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 eachc_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} <x^2 = \phi (x). 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.

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