## Algebra Solutions: Hungerford III.3

1) A nonzero ideal in a principle ideal domain is maximal if and only if it is prime.
Proof-
Let R be a principle ideal domain and I be a nonzero ideal in R.
$( \Rightarrow )$ Assume that I is maximal. Then $R/I$ is a field, which means that $R/I$ is an integral domain. Hence, I is prime.
$(\Leftarrow )$ Assume that I is a prime ideal. Since R is a PID, we can write $I = (a)$ for some nonzero $a\in R$. Then we know that the element a is prime and hence irreducible. Then since a is irreducible, $(a)$ is maximal in the set of all proper principle ideals of R. But since every ideal in R is principle, $(a)$ is maximal in R.

2) An integral domain R is a unique factorization domain if and only if every nonzero prime ideal in R contains a nonzero principle ideal that is prime.
Proof-
In order to prove this, we need a lemma: In a UFD, irreducible elements are prime.
Subproof- Let R be a UFD and $c\in R$ be irreducible. Then c is a nonzero nonunit. Now assume that $c | ab$ for some $a,b\in R$. Then we can write $cx = ab$ for some $x\in R$. If $a = 0$ or $b = 0$ then we are done trivially so assume that a, b are nonzero. If either a or b is a unit, then either $b = a^{-1}cx$ or $a = cxb^{-1}$ which gives that $c|b$ or $c|a$ so assume that a, b are nonunits as well. If $x=0$ then since R is an integral domain we would have either a or b is zero, a contradiction. If x is a unit then since c is irreducible, $c = abx^{-1} \Rightarrow$ a is a unit or $bx^{-1}$ is a unit. We know a is not a unit so that $bx^{-1}$ would have to be a unit. Then $a = c(bx^{-1})^{-1} \Rightarrow c|a$ so that c is prime. Now assume that x is not a unit. Being in a UFD means that we can now factor a, b, and x into a product of irreducibles. Let $a = a_1\cdots a_r, b = b_1 \cdots b_s, x = x_1 \cdots x_t$. We have that $a_1\cdots a_r b_1 \cdots b_s = c x_1 \cdots x_t$. Both sides are products of irreducibles and so by the definition of UFD we must have that $r + s = 1 + t$ and that any element on one side must be an associate to some element on the other side. Thus c is an associate of some $a_i$ or $b_j$. In the first case we get that $a_i = cu$ for some unit $u\in R$. Then $c|a_i \Rightarrow c|a$ and similarly if c is an associate of some $b_j$ then $c|b$. Thus c is prime. Now using the lemma we can do the proof:
$(\Rightarrow )$ Assume that R is a UFD. Let I be a nonzero prime ideal in R. Then there exists a nonzero element $c\in I$. This element cannot be a unit or else I would be all of R and this is a contradiction since prime ideals must be proper ideals. So then c is a nonzero nonunit. Since we are in a UFD we can write $c = c_1\cdots c_n$ where each $c_i$ is irreducible. Then by the lemma we have that each$c_i$ is in fact prime. Therefore $(c_i)$ is a prime ideal contained in I.
$(\Leftarrow )$ Assume that every nonzero ideal in R contains a nonzero principle ideal that is prime. Let S be the set of all units in R and all products of prime elements of R. Note that S is a multiplicative set since a product of units is a unit and a product of a product of primes is a product of primes. Also, if $p$ is prime and $u$ is a unit, then we need to know that $pu$ is prime or a unit as well. Suppose it is not a unit, then suppose $pu | ab \Rightarrow ab = pux\Rightarrow abu^{-1} = px \Rightarrow p |abu^{-1}$. But p is prime so then $p|a$ or $p|bu^{-1}$. In the first case we are done and if $p| bu^{-1} \Rightarrow bu^{-1} = py \Rightarrow b= pyu \Rightarrow p|b$. Therefore S is closed. We want to show that $S = R \backslash \{0\}$. This will give that every nonzero nonunit is a product of primes and hence a product of irreducibles since we are in an integral domain. Suppose to the contrary that there exists some nonzero nonunit $x\in R \backslash S$. Then we must have that $S \cap (x) = \emptyset$ since S is also closed under divisors, i.e. whenever a product $ab\in S \Rightarrow a,b \in S$. Next we need a few facts: 1) Since $(x)$ is an ideal, it is contained in a maximal ideal. 2) The set of all ideals of R which are disjoint from S contains a maximal element, and that element is prime. Therefore since $(x)$ is an ideal disjoint from S, it is contained in a maximal ideal, P, which is also a prime ideal and also disjoint from S. However, by assumption this prime ideal P contains a principle ideal which is prime, and therefore contains a prime element, say p. But S contains all prime elements so this contradicts the disjointness. Therefore S consists of units and irreducible elements and hence R is a UFD.

