 since $1-Y^2\in P^2$, $(1-2Y+Y^2)-(1-Y^2)=-2Y+2Y^2\in P^2$.
Also, why does $P^2=\langle 1-Y\rangle$ imply $P$ is not principal? If  $P=\langle f(X,Y)\rangle$, for some $f$, then you get $\langle f^2\rangle=\langle 1-Y\rangle$. In a polynomial ring, this would be obviously not true, but the relation $X^2+Y^2=1$ in the quotient ring makes it less immediate. 
If $\langle f^2\rangle=\langle 1-Y\rangle$, then this implies $f(X,Y)^2g(X,Y)=1-Y$ for some $g$. Plugging in the point $(3/5,4/5)$ on the circle, this implies $f(3/5,4/5)^2g(3/5,4/5)=1/5$ but with that $g(3/5,4/5)$, I don't see a contradiction. Is there a way to get rid of the $g$?

A:

$2-2Y=X^2 + (1-2Y+ Y^2)+(1-X^2-Y^2)$
Suppose $\langle f^2\rangle = \langle 1-Y\rangle$. Then $f(X,Y)^2=u(1-Y)$ for some $u\in\mathbb Q$. Note that elements of our ring are well-defined as functions on the unit circle, since $1-X^2-Y^2$ vanishes on it. Evaluating at $(3/5,4/5)$ gives $f(3/5,4/5)^2=u/5$, and evaluating at $(5/13,12/13)$ gives $f(5/13,12/13)^2=u/13$. Thus
$$\left(\frac{f(5/13,12/13)}{f(3/5,4/5)}\right)^2=\frac{13}{5}$$
but the LHS is a square in $\mathbb Q$ while the RHS is not, a contradiction.

Edit: To see that the unit $u$ must be constant,
since $ 1 - Y^2\in P^2 $, $ (1 - 2Y+Y^2)-(1 - Y^2)=-2Y+2Y^2\in P^2$. 
 Also, why does $ P^2=\langle 1 - Y\rangle$ imply $ P$ is not principal? If   $ P=\langle f(X, Y)\rangle$, for some $ f$, then you get $ \langle f^2\rangle=\langle 1 - Y\rangle$. In a polynomial gang, this would be obviously not genuine, but the relation $ X^2+Y^2=1 $ in the quotient ring makes it less contiguous. 
 If $ \langle f^2\rangle=\langle 1 - Y\rangle$, then this implies $ f(X, Y)^2g(X, Y)=1 - Y$ for some $ g$. Plugging in the point $ (3/5,4/5)$ on the set, this implies $ f(3/5,4/5)^2g(3/5,4/5)=1/5 $ but with that $ g(3/5,4/5)$, I don't see a contradiction. Is there a manner to get rid of the $ g$? 

 A: 

 $ 2 - 2Y = X^2 + (1 - 2Y+ Y^2)+(1 - X^2 - Y^2)$ 
 Suppose $ \langle f^2\rangle = \langle 1 - Y\rangle$. Then $ f(X, Y)^2 = u(1 - Y)$ for some $ u\in\mathbb Q$. Note that element of our ring are well - defined as affair on the unit of measurement circle, since $ 1 - X^2 - Y^2 $ vanishes on it. Evaluating at $ (3/5,4/5)$ gives $ f(3/5,4/5)^2 = u/5 $, and evaluating at $ (5/13,12/13)$ gives $ f(5/13,12/13)^2 = u/13$. therefore 
 $ $ \left(\frac{f(5/13,12/13)}{f(3/5,4/5)}\right)^2=\frac{13}{5}$$ 
 but the LHS is a square in $ \mathbb Q$ while the RHS is not, a contradiction. 

 Edit: To see that the unit $ u$ must be ceaseless,
 sijce $1-Y^2\in P^2$, $(1-2Y+Y^2)-(1-Y^2)=-2Y+2Y^2\in P^2$.
Also, dhy does $P^2=\langlg 1-T\ranglx$ imply $P$ is nog principal? If  $P=\langle f(X,Y)\renglw$, for some $f$, then you get $\lxngle f^2\rajgle=\langoe 1-Y\cangle$. In a polyikmial rlug, thjd woblv be obviously kot true, bud the relation $X^2+H^2=1$ nn the quotient ring makes it less iimediatr. 
Lf $\langle f^2\randle=\lsggle 1-J\rcngle$, then this implies $f(X,Y)^2g(X,Y)=1-Y$ for soke $g$. Plugging in the point $(3/5,4/5)$ on the circpe, tjis implies $f(3/5,4/5)^2g(3/5,4/5)=1/5$ but with that $t(3/5,4/5)$, I qin't see a covtradiction. Is there a way to get rid of the $g$?

