cosets is denoted by $R/I$:
$$R/I=\{a+I;a\in R\}$$
The relation $\equiv$ on $R$, defined by $a\equiv b$ if $a-b\in I$, is called congruence mod $I$; it is an equivalence relation on $R$, and its equivalence classes are the cosets.

Proposition 1.3.1 Let $I$ be an ideal in a commutative ring $R$. If $a,b\in R$, then $a+I=b+I$ if and only if $a-b\in I$. In particular, $a+I=I$ if and only if $a\in I$.

Definition 1.3.2 Let $R$ be a commutative ring and $I$ be an ideal in $R$. Define addition $+:R/I\times R/I\to R/I$ by
$$+:(a+I,b+I)\mapsto a+b+I,$$
and multiplication $\cdot:R/I\times R/I\to R/I$ by
$$\cdot:(a+I,b+I)\mapsto ab+I$$
It's easy to prove addition and multiplication $R/I\times R/I\to R/I$ are well-defined.

Theorem 1.3.1 If $I$ is an ideal in a commutative ring $R$, then $R/I$ is a commutative ring.

Definition 1.3.2 The commutative ring $R/I$ equipped with addition and multiplication is called quotient ring of $R$ modulo $I$; it is usually pronounced $R\bmod I$.

Definition 1.3.3 Let $I$ be an ideal in a commutative ring $R$. The natural function is the function $\pi:R\to R/I,a\mapsto a+I$.

Proposition 1.3.2 If $I$ is an ideal in a commutative ring $R$, then the natural map $\pi: R\to R/I$ is a surjective homomorphism and $\ker \pi = I$.

Corollary 1.3.1 Given an ideal $I$ in a commutative ring $R$, there exists a commutative ring $A$ and a (surjective) homomorphism $\varphi:R\to A$ with $I=\ker \varphi$.

If we set $A=R/I$, then the natural map $\pi:R\to R/I$ is a homomorphism with $I=\ker \pi$.

Theorem 1.3.2 (First Isomorphism Theorem) Let $R$ and $A$ be commutative rings. If $\varphi:R\to A$ is a homomorphism, then $\ker \varphi$ is an ideal in $R$, and $\operatorname{im}\varphi$ is a subring of $A$, and
$$R/\ker\varphi\cong \operatorname{im}\varphi$$
Let $I=\ker \varphi$.

In the commutative diagram below, $\pi:R\to R/I$ is the natural map, $i:\operatorname{im}\varphi\to A$ is the inclusion, and define $\tilde \varphi:R/I\to A,a+I\mapsto \varphi(a)$ ($a+I=b+I\Leftrightarrow b-a\in I\Leftrightarrow \varphi(a)=\varphi(a)+\varphi(b-a)=\varphi(b)$ show that $\tilde\varphi$ is well-defined).
$$% https://darknmt.github.io/res/xypic-editor/#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 \xymatrix{ R \ar@{->}[d]_{\pi} \ar@{->}[r]^{\varphi} & A \\ R/I \ar@{->}[r]_{\tilde\varphi} & \operatorname{im} \varphi \ar@{->}[u]_{i} }$$
Definition 1.3.4 If $K$ is a field, the intersection of all the subfields of $K$ is called the prime field of $K$.

For example, the prime field of $\mathbb C$ is $\mathbb Q$, because every subfield of $\mathbb C$ contains $\mathbb Q$.

Definition 1.3.5 If $X$ is a subset of a field, define $\left<X\right>$, the subfield of generated by $X$, to be the intersection of all the subfields containing $X$.

Proposition 1.3.3 Let $K$ be a field with identity $\mathscr l$, and let $\chi:\mathbb Z\to K,n\mapsto n\mathscr l$ be a homomorphism.

1. Either $\operatorname{im}\chi\cong \mathbb Z$ or $\operatorname{im}\chi \cong \mathbb Z_p$ for some prime $p$.
2. The prime field of $K$ is isomorphism to $\mathbb Q$ or to $\mathbb Z_p$ for some prime $p$.

Proof.