6) (a) If a and n are integers, $n> 0$, then there exist integers q and r such that $a = qn + r$ where $|r| \leq n/2$.
(b) the Gaussian integers $\mathbb{Z}[i]$ form a Euclidean domain with $\phi (a +bi) = a^2 + b^2$.
Proof- a)
Let $a,n\in \mathbb{Z}$ with $n>0$. By the division algorithm there exists $q,r \in \mathbb{Z}$ such that $a = qn + r$ where $0 \leq r \frac{n}{2}$. Then $n - r < \frac{n}{2}$. Pick $q' =q + 1$ and $r' = n-r$. Then $a = qn + r = qn + n-n+r = (q+1)n +r-n$, where $|r-n| = n -r <\frac{n}{2}$. This ends the problem

b) Consider $(\mathbb{Z}[i], \phi )$ where $\phi (a+bi) = a^2+b^2$. We must show that this is a Euclidean Ring. First assume that $x= a+bi, y = c+di \in \mathbb{Z}[i]$ and $xy \neq 0$. Then with some work we can show that $\phi (xy) = \phi (x)\phi (y) \geq \phi (x)$ since $\phi (y) \in \mathbb{N}$. Now let $y = a+bi$ and $x\in \mathbb{Z}$, $x >0$. Then we can find integers $q_1,q_2,r_1,r_2$ such that $a = q_1x + r_1, b = q_2x + r_2$, with $|r_1| < \frac{x}{2}, |r_2| <\frac{x}{2}$. Let $q = q_1 + q_2i, r = r_1+r_2i \Rightarrow y = qx + r$. If r is zero we are done so assume r is nonzero. Then $\phi (r) = r_1^2 + r_2^2 \leq \frac{x^2}{2} . So when $x\in \mathbb{Z}$ for $y \neq 0$ there exists $q,r \in \mathbb{Z}[i]$ such that either $r= 0$ or $r\neq 0$ and $\phi (r) 0$. Then just replace x with $x\bar{x}$ in the previous proof. Find $q,r_0 \in \mathbb{Z}[i]$ such that $y\bar{x} = q(x\bar{x}) + r_0$ with either $r_0 = 0$ or $r_0 \neq 0$ and $\phi (r_0) < \phi (x\bar{x})$. Set $r = y -qx$. Then $r\bar{x} = q(x\bar{x}) + r_0 - q(x\bar{x}) = r_0$. Then $\phi (r) \leq \phi (r_0) < \phi (x)$ so we have a Euclidean domain.

7) What are the units of Gaussian integers $\mathbb{Z}[i]$?
Proof –
Let $U(\mathbb{Z}[i]) = \{ x \in \mathbb{Z}[i] | x \text{ is a unit} \}$. We claim that $U(\mathbb{Z}[i]) = \{\pm 1, \pm i\}$ and show it by subset inclusion. Let $x = a+bi \in U(\mathbb{Z}[i])$. Then there is $y = c+di \in \mathbb{Z}[i]$ such that $xy= 1$. By exercise 6(b) we have that $\phi (x) \phi (y) = \phi (xy) = \phi (y) = 1$. Since $\phi (x), \phi (y) \in \mathbb{N}$ we can only have the case that $\phi (x) = 1$. But then $a^2 + b^2 = \phi (x) = 1$ says that either $a = \pm 1$ and $b = 0$ or $a = 0$ and $b = \pm 1$. Thus $x = \pm 1$ or $x = \pm i$. Therefore we have inclusion. The other inclusion is trivial to check.

