*26.* **a.** Statement: The equality modulo an ideal I of a ring R is an equivalence relation.

Proof: (Reflexivity)

An ideal by definition is nonempty and any number times 0 is 0 so $0 \in I$

(Symmetry)

0 is in the ideal and it is closed under subtraction.

(Transitivity)

The ideal is closed under addition.

**b.** Statement: Multiplication on the quotient ring $R/I$ is well defined: $(r+I)(s+I)= rs+I$.

Proof: Let $r, s \in I$, $r \sim_I r’ \rightarrow r-r’ \in I, s \sim_I s’ \rightarrow s-s’ \in I$, and $r+I=r’+I, s+I=s’+I$. Suppose $a=r-r’$ and $b=s-s’$. Then:

$(r + I)(s + I) = rs + I = (r' + a)(s' + b) + I = r's' + r'b + as' + ab + I$. It follows that, because I is an ideal $r'b + as' + ab \in I$ and because it is closed under addition $r'b + as' + ab + I = I$. Thus $r's' + r'b + as' + ab + I = r's' + I$.

*27.* **a.** Statement: Every unit of a ring $R$ has a unique inverse.

Proof: Let $u, v_1, v_2 \in R$ such that $v_1, v_2$ are inverses of $u$ of the form $uv_1 = 1, uv_2 = 1$. It follows then, that $uv_1=uv_2$. To show uniqueness, we begin by multiplying both sides of the equation by $v_1$.

We can then simplify the equation by using the fact that $uv_1$ exists on both sides and is equal to $1$.

(12)**b.** Statement: The units of $\mathbb{Z}/n\mathbb{Z}$ are precisely those $[a]$ such that $GCD(a,n)=1$.

Proof: Using the idea that $GCD(a,n)=1$, assume $\exists u, v \in \mathbb{Z}$ such that, by definition of the $GCD$, $au+nv = 1$. It follows that $[a][u] + [n][v] = [1]$ is the equation with the equivalence classes of $a, u, n, v, 1$ respectively. $[n][v] = 1$ as the equivalence class containing $n$ is 0. This makes the equation $[a][u] = [1] \rightarrow au=1$ which is the defintion of an inverse.

*28.* Statement: An ideal $I$ of a ring $R$ is maximal if and only if the quotient ring $R/I$ is a field.

Proof: Assume $R/I$ is a field. Let $J$ be an ideal such that $J \supsetneq I$. Since $J \supseteq I, \exists x \in J$ such that $x \notin I$. So $I + x \neq I + 0$. Since $R/I$ is a field, $\exists (I + y) \in R/I$ such that:

From the above, we get $xy \sim 1, xy-1 \in I \in J$.

$1 = xy - (xy - 1)$. Thus $\forall n \in R, n = n \cdot 1$, so $J = R$, meaning $I$ is maximal.

Assume $I$ is maximal. Let $I+a \in R/I$ such that $(I+a) \neq (I+0)$ and $a \notin I$. Consider $J=I+Ra = \{i+ra|i \in I, r \in R\}$, $\forall i \in I i = i +0a$, so $I \subseteq J$. Since $I$ is maximal, either $J=I$ or $J=R$. However, $a=0+1a$, so $a \in J$ and $a \notin I$, so $J=R$. $I+Ra = R$ and $1 \in R$, so $1=i+ra$ for some $i \in I$ and $r \in R$. $1-ra = i \in I$, so $(I+1)=(I+ra)=(I+r)(I+a)$ by multiplication in the quotient ring. Thus, $(I+a)$ is the inverse of $(I+r)$ and $(I+a), (I+r) \in R/I$. Thus, all non-zero elements in $R/I$ have an inverse which makes $R/I$ a field.

*29.* **a.** Statement: Consider the polynomial ring $\mathbb{Q}[x,y]$ in two variables with rational coefficients. $(x)$ is a prime ideal of $\mathbb{Q}[x, y]$ that is not maximal.

