Let $H = \langle(2, 2)\rangle = \lbrace (0, 0), (2, 2), (0, 4), (2, 6), (0, 8), (2, 10) \rbrace$. $(0, 1) + H$ is of order 4, so the factor group is not $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$. There are no elements of order 8. So it is $\mathbb Z_4 \times \mathbb Z_2$.
$\left\langle \begin{bmatrix}1 & 1 \ 0 & 1\end{bmatrix}\right\rangle$ generates an additive subgroup of real 2x2 matrices, but $\begin{bmatrix}1 & 1 \ 0 & 1\end{bmatrix}^2 = \begin{bmatrix}1 & 2 \ 0 & 1\end{bmatrix} \not\in \left\langle \begin{bmatrix}1 & 1 \ 0 & 1\end{bmatrix}\right\rangle$.
We use the fact that the subring properties hold if subtraction and multlipliation is closed.
(a) is a subring since the difference and product of even integers are even.
(b) is equivalent to the polynomials whose coefficient of $x$ is 0. This is a subring since it is closed under subtraction and multiplication. To see closure under multiplication, note that an $x$ term generate from a product requires multiplication of 2 terms with degree less than 1, which doesn’t exist.
(c) is not since it is not closed under multiplication: $x$ is in the subset but $x^2$ is not.
(a) In $\mathbb Z_p$ every element but $0$ has a multiplicative inverse. If $a = 0$ we are done. Otherwise $a^2 = a \implies a = 1$ (left multipying by $a^{-1}$).
(b) If $a \neq 0$ then $ab = 0 \implies b = 0$ by left-multiplication of $a^{-1}$. Likewise for $b \neq 0$.
(c) Left multiplication by $a^{-1}$.
$k\mathbb Z \subseteq \mathbb Z$ for $k \in \mathbb Z$ is a subring, since it is closed under subtraction and multiplication. To see this, let $ka. kb \in k\mathbb Z$. Then $ka - kb = k(a - b) \in k\mathbb Z$, and $(ka)(kb) = k(kab) \in k\mathbb Z$.
$4 \mid 2$ since $4 \cdot 2 = 8 \bmod 6 \equiv 2$.
$3 \mid 7$ since $3 \cdot 5 = 15 \bmod 8 \equiv 7$.
$9 \mid 12$ since $9 \cdot 3 = 27 \bmod 15 \equiv 12$.
The multiplicative inverse of $(1 - a)$ is $(1 + a + \dots + a^{n - 1})$, since $(1 - a)(1 + a + \dots + a^{n - 1}) = 1 + a^n = 1$.
Let $m \gt 0$ and suppose $a^m = 0$ for some $a$. If $m = n$ we are done. Suppose $m \lt n$. Then $n = qm + r$. So $a^n = a = a^{qm}a^r = 0$. If instead $n \lt m$ then choose $k$ such that $m \lt n^k$. Since $a^n = a$, $a^{n^2} = (a^n)^n = a^n = a$, and so on for $a^{n^k}$. So apply the $m \lt n$ case to $m \lt n^k$.
Since $a$ is a unit, $a = bc \implies 1 = a^{-1}bc$, so $c^{-1} = a^{-1}b$. Similarily, $b^{-1} = ca^{-1}$.
Note that $a + a = 0$ since $a + a = (a + a)^2 = a^2 + a^2 + a^2 + a^2 = a + a + a + a$. So $x + y = (x + y)^2 = x^2 + xy + yx + y^2 = x + y + xy + yx$, so $xy + yx = 0$. Thus $xy + xy + yx = xy$, but since $xy + xy = 0$, $yx = xy$.
Let $R$ be a commutative ring such that for $a, b, c \in R$ and $a \neq 0$, $ab = ac$ implies $b = c$. Then there are no zero-divisors. For if there were one, say $d$, with non-zero $x \in R$ such that $dx = 0$,
Define a ring isomorphism $f: \operatorname{Frac}(\mathbb Z[i]) \to \mathbb Q[i]$ as
\[f\left(\frac{a + bi}{c + di}\right) = \frac{ac + bd}{c^2 + d^2} + i\left(\frac{bc - ad}{c^2 + d^2}\right)\]and proceed to show that the isomorphism is structure-preserving.
Define a ring isomorphism $f: \operatorname{Frac}{F} \to F$ as the identity mapping.
$f(x)$ factored in $D$ factors into units in $F$.
Suppose it wasn’t primitive. Then Then its content is greater than 1, so it is not a unit in $\mathbb Z$, and can be factored out.
