Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I could not believe that my professor asked this on the exam especially since this is intro-level abstract algebra. My idea was to somehow find a norm function and use the inequality $N(r) < N(d)$ in $m = qd + r$. I erased that and said: "since we know that $\mathbb{R}$ is an UFD we can show that $\mathbb{R[x]}$ is an UFD." We only had 30 minutes to do per problem and this was one of the unusually hard problems.

share|improve this question
I don't know what do you mean by "intro-level", but since you know that the existence of a norm function implies unique factorization, that is, any euclidean ring is an UFD, then you must know (to prove) that the ring of polynomials over a field is euclidean. In this case I won't blame your teacher at all! –  user26857 Dec 22 '12 at 9:05
add comment

3 Answers

up vote 5 down vote accepted

Here's a more general case which leads to proving the division algorithm for any polynomial ring $F[x]$, where $F$ is a field.

Let $f$ and $g\neq 0$ be polynomials in $R[x]$, for $R$ a commutative ring. Let $\deg(g)=m$, with $b_m$ its leading coefficient. I claim we can write $$ b_m^j f(x)=q(x)g(x)+r(x) $$ for $j\in\mathbb{N}$, $q,r\in R[x]$, and $\deg(r)<\deg(g)$. First, if $\deg(f)<\deg(g)$, then simply take $f(x)=0\cdot g+f$, and you're done. Otherwise, suppose $\deg(f)=n\geq m=\deg(g)$, and let $a_n$ be the leading coefficient of $f$. Set $$ b_mf(x)-a_nx^{n-m}g(x)=f_1(x). $$ Since the coefficient of $x_n$ in $b_mf(x)$ and $a_nx^{n-m}g(x)$ is $a_nb_m$ in both cases, $\deg(f_1)<\deg(f)$. So by induction on the degree of $f(x)$, one can find $j_1\in\mathbb{N},q_1(x),r(x)\in R[x]$ with $\deg(r)<\deg(g)$ such that $$ b_m^{j_1}f_1(x)=g(x)q_1(x)+r(x). $$ Combining these, you find $$ b_m^{j_1+1}f(x)=b_m^{j_1}a_nx^{n-m}g(x)+g(x)q_1(x)+r(x)=g(x)q(x)+r(x) $$ where $q(x)=b_m^{k_1}a_n^{n-m}+q_1(x)$.

So in particular, if $b_m$ is a unit, you can multiply both sides of the above equation by $b_m^{-1}$ appropriately many times to find $$ f(x)=q(x)g(x)+r(x) $$ for slightly different $q(x)$ and $r(x)$, but still $\deg(r)<\deg(g)$. Thus the division algorithm of course holds for $F[x]$ where $F$ is a field. The following shows how to use the division algorithm to show $F[x]$ is in fact a PID, and BenjaLim's very nice answer shows how to then prove a PID is a UFD.

Let $I$ be any nonzero ideal. Then there exists a nonzero polynomial $g(x)$ of minimal degree among the nonzero elements of $I$. Then for any polynomial $f(x)\in I$, we can apply the division algorithm to find $$ f(x)=q(x)g(x)+r(x) $$ where $r(x)$ has lesser degree than $g(x)$. But then $$ r(x)=f(x)-q(x)g(x)\in I $$ so necessarily $r(x)=0$ lest we contradict the minimality of the degree of $g(x)$. So it follows that $I=(g(x))$. Since the zero ideal is obviously principal, and since $\mathbb{R}$ is a domain, so is $\mathbb{R}[x]$, it follows that $\mathbb{R}[x]$ is a PID, thus a UFD.

Note that this holds generally for $F[x]$ for any field $F$. (But only in the case of one indeterminate!)

share|improve this answer
You're already assuming that $\mathbb{R}[x]$ is an ED (where you use the division algorithm). ED $\Rightarrow$ PID, so your work here is superfluous. –  andybenji Dec 22 '12 at 15:12
@andybenji Thank you for your criticism. I wouldn't call it superfluous, as it shows why the general implication $ED\implies PID$ holds in this case, rather than just stating the result. I included justification for the division algorithm. –  Ben West Dec 22 '12 at 23:10
@BDub +1 For your nice answer. –  fpqc Dec 23 '12 at 0:09
@Bdub Very nice edits. –  andybenji Dec 24 '12 at 0:04
add comment

You may prove that $\mathcal{N}(f) = \deg(f)$ is a Euclidean norm on $\mathbb{R}[x]$ (more generally $k[x]$ for any field $k$), thus $\mathbb{R}[x]$ is a Euclidean domain $\Rightarrow$ PID $\Rightarrow$ UFD.

And the logic you wrote down is correct: A field is (trivially) a UFD, since all elements are units, and it is usually proved in classes that $R[x]$ is a UFD if $R$ is a UFD, so your teacher should have given you points for that.

share|improve this answer
add comment

I think we can bypass the norm and just prove that $\Bbb{R}[x]$ being a PID implies that is a UFD. The fact that it is a PID is clear because of the existence of the division algorithm. We now prove that it is a UFD. Now first of all we claim it is a factorisation domain; that is to say that any $f \in \Bbb{R}[x]$ can be written as a unit times a finite product of irreducibles polynomials. For suppose there is an $f$ in such a way. Then in particular because $f = 1\cdot f$ this would mean that $f$ itself cannot be irreducible and hence we can write $f = f_1f_2$ for some polynomials $f_1$ and $f_2$. These cannot be irreducible and so continuing this process we get an ascending chain of ideals

$$(f) \subseteq (f_1) \subseteq (f_3) \subseteq \ldots $$

that does not termiminate contradicting $\Bbb{R}[x]$ being a PID. To prove that we have uniqueness of factorisations, suppose

$$f = f_1\ldots f_k = q_1\ldots q_n$$

where the $f_i's$ and $q_m'$ are all irreducible. Now we have $(f_1\ldots f_k) = (f_1)\ldots (f_k) \subseteq (q_1).$

Because $\Bbb{R}[x]$ is a PID we know TFAE:

  1. $q_1$ irreducible
  2. $(q_1)$ is a prime ideal.
  3. $(q_1)$ is a maximal ideal.

It will now follow that at least one of the ideals in the product $(f_1)\ldots (f_k)$ say $(f_1)$ is contained in $(q_1)$. By maximality

$$(f_1) = (q_1)$$

and so $f_1 = q_1$. Continuing this way all the $f_i$ are equal to the $q_j$ and so we have proved uniqueness.

share|improve this answer
Like the above answer, you're assuming the division algorithm, where the division algorithm (a) needs to be proved to hold, and (b) shows that $\mathbb{R}[x]$ is a ED, which is a stronger result. –  andybenji Dec 22 '12 at 15:14
@andybenji I don't know of an abstract algebra class that doesn't cover the division algorithm. Besides, this is just ordinary polynomial long division as in high school. You mean to say the OP can't use results in class in an exam? The division algorithm shows that $\Bbb{R}[x]$ is a PID, and in my answer above I proved that a PID is always a UFD. –  fpqc Dec 22 '12 at 23:43
False. The division algorithm shows that $\mathbb{R}[x]$ is a ED. It has probably been proved in his/her class (or on a homework assignment) that ED implies PID implies UFD, so I would only bother showing ED. I doubt the OP proved the existence of a division algorithm for $\mathbb{R}[x]$ in high school, so if I was his/her teacher, no, he/she would not be allowed to use it on a test. And an although an algebra class will cover the division algorithm, it doesn't exist everywhere, so no, you cannot just assume it does exist. In particular, DA only exists in $R[x]$ when $R$ is a field. –  andybenji Dec 24 '12 at 0:01
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.