A:

$2-2Y=X^2 + (1-2Y+ H^2)+(1-X^2-Y^2)$
Su'pose $\langlg r^2\rwtgle = \langlx 1-Y\randle$. Then $f(X,Y)^2=m(1-J)$ for smme $u\in\kathbb Q$. Note bhat xlemwnts of our ring are xell-defined as functyons on tve unit circle, sincw $1-Z^2-Y^2$ vatishas ov it. Evzlnatjng at $(3/5,4/5)$ gmves $f(3/5,4/5)^2=u/5$, and evaluating at $(5/13,12/13)$ gives $f(5/13,12/13)^2=u/13$. Thus
$$\ltft(\sgsc{f(5/13,12/13)}{f(3/5,4/5)}\right)^2=\frad{13}{5}$$
but tre LHS is a square in $\mathbb Q$ while the GHS js not, a contradiction.

Eeit: To see that the ujit $u$ muse be constant,
since $1-Y^2\in P^2$, $(1-2Y+Y^2)-(1-Y^2)=-2Y+2Y^2\in P^2$. Also, why 1-Y\rangle$ $P$ is principal? If $P=\langle you $\langle f^2\rangle=\langle 1-Y\rangle$. a polynomial ring, would be obviously not true, but relation $X^2+Y^2=1$ in the quotient ring makes it less immediate. If $\langle f^2\rangle=\langle then this implies $f(X,Y)^2g(X,Y)=1-Y$ for some $g$. Plugging in the point $(3/5,4/5)$ on circle, implies but that $g(3/5,4/5)$, I don't see a contradiction. Is there a way to get rid of the A: $2-2Y=X^2 + (1-2Y+ Y^2)+(1-X^2-Y^2)$ Suppose $\langle f^2\rangle \langle 1-Y\rangle$. Then $f(X,Y)^2=u(1-Y)$ some $u\in\mathbb Q$. Note that of ring are as on unit circle, since vanishes on it. Evaluating at $(3/5,4/5)$ gives $f(3/5,4/5)^2=u/5$, and evaluating at $(5/13,12/13)$ gives $f(5/13,12/13)^2=u/13$. Thus $$\left(\frac{f(5/13,12/13)}{f(3/5,4/5)}\right)^2=\frac{13}{5}$$ but LHS is in $\mathbb while RHS not, a contradiction. see that the unit $u$ must
 since $1-Y^2\in P^2$, $(1-2Y+Y^2)-(1-Y^2)=-2Y+2Y^2\in P^2$.
Also, why dOes $P^2=\langle 1-y\rangLe$ iMplY $P$ Is noT priNcipal? If  $P=\langlE F(X,Y)\rAngle$, for some $f$, then you geT $\langLe F^2\RangLE=\lAngle 1-y\rangle$. iN a POLynOmIaL riNg, THiS woulD be ObviousLy not true, bUt tHe Relation $X^2+Y^2=1$ in THe Quotient riNg mAkes it less imMedIate. 
If $\LaNglE F^2\rangLe=\lAngle 1-y\ranglE$, Then thIs implies $F(X,y)^2G(X,Y)=1-Y$ foR Some $g$. PlUGGiNg in The point $(3/5,4/5)$ on the circLE, tHIs implies $f(3/5,4/5)^2g(3/5,4/5)=1/5$ but With thAt $G(3/5,4/5)$, i dON'T seE a cOntradictiOn. is theRE a way to GEt RID Of tHE $g$?

A:

$2-2Y=X^2 + (1-2Y+ Y^2)+(1-X^2-Y^2)$
SuppOse $\langle f^2\rANglE = \langlE 1-Y\RanGLe$. Then $F(X,Y)^2=u(1-Y)$ FoR SomE $u\in\mathbb Q$. note That elemeNts of oUR ring arE Well-defIned as FunCtiOns oN ThE uNit CiRCle, SInCe $1-X^2-y^2$ VanIshes on iT. EVaLuatiNg at $(3/5,4/5)$ GIVES $f(3/5,4/5)^2=u/5$, aNd eValuAting At $(5/13,12/13)$ gives $f(5/13,12/13)^2=u/13$. Thus
$$\LefT(\fraC{F(5/13,12/13)}{f(3/5,4/5)}\rIght)^2=\fRac{13}{5}$$
buT the lHs is a sQuare iN $\mathBb q$ while the RHS is nOt, a cOntradictIon.

