Field (Algebra)

up:: 021 MOC Algebra

A field is a triple $(\mathbb{K},+,\cdot )$, where $\mathbb{K}$ is a (non-empty) set with operations $+$ and $\cdot$ which satisfy

1. Addition is commutative, associative, has neutral element $0_{\oplus }$ and inverses
2. Multiplication is commutative, associative, has neutral element $1_{\odot }$ and inverses for all elements in $\mathbb{K}\setminus \{0\}$
3. Addition and multiplication are distributive: $(a+b)\cdot c=a\cdot c+b\cdot c$

Properties

Note that a field, then, is an Abelian Group under $(\mathbb{K},+)$, as well as under $(\mathbb{K}\setminus \{0\},\cdot )$, alongside the distributive property for both operations.

This implies that, since A group’s identity is unique and All elements of a group have a unique inverse, $0_{\oplus }$ and $1_{\odot }$ are unique, as well as its inverses under $+$ and $\cdot$.

Additive identity = multiplicative $0$

Note that one can identify the $0$‘s of $+$ and $\cdot$ with the field’s $0$. Denote them as $0_{\oplus }$ and $0_{\odot }$. Then, since for any $k\in \mathbb{K}$,

$$0_{\oplus }=k-k=1\cdot k-1\cdot k=(1-1)\cdot k=0_{\odot }\cdot k$$

it follows that $0=0_{\oplus }⟺0=0_{\odot }$. Thus, we can write it unambiguously as $0$.

Multiplicative identity = additive $1$

Analogously, the $1$‘s of $+$ and $\cdot$ are the same as the field’s $1$. Denote them as $1_{\oplus }$ and $1_{\odot }$. Then, since for any $k\in \mathbb{K}$,

$$k=1_{\odot }\cdot k=(0_{\oplus }+1_{\oplus })\cdot k=1_{\odot }\cdot k$$

$1_{\oplus }=1_{\oplus }+0=1_{\odot }\cdot 1_{\oplus }$

it follows that $1=1_{\odot }⟺1=1_{\oplus }$. Thus, we can write it unambiguously as $1$.

Only $0$ makes a multiplication go to $0$

Given any elements $a,b\in \mathbb{K}$, one has that

$$a\cdot b=0⟺(a=0)\vee (b=0)$$

The converse is trivial; the direct implication comes from the following considerations:

• If both $a$ and $b$ are equal to $0$, then the implication holds
• If $a\ne 0$, then it has an inverse $a^{-1}$, from which we can multiply on both sides to yield $b=0$
• Analogously if $b\ne 0$, then it has an inverse $b^{-1}$, from which we can multiply on both sides to yield $a=0$

Thus, this is a direct consequence from the fact that $0$ has no multiplicative inverse.

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References

• Um Curso de Álgebra Linear, Flávio Ulhoa Coelho, Mary Lilian Lourenço. Editora EDUSP.