Fields

Let R be a ring. We have a map $c : \mathbb{Z} \rightarrow R$ defined as $n_{> 0 } \mapsto n\times 1_R, n_{< 0} \mapsto n \times (-1_R)$. c is a ring homomorphism and ker(c) = n for some $n \in \mathbb{Z}_{\geq 0}$. n is called the characteristic of R.

Proposition: let F be a field. Then char(F) is prime or 0.

Proof: let $c : \mathbb{Z} \rightarrow Im(c) \leq F$. By the first isomorphism theorem $\mathbb{Z}/(n) \cong Im(c) \leq F$ so (n) is a prime ideal therefore n = 0 or $n >0$. Since $\mathbb{Z}$ is a UFD, n is prime (if $n > 0$) or n = 0.

The prime subfield of a field F is the subfield of G generated by the multiplicative identity $1_F$ of F. It is isomorphic to $\mathbb{Q}$ if char(F) = 0 or $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ if char(F) = p.

If $F \leq K$ is a subring of K *where F and K are fields) we say that K is a field extension of F and denote it $K \vert F$.

Let F be a field. A F vector space is a set V with 2 operations: $+ : V^2 \rightarrow V$, $\cdot : F \rightarrow V$ that satisfies the following axioms: 1) $(V, +)$ is an abelian group 2) for all $\lambda, \mu \in F$ and $x, y \in V, (\lambda + \mu) \cdot x =$ $\lambda \cdot x + \mu \cdot x$ $\lambda \cdot (x+y) = \lambda \cdot x + \mu \cdot y$ $(\lambda \mu) \cdot x = \lambda \cdot (\mu \cdot x)$, $1_f \cdot x = x$

Lemma: if $K \vert F$ is a field extension then $(K, +, \cdot)$ is an F vector space.

dim(K) is the degree of the extension $K \vert F$ and is denoted $[K :F]$ (note that although this notation is the same notation we use for group indexes, it is not the index).

Theorem: ket F be a field, $P \in F[x]$ an irreducible polynomial There exists a field extension $K \vert F$ such that P has a root in K.