If $a^2 = b^2$ in a field, then $a = b$ or $a = -b$

I'm trying to prove that if $a^2 = b^2$ in a field, then $a = b$ or $a = -b$

I know that a field is a commutative, division ring by definition.

Hence if $a^2 = b^2$ in a field, then we have $a^2 - b^2 = 0$ where we can say $a^2 - b^2 = (a - b)(a+b)$ because $$(a-b)(a+b) = a^2 +ab - ba - b^2 = a^2 - b^2$$ where $ab = ba$ by commutativity.

Hence we can say that $(a - b)(a + b) = 0$, which is only true if $a - b = 0$ or $a + b = 0$. If the first is true, then $a = b$ and if the second is true $a = -b$. Hence proved.

Is this the correct approach?

Solution 1:

Yes, this is correct. Indeed, your argument goes through in (commutative) integral domains, as well as in fields. (Recall that a field is an integral domain because if $xy=0$ and $x\neq 0$, we may multiply both sides by $x^{-1}$ to see $y=0$.)

[ NB: this answer is a compilation of the comments, to help remove this question from the unanswered queue. ]