Math 120B HW 7 Solutions

1. Let $R$ be a commutative ring with unity and let $a \in R$. Define $aR = \{ar : r \in R\}$.

   1. $aR$ is an ideal of $R$:

      • Consider $ar_1, ar_2 \in aR$. Then $ar_1 + ar_2 = a(r_1 + r_2) \in aR$.

      • $0 = a0 \in aR$.

      • Consider $ar \in aR$. Then $-(ar) = a(-r) \in aR$.

      • Consider $ar \in aR$ and $b \in R$. Then $(ar)b = a(rb) \in aR$ and $b(ar) = a(br) \in aR$. (Note that the latter statement uses commutativity.)

   2. We have $a = a1 \in aR$. Let $N$ be an ideal of $R$ such that $a \in N$. Then $aR \subseteq N$: Consider an element $ar \in aR$. Then $ar$ is the product of an element of $N$ (namely $a$) with an element of $R$, so it is in $N$.

2. The answer is no. Let $N$ be the principal ideal of $\mathbb{Q}[x]$ generated by $x^2 - 1$. Let $\phi : \mathbb{Q}[x] \to \mathbb{R}$ be a homomorphism and suppose toward a contradiction that $\ker(\phi) = N$. Note that $x+1$ and $x-1$ are not in $N$ because their degrees are 1 whereas every nonzero element of $N$ has degree at least 2. So $\phi(x+1)$ and $\phi(x-1)$ are nonzero, and because $\mathbb{R}$ has no zero divisors, we have $\phi(x^2-1) = \phi((x+1)(x-1)) = \phi(x+1)\phi(x-1) \ne 0$, so $x^2-1 \notin \ker(\phi)$. This is a contradiction, because $x^2-1 \in N$.

3. Define $\phi : (R_1 \times R_2)/(N_1 \times N_2) \to (R_1 / N_1) \times (R_2 / N_2)$ by $$\phi( (a_1,a_2) + (N_1 \times N_2)) = (a_1 + N_1, a_2 + N_2).$$ This equation defines a function (i.e. "$\phi$ is well-defined"):

   Let $(a_1,a_2),(b_1,b_2) \in R_1 \times R_2$ and assume that $(a_1,a_2) + (N_1 \times N_2) = (b_1,b_2) + (N_1 \times N_2)$. We have $(a_1,a_2) + (N_1 \times N_2) = (a_1 + N_1) \times (a_2 + N_2)$ and $(b_1,b_2) + (N_1 \times N_2) = (b_1 + N_1) \times (b_2 + N_2)$, so $(a_1 + N_1) \times (a_2 + N_2) = (b_1 + N_1) \times (b_2 + N_2)$. All cosets are nonempty, so by a general fact about Cartesian products of nonempty sets, we have $a_1 + N_1 = b_1 + N_1$ and $a_2 + N_2 = b_2 + N_2$. Therefore $(a_1 + N_1, a_2 + N_2) = (b_1 + N_1, b_2 + N_2)$, as desired.

   (Moreover, one can show that this function $\phi$ is an isomorphism.)

4. For each of the following ideals of $\mathbb{Z} \times \mathbb{Z}$, say whether or not it is prime. Justify any "no" answers.

   1. $2 \mathbb{Z} \times 3 \mathbb{Z}$ is not prime because it does not contain $(1,3)$ or $(2,1)$ but it contains their product $(2,3)$.
   2. $\{0\} \times \{0\}$ is not prime because it does not contain $(1,0)$ or $(0,1)$ but it contains their product $(0,0)$.
   3. $\mathbb{Z} \times \mathbb{Z}$ is not prime because it is improper.
   4. $\mathbb{Z} \times 3 \mathbb{Z}$ is prime.
   5. $\mathbb{Z} \times 4 \mathbb{Z}$ is not prime because it does not contain $(0,2)$ but it contains its square $(0,4)$.
   6. $\{0\} \times \mathbb{Z}$ is prime.

5. For each of the following ideals of $\mathbb{Z} \times \mathbb{Z}$, say whether or not it is maximal. Justify any "no" answers.

   1. $2 \mathbb{Z} \times 3 \mathbb{Z}$ is not maximal: $2 \mathbb{Z} \times 3 \mathbb{Z} \subsetneq 2\mathbb{Z} \times \mathbb{Z} \subsetneq \mathbb{Z} \times \mathbb{Z}$.
   2. $\{0\} \times \{0\}$ is not maximal: $\{0\} \times \{0\} \subsetneq \{0\} \times \mathbb{Z} \subsetneq \mathbb{Z} \times \mathbb{Z}$.
   3. $\mathbb{Z} \times \mathbb{Z}$ is not maximal because it is improper.
   4. $\mathbb{Z} \times 3 \mathbb{Z}$ is maximal.
   5. $\mathbb{Z} \times 4 \mathbb{Z}$ is not maximal because $\mathbb{Z} \times 4 \mathbb{Z} \subsetneq \mathbb{Z} \times 2 \mathbb{Z} \subsetneq \mathbb{Z} \times \mathbb{Z}$.
   6. $\{0\} \times \mathbb{Z}$ is not maximal because $\{0\} \times \mathbb{Z} \subsetneq 2\mathbb{Z} \times \mathbb{Z} \subsetneq \mathbb{Z} \times \mathbb{Z}$.

6. Let $R$ be a commutative ring with unity $1 \ne 0$. Let $N$ be a maximal ideal of $R$. We want to show that $N$ is prime. Let $a,b \in R$ with $ab \in N$ and $a \notin N$. We want to show that $b \in N$.

   Consider the ideal $A = \{x + ar : x \in N \text{ and } r \in R\}$.

   1. Note that $N \subseteq A$ because $x = x+ a0$. Moreover, $N \subsetneq A$ because $a = 0 + a1 \in A$ but $a \notin N$. So because $N$ is a maximal ideal and $A$ is an ideal, we have $A = R$. Therefore $1 \in A$.

   2. By part a we have $1 = x+ar$ for some $x \in N$ and $r \in R$. Therefore $b = 1b = (x+ar)b = xb + (ar)b = xb + (ab)r \in N$ because $x \in N$ and $ab \in N$.