Practice: Sets and Fields

(a) ${\mathbb{Z}}_{10}$ — not a field. Ten is composite, $10=2\cdot 5$, so the prime criterion predicts failure. The witness is immediate:

with both factors nonzero. Failing axiom: multiplicative inverses. The argument that neither factor can have an inverse is the part that matters. Suppose $2^{-1}$ existed. Multiplying $2\cdot 5=0$ through by it gives

false in ${\mathbb{Z}}_{10}$. So no such element exists, not merely that none was found. The same argument rules out $5^{-1}$. (The non-invertible elements are exactly $2,4,5,6,8$, those sharing a factor with 10.) (b) ${\mathbb{Z}}_{13}$ — a field. Thirteen is prime, so every nonzero $a$ satisfies $gcd(a,13)=1$, and Bézout gives integers $x,y$ with $ax+13y=1$. Reducing modulo 13, $ax\equiv 1$, so $x=a^{-1}$. Inverses exist for all twelve nonzero elements, for instance $2\cdot 7=14=1$, $3\cdot 9=27=26+1=1$, $5\cdot 8=40=39+1=1$, and $12\cdot 12=144=143+1=1$. The remaining axioms are inherited from integer arithmetic, and $1\ne 0$ since . (c) $D=\{a+b\sqrt{3}:a,b\in \mathbb{Q}\}$ — a field. Closure under multiplication:

which has the required form because $\mathbb{Q}$ is closed. Addition is componentwise, and $0=0+0\sqrt{3}$ and $1=1+0\sqrt{3}$ both lie in $D$. Inverses come from the conjugate $a-b\sqrt{3}$, whose product with $a+b\sqrt{3}$ is the rational norm $a^2-3b^2$:

This requires the norm to be nonzero for nonzero elements. If $a^2=3b^2$ with $b\ne 0$, then $(a/b{)}^2=3$, making $\sqrt{3}$ rational, which it is not. If $b=0$ then $a\ne 0$ and the norm is $a^2\ne 0$. So the norm never vanishes and every nonzero element is invertible. Example: $(2+\sqrt{3})$ has norm $4-3=1$, so its inverse is $2-\sqrt{3}$, and indeed $(2+\sqrt{3})(2-\sqrt{3})=4-3=1$. (d) $T=\{x\in \mathbb{Q}:x\ge 0\}$ — not a field. Closure holds: sums and products of nonnegative rationals are nonnegative. Both identities are present, $0,1\in T$, and multiplicative inverses exist for every nonzero element, since implies . Failing axiom: additive inverses. The witness is $1$: there is no $y\in T$ with $1+y=0$, because any such $y$ would be , outside $T$. This is the case where the multiplicative side is fine and the additive side fails. The reverse of $\mathbb{Z}$. (e) Why the argument form matters. "I checked every element and found no inverse" is a search report. For a finite structure it can be made rigorous by exhausting the cases, but it establishes nothing about infinite structures such as $\mathbb{Z}$ or $T$, where the candidates cannot be enumerated. Worse, it gives no reason, so it does not transfer: it would have to be repeated for ${\mathbb{Z}}_{10}$, ${\mathbb{Z}}_{12}$, ${\mathbb{Z}}_{15}$ separately. The argument in (a) instead derives a contradiction from the assumption that an inverse exists, which settles all candidates at once and explains the cause. The general principle. In any structure where the axioms of associativity and identity hold, a zero divisor cannot be invertible: > If $ab=0$ with $a,b\ne 0$, and $a^{-1}$ existed, then $b=(a^{-1}a)b=a^{-1}(ab)=a^{-1}\cdot 0=0$, contradicting $b\ne 0$. So exhibiting one zero divisor disqualifies the structure from being a field, and in ${\mathbb{Z}}_n$ a composite $n=ab$ supplies its own factors as the witnesses. Conversely a field has no zero divisors, which is what makes cancellation valid, and cancellation is used silently throughout linear algebra whenever a common nonzero factor is removed.