Math 361, Spring 2017, Assignment 6

Carefully define the following terms, then give one example and one non-example of each:

1. Princpial ideal (generated by an element $a$ in a commutative ring $R$).
2. Principal ideal domain (a.k.a. PID).
3. Standard representative (of a coset in the quotient ring $F[x]/\langle m\rangle$ where $F$ is a field and $m$ is a polynomial with coefficients in $F$).
4. Prime ideal.
5. Maximal ideal.

Carefully state the following theorems (you do not need to prove them):

1. Fundamental theorem of ring homomorphisms.
2. Classification of ideals in $F[x]$ ("Every ideal of $F[x]$ is...").
3. Theorem relating prime ideals to integral domains.
4. Theorem relating maximal ideals to fields.

Solve the following problems:

1. Section 27, problems 1, 2, 3, 10, 11, 15, 16, and 17.

--------------------End of assignment--------------------

Questions

I having trouble with Section 27, problem 16). Not quite sure where to start. The question is find a prime ideal of $\mathbb{Z} \times \mathbb{Z}$ that is not maximal. I was under the assumption that every prime ideal was also maximal. In problem 1 of Section 27 say find all the prime ideals and all maximal ideals of $\mathbb{Z}_6$ so...the work is... $ 1 = \mathbb{Z_6}$

$(2) = \{0,2,4\}$

$(3) =\{ 0,3\}$

$(6)= \{0\}$

$(2) = \{0,2,4\}$ Are both prime so they are both prime and maximal. $(3) =\{ 0,3\}$

and $(6)= \{0\}$ is neither prime nor maximal.

Every maximal ideal is prime, but not the other way around. In $\mathbb{Z}\times\mathbb{Z}$, take $I=\{(0,j)|j\in\mathbb{Z}\}$. This is an ideal because it is the kernel of the homomorphism $(i,j)\mapsto i$. To see that it is prime, apply the Fundamental Theorem of Homomorphisms to the map $(i,j)\mapsto i$ and conclude that the corresponding quotient ring is isomorphic to $\mathbb{Z}$, an integral domain. The ideal is not maximal since the quotient ring is not a field.

In $\mathbb{Z}_6$, both $\langle2\rangle$ and $\langle3\rangle$ are maximal (hence also prime) since their quotient rings are isomorphic to $\mathbb{Z}_2$ and $\mathbb{Z}_3$, respectively, which are both fields. The improper ideal is never maximal or prime, by definition, while $\langle0\rangle$ is not prime (hence not maximal either) because $\mathbb{Z}_6$ itself is not an integral domain. -Steven.Jackson (talk) 09:38, 10 April 2017 (EDT)