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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.

Thanks!

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1 Answer

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 $$2=(\sqrt2)^2=(\root4\of2)^4=(\root8\of2)^8=\dots$$

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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
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