Understanding proof of proposition $x \in F \to x \cdot 0 = 0$ where $F$ is a field.

Question

Axioms:

• F1. addition is commutaitve: $x+y = y+x$, for all $x,y \in F$.
• F3. existence of additive identity: there is a unique element $0 \in F$ such that $x + 0 = x$, for all $x \in F$.
• F9. distributivity: $x(y+z)=xy+xz$, for all $x,y,z \in F$.

Proof: By Axiom F3, $x0=x(0+0)$. By distributivity (Axiom F9), $x(0+0)=x0+x0$. By Axiom F3 again, $$0+x0=x0+x0,$$ and by Axiom F1 $$x0+0=x0+x0.$$ Hence, $0=x0$ by the preceding proposition.

How is the author instantiating the axioms in each justification ?

In the first justification, it seems he used Axiom F3 with $0$, reaching $$0=0+0$$ and then replacing in the equality $x0=x0$ but I do not see the steps clearly in the others.

EDIT:

When the author says "preceding proposition", he is referring to:

Proposition. If $F$ is a field and $x,y,z \in F$, then $$x+z=y+z \to x=y.$$

Answer 1

Before we start, I want to re-state your claim as follows: if $F$ is a field, then $$ t\in F \rightarrow t \cdot 0 = 0. $$

All I've done is replace your "x" with "t". I've done so because $x$ is used in many of the axioms, and I don't want name-collisions to cause confusion.

Proof: By Axiom F3, $x0=x(0+0)$.

Let's expand that. First, by axiom 3, instantiated with $x = 0$, we get $$ 0 + 0 = 0. $$ Then we apply the multiplication operation to both sides, multiplying by $t$. (I'm using here the fact that if $a = b$, then $ta = tb$, i.e., that multiplication is well-defined!). This gives us $$ t\cdot (0+0) = t\cdot 0. $$ (where $t \in F$ is the generic element of the field $F$ given in the hypotheses of this theorem).

By distributivity (Axiom F9), $x(0+0)=x0+x0$.

In our version, this is the observation that the left hand side and be rewritten by Axiom F9, using $x = t, y = 0, z = 0$, to get $$ t(0 + 0) = t \cdot 0 + t \cdot 0. $$ So the whole equality becomes $$ t \cdot 0 + t \cdot 0 = t \cdot 0. $$

By Axiom F3 again, $$0+x0=x0+x0,$$

The author has implicitly exchanged the two sides to write $$ t \cdot 0 = t \cdot 0 + t \cdot 0 \tag{1} $$ and then has applied axiom 3 to substitute for the left hand side, using $x = t \cdot 0$ in the axiom to conclude that $$ t \cdot 0 = 0 + t \cdot 0 \tag{2} $$ and then replacing the left hand side of Equation 2 using Equation 1 to get $$ 0 + t \cdot 0 = t \cdot 0 + t \cdot 0. \tag{3} $$

and by Axiom F1 $$x0+0=x0+x0.$$

He's applying Axiom F1 to the left hand side of Equation 3, with $x = 0, y = t \cdot 0$, to get that $$ 0 + t \cdot 0 = t \cdot 0 + 0, $$ and then substituting this in Equation 3 to get $$ t \cdot 0 + 0 = t \cdot 0 + t \cdot 0 $$

Hence, $0=x0$ by the preceding proposition.

I'm assuming that the prior proposition says that if (in a field) we have $$ a + b = a + c $$ then $b = 0$. The author's applying this to $a = t\cdot 0, b = 0, c = t \cdot 0$.

Answer 2

By $F_{3}$ we have $x0 = x(0 + 0)$. Then, by $F_{9}$, we distribute and have $x(0 + 0) = x0 + x0$. Thus, together we have $x0 = x(0 + 0) = x0 + x0$. By $F_{3}$ we have $0 + x0 = x0$. This together with the previous yields $0 + x0 = x0 = x(0 + 0) = x0 + x0$. So we have $0 + x0 = x0 + x0$. Now, using $F_{1}$, we commute $0 + x0$ such that $0 + x0 = x0 + 0$. So we arrive at $x0 + 0 = x0 + x0$. By the previous proposition (left cancellation) we now have $0 = x0$.