Rings, ideals and fields

Answer key — Rings, ideals and fields

1. 1.

   $a$ is a unit modulo $n$ iff $\gcd(a, n) = 1$: if $ua + vn = 1$ then $ua \equiv 1$; conversely $ab \equiv 1$ gives $ab - kn = 1$, so $\gcd(a, n) = 1$. The residues coprime to $12$ are $1, 5, 7, 11$, so there are $\varphi(12) = 4$ units. Each is its own inverse: $5^2 = 25 \equiv 1$, $7^2 = 49 \equiv 1$, $11^2 = 121 \equiv 1$. The other seven non-zero residues are zero divisors.

2. 2.

   The domains are $\Z/7\Z$, $\Z[i]$ and $\R[x]$. $\Z/8\Z$ has $2 \cdot 4 = 0$; $\Z \times \Z$ has $(1,0)(0,1) = 0$; $M_2(\R)$ is not commutative and has nilpotent matrices. Subrings of fields are always domains, and $R[x]$ is a domain whenever $R$ is (compare leading coefficients).

3. 3.

   $(x)$ and $\{f \mid f(0) \text{ even}\} = (2, x)$ are ideals: they are the kernels of $f \mapsto f(0) \in \Z$ and of $f \mapsto f(0) \bmod 2 \in \Z/2\Z$. The constants form a subring but not an ideal, and polynomials of degree at most $1$ are not closed under multiplication by $x$. A proper ideal never contains $1$, so it is never a subring in the sense of this lesson (where subrings contain $1$).

4. 4.

   In a PID the ideal $(a, b)$ is generated by $\gcd(a, b)$. Here $(x^2 - 1) - (x^2 - 3x + 2) = 3x - 3$, so $x - 1 = \frac13(3x - 3) \in I$. Conversely both generators are multiples of $x - 1$, so $I \subseteq (x - 1)$. Hence $I = (x - 1)$. Every element of $I$ vanishes at $1$, which shows $I \ne \Q[x]$.

5. 5.

   $r = a - qb$ with $a \in I$ and $qb \in I$ (an ideal absorbs multiplication by ring elements), so $r \in I$. If $r \ne 0$, then $\delta(r) \lt \delta(b)$ contradicts the minimality of $b$. Hence $r = 0$, $a = qb$, and $I = (b)$. The ideal property is used essentially: the same argument fails for a mere additive subgroup that is not closed under multiplication by $q$.

6. 6.

   Both (a) and (b). $3 \mid 6$ and $3 \mid 9$, but $3$ divides an element $x + y\sqrt{-5}$ only when $3 \mid x$ and $3 \mid y$, which fails for $1 \pm \sqrt{-5}$ and $2 \pm \sqrt{-5}$. So $9 = 3 \cdot 3 = (2 + \sqrt{-5})(2 - \sqrt{-5})$ is another failure of unique factorisation. In the language of ideals, $(3) = (3, 1 + \sqrt{-5})(3, 1 - \sqrt{-5})$ is a product of two non-principal prime ideals.

7. 7.

   $25 = 25 + 0 = 0 + 25 = 9 + 16 = 16 + 9$. This gives $\pm 5$, $\pm 5i$ (four elements) and $\pm 3 \pm 4i$, $\pm 4 \pm 3i$ (eight elements): $12$ in all. Unique factorisation explains the count: $25 = (2+i)^2(2-i)^2$ up to units, and an element of norm $25$ is a unit ($4$ choices) times $(2+i)^2$, $(2+i)(2-i) = 5$ or $(2-i)^2$ ($3$ choices).

8. 8.

   Option (a). Dropping IsDomain fails for ZMod 6, dropping Finite fails for ℤ, and (b) is the converse. The proof (finite_domain_inverse) is that multiplication by a non-zero element is injective, hence surjective on a finite set, so it hits $1$.

9. 9.

   def classify(n):
       units = [a for a in range(n) if any(a * b % n == 1 for b in range(n))]
       zero_divisors = [a for a in range(1, n)
                        if any(a * b % n == 0 for b in range(1, n))]
       return units, zero_divisors
   

   The last test confirms, for $n \lt 40$, that every non-zero residue is a unit or a zero divisor and never both. The proof: if $d = \gcd(a, n) = 1$, Bézout gives an inverse; if $d \gt 1$, then $a \cdot \frac{n}{d} = \frac{a}{d} n \equiv 0$ with $\frac{n}{d} \not\equiv 0$. And a unit is never a zero divisor (unit_mul_eq_zero in the lesson).