Definition 6.1. A field is a triple $(K,+,\cdot)$ consisting of a set $K$ such that $(K,+)$ is an additive abelian group and $(K\backslash\{0\},\cdot)$ is an abelian multiplicative group.
Example 6.2. $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$, $\mathbb{Q}_{p}$ are all fields. In particular, they are fields of characteristic 0.
Definition 6.3. A homomorphism of fields $f:K\rightarrow L$ is a function $K\rightarrow L$ such that $f(x+y)=f(x)+f(y)$, $f(xy)=f(x)f(y)$, and $f(1)=1$ for all $x,y\in K$. A homomorphism of fields is an isomorphism if it is bijective.
Proposition 6.4. Every homomorphism of fields is injective.
Proof. Since there is no non-zero proper ideal of a field.
Definition 6.5. Let $K$ be a field. A field extension of $K$ is a field $E$ such that $K$ is a subfield of $E$. We denote this by $K\subset E$.
Example 6.6. $\mathbb{Q}\subseteq\mathbb{R}$, $\mathbb{R}\subseteq\mathbb{C}$, $\mathbb{Q}\subseteq\mathbb{Q}_{p}$ are field extensions.
Definition 6.7. Let $K\subset E$ and $K\subset L$ be two field extensions. A $K$-homomorphism $f:E\rightarrow L$ is a field homomorphism $E\rightarrow L$ such that $f(x)=x$ for all $x\in K$. A $K$-isomorphism is a $K$-homomorphism that is a field isomorphism.
A subring of a domain is clearly a domain. And we want to show that every domain is a subring of a field. This motivates us to construct the fraction field of a domain by ``adjoining inverses''. Let $R$ be a domain. We define a relation on $R\times R\backslash\{0\}$ by $$ (x,y)\sim(z,h)\textrm{ if and only if }xz=yh. $$
It is easy to check that the relation is an equivalence relation. Then we write $\frac{x}{y}$ or $x/y$ for an equivalence class of $(x,y)\in R\times R\backslash\{0\}$ and called it fraction. We denote by $\textrm{Frac}(R)$ the set of all equivalence classes. Next, we define addition and multiplication on $\textrm{Frac}(R)$ by $$ \frac{x}{y}+\frac{z}{h}=\frac{xh+zy}{yh}, \ \ \ \frac{x}{y}\frac{z}{h}=\frac{xz}{yh}, \ \ x,z\in R, y,h\in R\backslash\{0\}. $$
Definition/Proposition 6.8. Let $R$ be a domain. Then ${\rm{Frac}}(R)$ with the operations defined above is a field, which is called the fraction field of $R$. And there is a canonical injective homomorphism $$ R\rightarrow {\rm{Frac}}(R),\ \ x\mapsto \frac{x}{1}, $$ which shows that $R$ is a subring of the field ${\rm{Frac}}(R)$.
Remark 6.9. If $R$ is not a domain, but a commutative ring with identity, then we can construct the fraction ring in a similar process.
Example 6.10. If $R=\mathbb{Z}$, then $\textrm{Frac}(\mathbb{Z})=\mathbb{Q}$. Moreover, if $R=\mathscr{O}_{K}$ is the valuation ring of the field $K$, then $\textrm{Frac}(\mathscr{O}_{K})=K$. In particular, if $R=\mathbb{Z}_{p}$, then $\textrm{Frac}(\mathbb{Z}_{p})=\mathbb{Q}_{p}$.
Proposition 6.11. Let $R$ be a domain. The fraction field ${\rm{Frac}}(R)$ corresponds to a field $K$ whose elements are of the form $xy^{-1}$ for $x,y\in R$, i.e. there is an isomorphism $$ {\rm{Frac}}(R)\xrightarrow{\sim} K, \ \ \frac{x}{y}\mapsto xy^{-1}. $$
Proof. Let $K$ be a field containing $R$. So it must contain all elements of the form $xy^{-1}$ for $x,y\in R$. Clearly, the map ${\rm{Frac}}(R)\rightarrow K$ is a homomorphism. If every element in $K$ can be written in the form $xy^{-1}$, then we obtain the isomorphism as desired.
Let $f:R\rightarrow R'$ be a homomorphism of domains. Then we have a homomorphism of fraction fields $$\textrm{Frac}(f):\textrm{Frac}(R)\rightarrow \textrm{Frac}(R'),\ \ \frac{x}{y}\mapsto\frac{f(x)}{f(y)}.$$
Thus we can view Frac(-) as a functor, which assigns to each domain $R$ its fraction field Frac$(R)$ and to each homomorphism of domains $f$ a homomorphism of fraction fields Frac$(f)$, i.e. Frac(-) is a functor from the category of domains to the category of fraction fields $$\textrm{Frac}: Domains\rightarrow Fraction-Fields.$$
Definition 6.12. Let $K\subset E$ be a field extension. An element $\alpha\in E$ is algebraic over $K$ if there exists $f\in K[X]$ such that $f(\alpha)=0$. And $\alpha$ is transcendental over $K$ if $f(\alpha)\neq0$ for all $f\in K[X]$. An algebraic extension of $K$ is a field extension $K\subset L$ whose elements are all algebraic over $K$. A transcendental extension of $K$ is a field extension $K\subset L$ if there exists $\alpha\in L$ that is transcendental over $K$.
Definition 6.13. A field $K$ is algebraically closed if every non-constant polynomial over $K$ has a root in $K$.
Example 6.14. $\mathbb{C}$ is an algebraically closed field.
Definition 6.15. Let $K$ be a field. The algebraic closure of $K$, denoted by $\overline{K}$, is the smallest algebraically closed field that contains $K$.
Remark 6.16. The algebraic closure of $K$ is indeed not unique. But since all algebraic closures are $K$-isomorphic, we shall call "the algebraic closure" of $K$ rather than "an algebraic closure".
Example 6.17. $\mathbb{C}=\overline{\mathbb{R}}$ is the algebraic closure of $\mathbb{R}$