## Math 120B HW 1 Solutions

1. For each part, say whether or not the given structure is a ring. Justify your answers.
1. $(\mathbb{Z},+,\max)$ is not a ring because $\max$ does not distribute over addition: for example, $\max(1,0+0) = 1$ but $\max(1,0) + \max(1,0) = 2$.
2. $(\mathbb{Z},+,\ast)$, where $a \ast b = 0$ for all $a,b \in \mathbb{Z}$, is a ring. We know that $(\mathbb{Z},+)$ is an abelian group. The operation $\ast$ is commutative because $a \ast b = 0 = b \ast a$ for all $a,b \in \mathbb{Z}$. The distributivity laws hold for the structure: $a \ast (b + c) = 0 = 0 + 0 = (a \ast b) + (a \ast c)$ and $(b + c) \ast a = 0 = 0 + 0 = (b \ast a) + (c \ast a)$ for all $a,b,c \in \mathbb{Z}$. (We only really need to verify one of the distributivity laws, because we showed that $\ast$ is commutative.)
3. $(\mathbb{Z},\ast,\ast)$, where $\ast$ denotes ordinary addition, is not a ring because the distributivity laws fail: addition does not distribute over itself. For example, $1 + (0 + 0) = 1$ but $(1 + 0) + (1 + 0) = 2$.
2. Let $n \in \mathbb{Z}^+$. Recall that $\mathbb{Z}_n$ denotes the ring with elements $\{0,\ldots,n-1\}$ under the operations of addition modulo $n$ and multiplication modulo $n$.
1. Assume that $n$ is prime and let $a,b \in \mathbb{Z}_n$. If $ab = 0$ in $\mathbb{Z}_n$ then $n \mid ab$ (now we are using $ab$ to denote ordinary multiplication in $\mathbb{Z}$.) Because $n$ is prime, we have $n \mid a$ or $n \mid b$. (This important property of prime numbers can be proved using the fundamental theorem of arithmetic.) Because $a,b \in \mathbb{Z}_n$, this means $a = 0$ or $b = 0$.
2. Assume that $n$ is composite, say $n = ab$ where $a,b \in \{1,\ldots,n-1\}$. Then $a,b \in \mathbb{Z}_n$ and $ab = 0$ in $\mathbb{Z}_n$.
(Note that the number $1 \in \mathbb{Z}^+$ is neither prime nor composite, so neither case applies to it. However, we can say that $\mathbb{Z}_1$ is not an integral domain because it satisfies $1 = 0$, meaning that its multiplicative identity element equals its additive identity element.)
3. For each part, say whether or not the given function is a homomorphism (of rings.) Justify your answers.
1. The function $\phi : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ defined by $\phi((a,b)) = a+b$ is not a homomorphism because it does not preserve multiplication. For example, $\phi((1,2)(3,4)) = \phi((3,8)) = 11$ but $\phi((1,2))\phi((3,4)) = 3\cdot 7 = 21$.
2. The function $\phi : M_2(\mathbb{R}) \to \mathbb{R}$ defined by $\phi(A) = \det(A)$ is not a homomorphism because it does not preserve addition. For example, letting $I$ denote the unity of $M_2(\mathbb{R})$ (the identity matrix) we have $\det(I+I) = 4$ but $\det(I) + \det(I) = 2$.
3. There is no function $\phi : \mathbb{Q} \to \mathbb{Z}$ defined by $\phi(a/b) = a$. (Some people would say "the function $\phi$ is not well defined" but this is misleading because there is no function $\phi$ on $\mathbb{Q}$ to speak of.) This is because (for example) $1/3 = 2/6$ but $1 \ne 2$. Every homomorphism is a function, so there is no homomorphism $\phi : \mathbb{Q} \to \mathbb{Z}$ defined by $\phi(a/b) = a$.
4. Let $\phi:\mathbb{Z} \to \mathbb{Z}$ be a homomorphism. Assume that $\phi$ is not trivial. We will show that $\phi$ is the identity function. Take $b \in \mathbb{Z}$ such that $\phi(b) \ne 0$. Let $a \in \mathbb{Z}$. Then we have \begin{align*} \phi(ab) &= \phi(a)\phi(b),\text{ and also}\\ \phi(ab) &= \phi(\underbrace{b+\cdots +b}_{a\text{ times}}) = \underbrace{\phi(b) + \cdots + \phi(b)}_{a\text{ times}} = a \phi(b). \end{align*} (Note that the first equation only makes sense because $a$ is a ring element, and the second equation only makes sense because $a$ is an integer.) Therefore $\phi(a)\phi(b) = a \phi(b)$, and because $\phi(b) \ne 0$ we have $\phi(a) = a$ by cancellation in $\mathbb{Z}$. Therefore $\phi$ is the identity function. (Note that $a$ was arbitrary, whereas $b$ was fixed.)
5. Let $(F,+,\cdot)$ be a field and let $F^*$ denote the set of all nonzero elements of $F$. Then $(F^*, \cdot)$ is an abelian group:
• $F^*$ is closed under multiplication because if $a,b \in F^*$ then $ab \ne 0$; otherwise, mutiplying both sides by $a^{-1}$ (which exists because $F$ is a field) we get $b = 0$, a contradiction. (Or we can use the fact that every field is an integral domain, so $F$ has no zero divisors.)
• The multiplication operation on $F^*$ is associative because it is induced from the multiplication operation on $F$, which is associative.
• $1 \in F^*$ because $1 \ne 0$ in $F$. (And because $1$ is a multiplicative identity element for $F$, it is a multiplicative identity element for the subset $F^*$.)
• Every element $a \in F^*$ has a multiplicative inverse $a^{-1}$ in $F$ because $F$ is a field. We have $a^{-1} \in F^*$ because otherwise if $a^{-1} = 0$ then $1 = aa^{-1} = a0 = 0$, a contradiction.
6. In $\mathbb{Z}_{11}$ we have $1^{-1} = 1$, $2^{-1} = 6$, $3^{-1} = 4$, $4^{-1} = 3$, $5^{-1} = 9$, $6^{-1} = 2$, $7^{-1} = 8$, $8^{-1} = 7$, $9^{-1} = 5$, and $10^{-1} = 10$.

In $\mathbb{Z}_{12}$ the units are $1$, $5$, $7$, and $11$, and we have $1^{-1} = 1$, $5^{-1} = 5$, $7^{-1} = 7$, and $11^{-1} = 11$.