II.2 · Fields & Order — Problem 2 #

← Back to II. Fields & Order, Problem 2

Problem. Let $(F,+,\cdot)$ be a field and $x,y,z\in F$. Show that (a) if $x\neq0$ and $xy=xz$, then $y=z$ (cancellation); and (b) if $x\neq0$ and $xy=x$, then $y=1$.

Hint #

Because $x\neq 0$, it has a multiplicative inverse $x^{-1}$. Multiply both sides of the hypothesis by $x^{-1}$ and use associativity together with $x^{-1}x=1$. Part (b) is the special case $z=1$ of part (a).

Solution #

(a) Assume $xy=xz$ and note that since $x\neq 0$, it has a multiplicative inverse $1/x$. Then

$$y = 1\cdot y = (1/x)\cdot(x\cdot y) = (1/x)\cdot(x\cdot z) = 1\cdot z = z,$$

as desired.

(b) This follows from the previous part by letting $z=1$: the hypothesis $xy=x=x\cdot 1$ is $xy=xz$ with $z=1$, so $y=1$.

Long solution #

Strategy. A field has no primitive operation called “division”: to divide by $x$ means to multiply by the inverse $x^{-1}$, which exists only when $x\neq 0$. That is the whole proof, and the whole reason the hypothesis $x\neq 0$ appears. Multiply $xy=xz$ by $x^{-1}$, regroup with associativity so that $x^{-1}x$ collapses to $1$, and remove the $1$. Part (b) is part (a) with $z=1$.

Axioms used.

M2 (associativity): $(uv)w=u(vw)$.

M3 (commutativity): $uv=vu$.

M4 (identity and inverses): $1\cdot u=u$ for all $u$, and every $x\neq 0$ has $x^{-1}\in F$ with $x\,x^{-1}=1$.

We also use that multiplication is a function: if $u=v$ then $wu=wv$. “Multiplying both sides by the same thing” is exactly this.

(a) Cancellation. Suppose $x\neq 0$ and $xy=xz$. By M4 there is $x^{-1}\in F$ with $x\,x^{-1}=1$, and by M3 also $x^{-1}x=1$. Then

$$y \overset{1}{=} 1\cdot y \overset{2}{=} (x^{-1}x)\,y \overset{3}{=} x^{-1}(xy) \overset{4}{=} x^{-1}(xz) \overset{5}{=} (x^{-1}x)\,z \overset{6}{=} 1\cdot z \overset{7}{=} z.$$

The justifications: (1) and (7) are the identity law M4; (2) and (6) are $x^{-1}x=1$; (3) and (5) are associativity M2; (4) is the hypothesis $xy=xz$, multiplied on the left by $x^{-1}$. $\blacksquare$

(b) If $xy=x$ then $y=1$. Suppose $x\neq 0$ and $xy=x$. By the identity law, $x=x\cdot 1$, so the hypothesis reads $xy=x\cdot 1$. This is the hypothesis of part (a) with $z=1$, so part (a) gives $y=1$. $\blacksquare$

Remark (why $x\neq 0$ is essential). For $x=0$ cancellation fails completely: $0\cdot y=0=0\cdot z$ for every pair $y,z$, so nothing can be deduced. Cancellation holds in a field precisely because nonzero elements are invertible. The phrase “divide both sides by $x$” is shorthand for the argument above.