Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There is this example in which the first part of the definition of UFD (that is the existence of factorisation) fails to hold that I don't quite understand.

Let $R=\mathbb{R}[X_1,X_2,\dots]$, and let $I\triangleleft R$ be the ideal generated by the set $\{X_2^2-X_1,X_3^2-X_2,X_4^2-X_3,\dots\}\subset R$.

Then in $R/I$ the element $X_1+I$ has no factorisation as a product of irreducibles (and is not a unit): $$X_1+I=(X_2+I)(X_2+I)=(X_3+I)(X_3+I)(X_3+I)(X_3+I)=\dots$$

My questions are:
1. First of all, why is $X_1+I$ is an element in $R/I$? By using the definition of quotient ring, $X_1\in R$, right? But if we use the definition of polynomial ring $R=\mathbb{R}[X_1,X_2,\dots]=((\mathbb{R}[X_1])[X_2])[X_3]\dots$, how can we show $X_1\in R$? These multilayers seems a bit confusing for me..
2. I also have little clue how can we derive $X_1+I=(X_2+I)(X_2+I)$? I know that somehow we need to use the definition of $I$ as the generating set of $\{X_2^2-X_1,X_3^2-X_2,X_4^2-X_3,\dots\}$, but not entirely sure how to proceed.

It will be really appreciated if anyone could help me out on this.


share|cite|improve this question
up vote 1 down vote accepted

$\bf R$ is the reals, and the number $1$ is in $\bf R$, so $X_1$ is in ${\bf R}[X_1]$, so I'm not sure where the difficulty is in seeing that it's in $R$.

For the second question, the way multiplication of cosets goes, $(X_2+I)(X_2+I)$ is $X_2^2+I$, the coset containing $X_2^2$. But $X_2^2-X_1$ is in $I$, so $X_2^2+I=X_1+I$.

Another example along the same lines is that the set of all algebraic numbers is not a UFD because $$2=(\sqrt2)^2=(\root4\of2)^4=(\root8\of2)^8=\dots$$

share|cite|improve this answer
Thanks for your answer. For the first question, I know that $X_1\in\mathbb{R}[X_1]$ but don't know why $X_1\in R$. Here $R$ is different to $\mathbb{R}$, $\mathbb{R}$ is the reals whereas $R$ is the polynomial ring $\mathbb{R}[X_1,X_2,\dots]$, hopefully that clarifies my question. – user71346 Jun 8 '13 at 1:53
Do you understand why $x$ is in ${\bf R}[x,y]$? ${\bf R}[x,y]$ is polynomials in $x$ and $y$. If $f$ is a polynomial in $x$ then it is also a polynomial in $x$ and $y$. I'm having trouble understanding exactly what the difficulty is. – Gerry Myerson Jun 8 '13 at 3:36
Sorry, now I understand it. What is written in my notes is quite confusing. But now its clear. Thanks so much! – user71346 Jun 8 '13 at 5:45

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.