9) Let R be a UFD and d a nonzero element of R. There are only a finite number of distinct principal ideals that contain the ideal $(d)$.
Proof-
If d is a unit then we are done trivially, so assume d is a nonunit. Then since we are in a UFD we can write $d=d_1 \cdots d_l$ where $d_i$ is irreducible. Let $k\in R$ be any element satisfying $(d) \subset (k)$. Then k cannot be 0 and if k is a unit then it generates all of R so that $(k) = R$ is the only principle ideal which contains (d). So assume that k is a nonunit. Then we can write $k = k_1 \cdots k_m$ with each $k_j$ irreducible. Now $(d) \subset (k) \Rightarrow k|d$ so that $d = kx$ for some $x\in R$. If $x = 0$ then d would be zero also, which is a contradiction. If x is a unit then $k = dx^{-1} \Rightarrow kr = dx^{-1}r \forall r \in R$ which tells us that $(k) = (d)$ is the only principle ideal containing $(d)$. Therefore we can assume x is a nonunit. Then we can write $x = x_1\cdots x_n$ for some irreducibles $x_k$. Then we have $d_1 \cdots d_l = k_l\cdots k_m x_1 \cdots x_n \Rightarrow l = m+n$ since we are in a UFD. Note that for each $1\leq i \leq m$ we have $(k) \subset (k_i)$ and each $k_i$ is irreducible so that each $(k_i)$ is maximal. Thus $(k)$ is contained in m maximal ideals. If there were infinitely many such principal ideals $(k)$ we would have to have infinitely many m’s which satisfy $m +n = l$. But $l,m,n \in \mathbb{N}$  and l is fixed so this is impossible. Hence there are only finitely many k satisfying the condition.

10) If R is a UFD and $a,b \in$ are relatively prime and $a|bc$ then $a|c$.
Proof-
First note that if $gcd(a,b) = 1$ then $a,b$ have no irreducible factors in common. This means that if $a = a_1 \cdots a_m, b = b_1\cdots b_n$ with $a_i,b_j$ irreducible, then no $a_i$ is an associate of $b_j$. For if they were associates, then we could say $a_i = b_ju$ where u is a unit, then $a = a_1\cdots a_{i-1}b_jua_{i+1}\cdots a_m \Rightarrow b_j |a$. But since $b_j |b$ and gcd(a,b) = 1 this would give that $b_j |1$ which means that $b_j$ is a unit, contradicting that $b_j$ is irreducible. Finally, assume that $a|bc$. Then $bc = ax$for some $x\in R$. Factoring into products of irreducibles we have that $b_1 \cdots b_n c_1 \cdots c_l = a_1 \cdots a_m x$. Now $a_i$ cannot be associate of any $b_j$ so each $a_i$ must be an associate of some $c_k$. Therefore $l\geq m$ so without loss of generality say that $a_i$ is an associate of $c_i$ for $1 \leq i \leq m$. This tells us that $a_i | c_i$ for $1 \leq i \leq m$. Thus we can say $c_1 = a_1 y_1, c_2 = a_2y_2, \ldots c_m = a_m y_m \Rightarrow c_1 \cdots c_m = a_1\cdots a_m y_1 \cdots y_m =a y_1 \cdots y_m$. But then we can multiply by what we need to get all of c on one side, so that $c = c_1 \cdots c_m c_{m+1} \cdots c_l = ay_1\cdots y_m c_{m+1}\cdots c_l \Rightarrow a |c$.

11) Let R be a Euclidean ring and $a\in R$. Then a is a unit iff $\phi (a ) = \phi (1_R)$.
Proof-
$( \Rightarrow )$ Assume that a is a unit. Then $\phi(a) \leq \phi(aa^{-1} )= \phi (1) \leq \phi (1a) = \phi (a)$. Therefore $\phi (a) = \phi (1)$.
$(\Leftarrow )$ Assume now that $\phi (a) = \phi (1)$. There must exist $q,r \in R$ such that $1 = aq +r$ where either r = 0 or $\phi (r) < \phi (a)$. Note that $\phi (1) \leq \phi (1r) = \phi (r)$, so we cannot have that $\phi (r) < \phi (a) = \phi (1)$. Therefore r = 0 and hence $q = a^{-1}$ so that a is a unit.