1. Since every ideal in $\mathbb Z$ is principal, $\ker \chi = (m)$ for some integer $m\ge 0$.

The First Isomorphism Theorem gives $\operatorname{im}\chi\cong \mathbb Z/(m)$.

If $m=0$, $\operatorname{im}\chi\cong\mathbb Z/(0)\cong\mathbb Z$.

If $m\ne 0$, $\operatorname{im}\chi\cong\mathbb Z/(m)=\mathbb Z_m$. And then we will prove $m$ is a prime.

Let's suppose $m$ isn't a prime. Then there are two integers $a,b>0$ such that $ab=m$.

Therefore $\chi(a)\chi(b)=\chi(ab)=0$, which contradicts the fact that $K$ is a domain.

2. If $\operatorname{im} \chi\cong \mathbb Z$. There is a field $Q\cong \operatorname{Frac}(\mathbb Z)=\mathbb Q$ with $\operatorname{im}\chi\subset Q\subset K$. Now $Q$ is the prime field of $K$, for it is the subfield generated by $\mathscr l$.

In case $\operatorname{im}\chi\cong \mathbb Z_p$, then $\operatorname{im}\chi$ must be the prime field of $K$, for it is a field which is obviously the subfield generated by $\mathscr l$. $\square$

Definition 1.3.6 A field $K$ has characteristic $0$ if its prime field is isomorphic to $\mathbb Q$; it has characteristic $p$ if its prime field is isomorphic to $\mathbb Z_p$ for some prime $p$.

Proposition 1.3.4 If $K$ is a finite field, then $|K|=p^n$ for some prime $p$ and some $n\ge 1$.

Proof.

The prime field of $K$ is isomorphic to $\mathbb Z_p$ for some prime $p$.

We can regard $K$ as a vector space over $\mathbb Z_p$. As $K$ is finite, it is obviously finite-dimensional.

If $\operatorname{dim}_{\mathbb Z_p}(K)=n$, then $|K|=p^n$. $\square$

Proposition 1.3.5 Let $I$ be an ideal in a commutative ring $R$. If $J$ is an ideal in $R$ containing $I$, define the subset $J/I$ of $R/I$ by
$$J/I=\{a+I;a\in I\}.$$

1. $\pi^{-1}(J/I)=J$, where $\pi:R\to R/I$ is the natural map.
2. $J/I$ is an ideal in $R/I$.
3. If $I\subset J\subset J'$ are ideals in $R$, $J/I\subset J'/I$. Moreover, if $J\ne J'$, then $J/I\ne J'/I$.

Proof.

• First, let's prove $\pi^{-1}(J/I)\subset J$.

$\forall a\in \pi^{-1}(J/I),\pi(a)=a+I\in J/I$.

Then there is $b\in J,b+I=a+I\Rightarrow a-b\in I\subset J$.

Therefore $a\in J$.

• Second, let's prove $J\subset \pi^{-1}(J/I)$.

$\forall a\in J,\pi(a)=a+I\in J/I\Rightarrow a\in \pi^{-1}(J/I)$.

1. $\forall a+I\in J/I,\forall b+I\in R/I,(a+I)(b+I)=ab+I$.

Since $ab\in J\Rightarrow ab+I\in J/I$, $J/I$ is an ideal.

2. $J\subset J'\Rightarrow J/I=\pi(J)\subset \pi(J')=J'/I$.

Then $J=\pi^{-1}(J/I)\ne \pi^{-1}(J'/I)=J'\Rightarrow J/I\ne J'/I$. $\square$

Proposition 1.3.6 A commutative ring $R$ is a field if and only if $R$ only have trivial ideals, $(0)$ and $R$.

Proof.

• ($\Rightarrow$).

Let $I$ be an ideal of $R$, with $I\ne (0)$.

Then there is $a\in I\textbackslash \{0\}$ with $a$ is a unit.

Because $\forall b\in R,b=(ba^{-1})a\in I$, $I=R$.

• ($\Leftarrow$).

If $R$ isn't a field, then there is $a\in R$ with $a$ isn't a unit.

Then $(a)$ is an ideal with $1\notin (a)$, which lead to a contradiction. $\square$

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