Why Do Functions \(X + UY\) and \(X - UY\) Have Zeroes Only at \(t = \pm 1\)?

[mathmari]

Hey!

I am reading the proof of the following lemma but I got stuck at some points... Let $R$ be any integral domain of characteristic zero.

We consider the Pell equation $$X^2-(T^2-1)Y^2=1 \tag 1$$ over $R[T]$.

Let $U$ be an element in the algebraic closure of $R[T]$ satisfying $$U^2=T^2-1 \tag 2$$

Define two sequences $X_n, Y_n, n=0, 1, 2, \dots$, of polynomials in $\mathbb{Z}[T]$, by setting $$X_n+UY_n=(T+U)^n\tag 3$$

Lemma.

The solutions of $(1)$ in $R[T]$ are given precisely by $$X=\pm X_n, Y=\pm Y_n, n=0, 1, 2, \dots$$

Proof.

$(1)$ is equivalent too $$(X-UY)(X+UY)=1 \tag 4$$

From $(3)$ and $(2)$ follows $$X_n-UY_n=(T-U)^n=(T+U)^{-n}$$

Hence the $X_n, Y_n$ are solutions of $(1)$.

Conversely, suppose $X$ and $Y$ in $R[T]$ satisfy $(1)$.

Let us parametrise the curve $(2)$ by $$T=\frac{t^2+1}{t^2-1}\ \ ,\ \ U=\frac{2t}{t^2-1}$$

The rational functions $X+UY$ and $X-UY$ in $t$ have poles only at $t=\pm 1$.

Moreover $(4)$ implies they have zeroes only at $t=\pm 1$.

Hence $$X+UY=c\left ( \frac{t+1}{t-1}\right )^m=c(T+U)^m, c \in R, m \in \mathbb{Z}$$

Thus also $X-UY=c(T-U)^m$.

But substistuting this in $(4)$ gives $c^2=1$, which proves the lemma by $(3)$.

Could you explain to me the following part of the proof?
Moreover $(4)$ implies they have zeroes only at $t=\pm 1$.

Hence $$X+UY=c\left ( \frac{t+1}{t-1}\right )^m=c(T+U)^m, c \in R, m\in \mathbb{Z}$$

Thus also $X-UY=c(T-U)^m$.

But substistuting this in $(4)$ gives $c^2=1$, which proves the lemma by $(3)$.

Why do we have from the relation $(4)$ that the functions $X+UY$ and $X-UY$ have zeroes only at $t=\pm 1$ ?

How do we get to the equation $X+UY=c\left ( \frac{t+1}{t-1}\right )^m=c(T+U)^m$ ? How does $c$ appear? 11

 

[Ackbach]

Why do we have from the relation $(4)$ that the functions $X+UY$ and $X-UY$ have zeroes only at $t=\pm 1$ ?

Ok, so here's $(4)$:
$$(X-UY)(X+UY)=1.$$
Suppose $X-UY$ had a pole at $t=1$. That means it blows up at $t=1$. What's the only way the entire expression $(X-UY)(X+UY)$ could equal $1$ when $t=1$?

What's puzzling me, though, is that it's not entirely obvious to me (moving back a bit in the proof) why, if $T$ and $U$ have poles only at $t=\pm 1$, which is clear, that $(X\pm UY)$ should have poles only at $t=\pm 1$. *thinking out print here* If we examine $X^2-U^2Y^2=1$, and plug in the parametrization, we get
$$X^2-\frac{4t^2}{\left(t^2-1\right)^2} \, Y^2=1.$$
In order for this equation to hold, $Y$ would simply have to have a zero at $t=\pm 1$. Otherwise, we would have an $\infty-\infty$ situation, which is undefined. Because of this fact, it's not clear to me that $X\pm UY$ have poles at all. I'm not sure I buy this proof! It looks to me like the author is hammering $(X-UY)(X+UY)=1$ for all it's worth, but ignoring $X^2-U^2Y^2=1$ and its implications for poles and zeros.