Proof: Assume $pq \in (x)$, where $p,q \in \mathbb{Q}[x, y]$, and show that either $p \in (x) \wedge q \in (x)$. Suppose $p = a_nx^n + · · · + a_1x + a_0$ and $p = b_mx^m + · · · + b_1x + b_0$, where $a_n,...,a_0,b_m,...,b_0 \in \mathbb{Q}[y]$. Then $pq$ has one term with x-degree 0, the polynomial $a_0b_0 \in mathbb{Q}[y]$.

From the assumption $pq \in (x)$, we know there is some polynomial $s \in mathbb{Q}[x,y]$ such that $pq = sx$. Note that every non-zero term in $sx$ has $x$ as a factor, hence its x-degree is at least 1. Thus, every nonzero term in $pq$ has degree at least 1. It therefore follows from $pq = sx$ that $a_0b_0 = 0$. Now $\mathbb{Q}$ is an integral domain, so $\mathbb{Q}[y]$ is also an integral domain, so from $a_0b_0 = 0$ we conclude that either $a_0 = 0$ or $b_0 = 0$. If $a_0 = 0$ then every term of $p$ has $x$ as a factor, hence $p$ is equal to $x$ multiplied by some polynomial, so $p \in (x)$. Similarly, if $b_0 = 0$, then $q \in (x)$. This shows that $p$ or $q$ is in $(x)$, so $(x)$ is prime.

$(x)$ is not a maximal ideal because it is a proper subset of the maximal ideal $(x,y)$. We have $(x) ⊆ (x,y)$ since ${x} ⊆ {x,y}$. To show the inclusion is proper it is enough to show that $y \in (x,y)$ but $y \notin (x)$ because the y-degree of y is 1 but the y-degree of every polynomial in $(x)$ is 0.

**b.** An ideal $I$ of a ring $R$ is a *principal ideal* if there exists $x\in R$ such that $I=(x)$; that is, the ideal is generated by

a single element. Prove that every ideal of $\mathbb{Z}$ is principal.

Assume $I \neq (0)$. Let $x$ be the smallest positive element in $I$ which exists by the well-ordering principal. Then $(x) \subseteq I$. Conversely, let $a \in I$. By the division algorithm, we can write $a = xq + r$ with $0 \leq r < x$. Then $r = a − xq \in I$. As $x$ is the least element, $r$ must equal $0$, as otherwise $r$ is a smaller positive element of I. Therefore $a \in (x)$, so $I =(x)$ is principal.

**c.** Statement $\mathbb{Q}[x,y]$ is not a principal ideal domain because $(x,y)$ is not a principal ideal in $\mathbb{Q}[x,y]$.

Proof: Suppose towards contradiction that $(x,y) = (p)$ for some polynomial $p \in \mathbb{Q}[x,y]$. Then, since $x, y \in (x,y) = (p)$, there are some $s, t \in \mathbb{Q}[x,y]$ such that $x = sp$ and $y = tp$. Then $0 = deg_y (x) = deg_y (s) + deg_y (p)$, so $0 = deg_y (p)$ where $deg$ is the degree of the polynomial. Similarly, $0 = deg_x (y) = deg_x (t)+deg_x (p)$, so $0 = deg_x (p)$. From $0 = deg_y (p) = deg_x(p)$ we get $deg(p) = 0$ and $p \in \mathbb{Q}$. But $p \in (p) = (x,y)$, so $p = fx+gy$ for some $f,g \in \mathbb{Q}[x,y]$, so $deg(p) = deg(fx+gy) = min(deg(f)+deg(x),deg(g)+deg(y)) = min(deg(f)+1,deg(g)+1) ≥ 1$, contradicting $deg(p) = 0$.

Comments - Andrew Furash

You are doing the right thing for 26, but sort of backwards. For example, in part a you should say: for every x (3) and 0 \in I for every I so (1).

The rest looks good!