To avoid mixing up things, I wanted to collect properties a polynomial ring is inheriting from the coefficient ring and what property implies another.

Let $R$ be a ring (what else do I need at which point, so $1$, commutative, etc. -- our teacher is somehow not paying to much attention on these details).

A. It is $R^*=R[X]^*$ iff $R$ is a field. Why? What about $R[X][Y]^*=R[X,Y]^*=?$

B. $R$ factorial (or $R$ field) $\implies$ $R[X]$ is factorial $\implies$ $R[X,Y]$ is factorial

C. Digested property hierarchy:


euclidean ring (missing ? to a field?)

principal ideal domain (no euclidean function exists -- intuitive meaning?)

factorial ring here unique factorization domain (not every ideal is a principal ideal -- intuitive meaning?)

integral domain (no unique factorization of elements)

commutative ring

ring (not commutative)

Where to insert division ring, so the existence of inverses for every $a\in R\setminus\{0\}$? I'm missing the intuition about these terms, so why is it clear, that a principal ideal domain is a integral domain? Is it just my teacher being sloppy and actually a principal ideal domain is defined as a factorial ring where every ideal is a principal ideal and so forth? Since in my definition it just says "a commutative ring, with 1, where every ideal is a principal ideal, is a principal ideal domain".

D. Example. Let $R = \mathbb{Q}[X,Y]$.

Since $\mathbb{Q}$ is a field, it follows, that $R$ is factorial. Hence (going down the aforementioned hierarchy) $R$ is a integral domain, a commutative ring, a ring. Is it a division ring?

I guess it is $R[X,Y]^*=R^*=\mathbb{Q}\setminus\{0\}$.

My idea is that $R$ is a principal ideal domain, but I didn't find a very rigorous argument. I think there are basically only the trivial ideals in $R$, and for the trivial ideals it is clear, that they are principal ideals since $R=R1$ and $\{0\}=R\{0\}$.

I don't know how to decide on euclidean and field. Just my intuition is, that i can't have division with remainder and therefore $R$ shouldn't be a euclidean ring and therefore not a field. But how "to show" that there can't be a euclidean function on $R$? (And is there a way to directly argue that $R$ is not a field?).

  • $\begingroup$ So what is true about my example? $R$ is factorial, since $\mathbb{Q}$ is a field. So $R$ is factorial, a integral domain, commutative domain, a commutative ring, a ring. It is no division ring, since $X$ has no inverse, and hence $R$ is not a field, too. Is my conclusion about principal ideal domain correct? How to decide wether $R$ is an euclidean domain? $\endgroup$ – Frank H Jul 18 '13 at 19:44
  • $\begingroup$ $R$ isn't a PID because $(X,Y)$ isn't principal, and hence it isn't a Euclidean domain either (because all of those are PIDs!) Incidentally you have another proof it isn't a field. If it were a field, it would be a PID! $\endgroup$ – rschwieb Jul 18 '13 at 19:55
  • $\begingroup$ I'm sorry for annoying you with my confusion about all these terms! But how do I see, that a Euclidean domain is a PID? Is it again just by definition? $\endgroup$ – Frank H Jul 18 '13 at 19:58
  • $\begingroup$ No, question are fine :) If they were annoying I'd just stop answering. The fact that ED's are PIDs does not follow by definition (at least, not by normal definition!) This implication requires a proof. The proof goes along these lines: for your function $\eta$ and a nonzero ideal $I$, pick a nonzero $x\in I$ such that $\eta(x)$ is minimal. You can then argue that $(x)=I$. Once that is done presto every ED is a PID! $\endgroup$ – rschwieb Jul 18 '13 at 20:05
  • $\begingroup$ You can find that argument resketched here if you'd like to see it in someone else's words. $\endgroup$ – rschwieb Jul 18 '13 at 20:06

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 $1$?


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.

  • $\begingroup$ Ok, so I think for some parts, the problem was, that some details weren't in the definitions (PID is an ID+extra property). Is it the same reason, that a UFD is a integral domain, just a fortriori? What about the "more stringent structure" of euclidean domains, how to see euclidean domains in this hierarchy branch? How to decide wether $R$ in example D is a euclidean domain? (I just now, that one way is to give a Euclidean function, but to state that it is not a Euclidean ring, ...?) $\endgroup$ – Frank H Jul 18 '13 at 19:49
  • $\begingroup$ @FrankH The hierarchy helps you decide about $R$. If $R$ were an ED, it would be a PID. But it's not a PID since $(X,Y)\lhd R$ is not principal. Thus $R$ isn't an ED. $\endgroup$ – rschwieb Jul 18 '13 at 19:51
  • $\begingroup$ PS. So $R[X]$ is UFD, when $R$ is. And when $R$ is commutative this is even stronger: commutative $R$ is UFD, iff $R[X]$ is UFD ($R[X]$ is automatically commutative then, as well)? $\endgroup$ – Frank H Jul 18 '13 at 19:51
  • $\begingroup$ @FrankH Hmm, we've had some miscommunication maybe. If $R$ is commutative, $R[X]$ is commutative. Certainly you can't conclude $R[X]$ is a UFD if $R$ is just commutative. $\endgroup$ – rschwieb Jul 18 '13 at 19:52
  • $\begingroup$ Ok, so I think my problem for PID is, that my understanding of $(X,Y)$ and ideals is just wrong. How does $(X,Y)$ "look like"? $(X,Y) = \{rX,rY:r \in R[X,Y] \}$? So all polynomials with two indeterminates is multiplied with either $X$ and $Y$, to make sure, that for all $r\in R[x]$ lies in the ideal $(X,Y)$ (otherwise $(X,Y)$ wouldn't be an ideal). So is $(X,Y)$ just missing $1$? $\endgroup$ – Frank H Jul 18 '13 at 19:56

$R$ 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

  • 2
    $\begingroup$ Dear Hagen, the equality $R^\times =R[X]^\times$ is false as soon as $R$ has a non-zero element $r$ with $r^2=0$, since then $(1+rX)(1-rX)=1$ . $\endgroup$ – Georges Elencwajg Jul 18 '13 at 18:29
  • $\begingroup$ @GeorgesElencwajg Oops, you're absolutely right $\endgroup$ – Hagen von Eitzen Jul 18 '13 at 18:44
  • $\begingroup$ $R^\times$: all units, i.e. (multiplicative) invertible elements. $R^*$: all non-zero-divisors. So $R^\times = R^*$ for fields, but otherwise we are talking about different sets, no? Is $\operatorname{Quot}(R[X])$ a division ring, since $X$ has the Invverse $1/X$? I don't know the notation $(X,Y)$; what does it refer to and why isn't it a principal ideal? $\endgroup$ – Frank H Jul 18 '13 at 19:29
  • $\begingroup$ @FrankH $(X,Y)$ means "the smallest ideal containing $X$ and $Y$." or "the ideal generated by $X$ and $Y$". In order to find something that generates both $X$ and $Y$, you would have to find a common divisor of $X$ and $Y$ in the ring... but it should be clear that the only such things in that ring are units. $\endgroup$ – rschwieb Jul 18 '13 at 19:35
  • $\begingroup$ Please count me if I'm wrong: [...] and if you then try to generate the ideal from a unit, you would get $X$ and $Y$ but all the other units as well (which are not in $(X,Y)$), hence, you can't generate $(X,Y)$ from one element, and hence $(X,Y)$ is not a PID. $\endgroup$ – Frank H Jul 18 '13 at 20:21

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