Field Product with Zero

Theorem

Let $\struct {F, +, \times}$ be a field whose zero is $0_F$.

Let $a \in F$.


Then:

$a \times 0_F = 0_F$


That is, $0_F$ acts as a zero element of $F$, which justifies its name.


Proof

\(\ds a\) \(=\) \(\ds a \times 1_F\) Field Axiom $\text M3$: Identity for Product: $1_F$ is the unity of $F$
\(\ds \) \(=\) \(\ds a \times \paren {0_F + 1_F}\) Field Axiom $\text A3$: Identity for Addition
\(\ds \) \(=\) \(\ds a \times 0_F + a \times 1_F\) Field Axiom $\text D$: Distributivity of Product over Addition
\(\ds \) \(=\) \(\ds a \times 0_F + a\) Field Axiom $\text M3$: Identity for Product
\(\ds \leadsto \ \ \) \(\ds \paren {a \times 0_F + a} + \paren {-a}\) \(=\) \(\ds a + \paren {-a}\) adding $-a$ to both sides
\(\ds \leadsto \ \ \) \(\ds a \times 0_F + \paren {a + \paren {-a} }\) \(=\) \(\ds a + \paren {-a}\) Field Axiom $\text A1$: Associativity of Addition
\(\ds \leadsto \ \ \) \(\ds a \times 0_F + 0_F\) \(=\) \(\ds 0_F\) Field Axiom $\text A4$: Inverses for Addition
\(\ds \leadsto \ \ \) \(\ds a \times 0_F\) \(=\) \(\ds 0_F\) Field Axiom $\text A3$: Identity for Addition

$\blacksquare$


Sources

  • 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 87 \alpha$
  • 1973: C.R.J. Clapham: Introduction to Mathematical Analysis ... (previous) ... (next): Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Theorem $2 \ \text {(iii)}$