Polynomial rings -- Inherited properties from coefficient ring

Solution 1:

Throughout the post, I keep to the standard assumption that UFDs, PIDs, EDs and integral domains all refer to commutative rings. (But of course, there are noncommutative domains and PIDs and even some of the others, if you work hard enough ;) )

$R[X]$ is a UFD when $R$ is

For $B$, you can find in many commutative algebra texts that a commutative ring $R$ is a UFD iff $R[x]$ is. (For example, Corollary 16.20 in Isaacs Graduate Algebra)

Why is it clear, that a principal ideal domain is a integral domain?

Look back at your definitions: a principal ideal domain is just an integral domain with an extra property (having all ideals principal). A PID is a fortiori an integral domain.

After reading what you described about your definition, it sounds like maybe this didn't make it into your notes. A principal ideal ring is a ring in which all ideals are principal, but such a ring doesn't have to be a domain (For example, $\Bbb Z/\Bbb 4$ is a principal ideal ring, but not a domain, since $2^2=0$.) A (commutative) principal ideal domain is just a (commutative) principal ideal ring that is also a domain.

Hierarchy of properties

For $C$: Making a hierarchy like this is really good exercise. (In fact, I've embarked on pictures like that with dozens of ring types.) However, I hope you're not under the impression that you are going to organize all ring types linearly.

All of the domains you mentioned are subclasses of commutative rings, but the class of division rings is not contained in commutative rings. Out of all the rings you mentioned, there is one branch containing the domains:

$\text{field}\subseteq \text{Euclidean domain}\subseteq PID\subseteq UFD\subseteq\text{domain}\subseteq \text{commutative ring}\subseteq \text{ring}$

and then there is another branch

$\text{field}\subseteq\text{division ring}\subseteq\text{ring}$

You wrote that a PID "does not have a euclidean function" which is a bit like concluding that a rectangle does not have four equal side lengths. A PID does not necessarily have a euclidean function, but it might. Just like rectangles might have four equal sides, and hence be both squares and rectangles. You should just keep in mind that a Euclidean domain has more stringent structure than a PID, since they are a special subcase. Similar comments can be made about what you wrote about a UFD not having all ideals principal, etc.

$R[X]$ not a field

For $D$: To easily see that $R$ is not a division ring, just ask yourself if you can invert $X$ or not. When you multiply polynomials together, you're only going to get higher degrees of $X$. How will you get back down to $1$?

Inheritance

People have already pointed out how it's pretty easy to prove that $R$ is a domain iff $R[x]$ is, or the same for commutativity, and for the UFD property. Just in the last section we see that the case is not so for "being a field". Someone has also given an example that whlie $F[x]$ is a PID, $F[x][y]$ is not, so that property isn't preserved either. The same is also true for Euclidean domains since $F[x]$ is actually an example of a Euclidean domain.

Solution 2:

$R$ is a ring with $1$ iff $R[X]$ is a ring with $1$.

$R$ is commutative iff $R[X]$ is commutative.

$R$ has zero divisors iff $R[X]$ has zero divisors.

Your A is wrong. For example $\mathbb Z[X]^\times=\mathbb Z^\times=\{-1,1\}$ (and $\mathbb Z$ is not a field). If $R$ is any unital ring, we have $R^\times =R[X]^\times$ (at least as long as there are no zero divisors)

Note that $R[X]$ will never be a division ring as $X$ has no inverse.

Also note that $\mathbb Q[X,Y]$ is not pricipal (this observation is the fundament of algebraic geometry) since $(X,Y)$ is not a principal ideal