Since $r = p/q$ is a root, then
\[\frac{p^n}{q^n} + a_{n - 1}\frac{p^{n - 1}}{q^{n - 1}} + \dots + a_0 = 0\]Multiplying both sides by $q^n$ we have
\[p^n + a_{n - 1}qp^{n - 1} + \dots + q_na_0 = 0\]So $q \mid p^n$. Since $\gcd(p, q) = 1$, $\gcd(p^n, q) = 1$, so $q = 1$.
(a) If $f(x) = g(x)h(x)$, then $af(x) = ag(x)h(x)$, with $g(x), h(x)$ non-units. $\blacksquare$
(b) If $f(ax)$ is irreducible, then $f(ax) = g(ax)h(ax)$, where either $g$ or $h$ is a unit polynomial (constant). Then for $f(x) = g(x)h(x)$, either $g$ or $h$ is a unit polynomial. $\blacksquare$
(c) Same proof as above. $\blacksquare$
(d) Since $f(x + 1) = 8x^3 + 24x^2 + 18x + 3$, by Eisenstein’s Criterion with $p = 3$, $f(x + 1)$ is irreducible. So $f(x)$ is irreducible. $\blacksquare$
By the theorem that states $\mathbb Z_p[x] / \langle f(x) \rangle$ is a field with $p^n$ elements. where $f(x)$ is a polynomial of degree $n$, $\mathbb Z_5[x]/\langle x^2 + 1 \rangle$ is a field with 25 elements.
Similar to 11, $\mathbb Z_3[x]/\langle x^3 - 2 \rangle$.
(a) Irreducible by Eisenstein’s with $p = 3$.
(b) By Theorem 17.3 it suffices to check that this polynomial is irreducible over $\mathbb Z_2[x]$. $f(x) \neq 0 \pmod 2$ for $x = 0, 1$ so there are no linear factors. So both factors of $f(x)$ would have to be quadratic. But, none of the 4 quadratic polynomials in $\mathbb Z_2[x]$ factor $f(x)$ (by polynomial long division), so it is irreducible in $\mathbb Z_2[x]$.
(c) Irreducible by Eistestein’s with $p = 3$.
(d) Again, it suffices to check $x^5 + x^2 + 1 \in \mathbb Z_2[x]$ is irreducible.
(e) Multiply by 14 for $35x^5 + 63x^4 + 210 x^3 + 6x^2 + 84x + 3$, which is irreducible by Eisenstein’s with $p = 3$. Since $af(x)$ is irreducible, $f(x)$ is irreducible.
Done in class: use Eisenstein’s Criterion on $f(x + 1)$.
$f(1) = 0$ so $(x - 1)$ is a factor. Then by polynomial long division, $f(x) = (x - 1)(x^2 + x + 1)$. To verify,
\[(x - 1)(x^2 + x + 1) = x^3 + x^2 + x - x^2 - x - 1 = x^3 - 1 = x^3 + 6\]If $a = ub$ for some unit $u$, then $a \in (b)$ so $(a) \subseteq (b)$, likewise for the other direction. So $(a) = (b)$.
If $(a) = (b)$, then $a = pb$ and $b = qa$ for some $p, q \in D$. Then $a = pqa$, so $pq = 1$ and $p, q$ are units. $\blacksquare$
Consider $pu = ab$, where $p$ is an irreducible and $u$ is a unit, and $a, b$ are non-units. Then $p = (ab)u^{-1}$. WLOG, suppose $a$ is a non-unit. Then $bu^{-1}$ must be a unit. But then, since associates of units are also units $(bu^{-1})u = b$ is a unit; a contradiction. $\blacksquare$
$(\rightarrow)$ If $a$ is a unit, then $a^{-1}ab = b \in (ab)$, so $(b) \subseteq (ab)$, and $(ab)$ can’t be a proper subset of $(b)$.
$(\leftarrow)$ Suppose $a$ is not a unit. Suppose $(b) = (ab)$. Then $b = xab$ for some $x \in D$. But then, since $b \neq 0$, $xa = 1$. $\blacksquare$
Reflexive: $a = 1a$, and $1$ is a unit. Symmetric: If $a = ub$ for $u$ a unit, then $b = u^{-1}a$. Transitive: If $a = ub$ and $b = vc$ with $u, v$ units, Then $a = uvc$, and $uv$ is a unit since $(uv)(v^{-1}u^{-1}) = 1$.
$(\rightarrow)$ If $(p)$ is maximal, then it is prime, so $p$ is prime, so it is irreducible.
$(\leftarrow)$ If $p$ is irreducible, then $p$ is prime since $D$ is a PID, so $(p)$ is prime, so $(p)$ is maximal, since $D$ is a PID. $\blacksquare$