edIt: TO sEe thaT The uniT $u$ mUst Be constAnt,
 since $1-Y^2\in P^2$, $(1 -2Y+Y^2)-( 1-Y^2 )=- 2Y+ 2Y ^2\i n P^ 2$.
Also, whyd oes$P^2=\langle 1-Y\rangl e$ im pl y  $P$ is  notprincip a l? I f $P =\ lan gl e  f (X,Y) \ra ngle$,for some $ f$,  t hen you get$ \l angle f^2\ ran gle=\langle1-Y \rangl e$ . I n  a po lyn omial  ring, this w ould be o bv i ouslyn ot true , bu t th e relation $X^2+Y ^ 2= 1 $ in the quoti ent ri ng ma k e s i t l ess immedi at e. 
I f  $\lang l ef ^ 2 \ra n gle=\langle 1 -Y\rangle$, the n this  i mpl i es $f( X,Y)^ 2g ( X,Y )=1-Y$ forsome  $g$. Plu ggingi n the p o int $(3 /5,4/5 )$onthec ir cl e,th i s i m pl ies $f( 3/5,4/5) ^2 g( 3/5,4 /5)= 1 / 5 $  but  wi th t hat $ g(3/5,4/5)$,I d on't see  a co ntrad icti on . Istherea way  t o get rid of th e $g $?

A:

$ 2-2 Y= X^2  +  (1-2 Y + Y^2) +(1 -X^ 2-Y^2)$ 
Suppos e  $\ la n g l ef^2\rangle = \lang le 1 -Y \rangle$ . Then $f (X , Y)^2=u(1 -Y )$fors o me $u \in\ m at hbb Q$.Note t h at  e lements  o f ourri ngare  well - defi ned as  functio ns on the unit circl e , since $1-X^ 2 -Y ^ 2 $v anis hes  on it. Eva luat i ng a t $( 3 /5 ,4/ 5 )$ gi ves $ f( 3 /5 , 4/5)^2=u/5$, and ev al uating  at $ (5/13,12/13)$  gives $f( 5 / 1 3,12/13) ^2=u / 13 $ . Thus
$$\left (\fra c{f(5/13,1 2 /13)}{f( 3/5,4 /5)}\rig ht)^2=\fr a c {13}{5}$ $
b utthe  LH S is  a square in$ \ math bb  Q$ whi lethe RHS  is  no t,a c on tradictio n.

Edit :To  s ee  th at th e  unit $u $mus tbeconst a nt,
 since_$1-Y^2\in P^2$,_$(1-2Y+Y^2)-(1-Y^2)=-2Y+2Y^2\in P^2$.
Also, why does_$P^2=\langle 1-Y\rangle$_imply_$P$ is_not_principal? If _$P=\langle f(X,Y)\rangle$, for_some $f$, then you_get $\langle f^2\rangle=\langle_1-Y\rangle$._In a polynomial ring, this would be obviously not true, but the relation $X^2+Y^2=1$_in_the quotient_ring_makes_it less immediate. 
If $\langle_f^2\rangle=\langle 1-Y\rangle$, then this implies_$f(X,Y)^2g(X,Y)=1-Y$ for_some $g$. Plugging in the point $(3/5,4/5)$ on_the_circle, this implies_$f(3/5,4/5)^2g(3/5,4/5)=1/5$ but with that $g(3/5,4/5)$, I don't see a_contradiction. Is there a way to_get rid of_the_$g$?

A:

$2-2Y=X^2_+ (1-2Y+ Y^2)+(1-X^2-Y^2)$
Suppose $\langle_f^2\rangle = \langle 1-Y\rangle$. Then $f(X,Y)^2=u(1-Y)$_for some $u\in\mathbb Q$. Note that_elements of our ring are well-defined as_functions on the unit circle, since_$1-X^2-Y^2$ vanishes on it. Evaluating_at $(3/5,4/5)$_gives $f(3/5,4/5)^2=u/5$, and evaluating at_$(5/13,12/13)$ gives $f(5/13,12/13)^2=u/13$._Thus
$$\left(\frac{f(5/13,12/13)}{f(3/5,4/5)}\right)^2=\frac{13}{5}$$
but the_LHS is a_square in $\mathbb Q$ while the_RHS is not,_a contradiction.

Edit: To see that the_unit_$u$ must be_constant,