# Example in which the first part of the definition of UFD fails to hold

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