% List of Errata for Rational Points on Elliptic Curves
% by Joseph H. Silverman and John Tate
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\def\versiondate{July 5, 1994}
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\centerline{Errata List for {\it Rational Points on Elliptic Curves}}
\centerline{by Joseph H. Silverman and John Tate}
\centerline{Version \versionnumber --- \versiondate}
\bigskip
\noindent
The authors would like to thank the following individuals for their assistance
in compiling this errata sheet:
G.~Allison, % Brown 1992
P.~Berman, % Brown 1993
D.~Appleby, % Wellesley 1992
K.~Bender, % Brown 1992
G.~Bender, % University of Georgia, 1993
J.~Blumenstein, % Brown 1992
D.~Freeman, % Brown 1992
L.~Goldberg, % MSRI 1993
A.~Guth, % 1994
A.~Granville, % University of Georgia, 1993
J.~Kraft, % Ithaca College 1993
J.~Lipman, % Purdue
M.~Mossinghoff, % UTexas 1992
R.~Pennington, % Surrey, UK 1994
R.~Pries, % Brown 1992
K.~Ribet, % Berkeley 1993
H.~Rose, % Bristol, 1993
L.~G\'omez-S\'anchez % Venezuela, 1996
J.-P.~Serre, % College de France 1992
M.~Szydlo, % Boston University 1993
J.~Tobey, % Brown 1992
C.R.~Videla, % U. Los Andes, Bogota 1992
J.~Wendel. % 1993
\par
Material referred to as being on ``attached pages'' is not yet available
for distribution.
\bigskip
% -------------------------------------------------------------------------
\\vii\\Computer Packages\\
Remove the offer to send a formatted disk and give the ftp site for picking
up computer packages. Also include sites for pari, simath, and maple packages
as described in {\it The Arithmetic of Elliptic Curves} Volume~II.
\\1\\Footnote 2\\
Fermat's Last Theorem is now Wiles' Theorem (Summer 1993)! Then again,
maybe not (Spring 1994)! Yes, looks okay (Summer 1995)!
\\4--5\\Footnote\\
Replace ``$f(x,y)$'' with ``$f(x_1,x_2,\ldots,x_n)$'', since
this footnote deals with polynomials in many variables. Further, this footnote
is split at a very bad place between pages 4 and 5, since the part on page 4
alone is grammatically correct and gives a false statement.
\\7\\Line 1\\
``turns them'' \SB ``turns it''
\\11\\Figure 1.2\\
The point marked $(-1,t)$ \SB $(-1,0)$.
\\13\\Lines 11--12\\
Replace ``you take two relatively prime integers $m$ and $n$ and let'' with
``you take two relatively prime integers $m$ and $n$, one odd and one even, and
let''
\\15\\Line 3\\
After ``solution in integers'', add ``, not all zero,''.
\\17\\Paragraph 1\\
Reiterate that this is just a plausibility argument, not a proof, because the
linear conditions might not be independent.
\\22\\Line $-1$\\
After ``simple form.'', add the explanation ``(When we speak of the
``$X$ axis'' in $\PP^2$, we mean the line $X=0$, and similarly for the $Y$
and~$Z$ axes.)''
\\23\\Figure 1.10\\
The lines labeled $X,Y,Z$ \SB labeled $X=0$, $Y=0$, $Z=0$. The point labeled
$\cal O$ \SB labeled ${\cal O}=[1,0,0]$. The point where the $Z$-line hits $C$
\SB labeled $[0,1,0]$.
\\24\\Example\\
Add a more typical example, worked out in detail. (See attached pages for
$X^3+2Y^3+4Z^3=0$, ${\cal O}=[1,1,1]$.) Note that the $u^3+v^3=\alpha$ example
is referred to on the bottom of page 149.
\\26--27\\Singular curves\\
Since we're working over $\RR$, we should also include the ``non-split'' case.
In other words, it's possible to have distinct tangent directions which are
not defined over $\RR$. A typical equation is $y^2=x^2(x-1)$, and the picture
has an isolated point at $(0,0)$. So we should say that there are three
possible pictures for the singularity, and include a third picture. A good
exercise would be to show that if $y^2=f(x)$ is singular, then there is a
change of variables (over $\RR$) which puts the curve into one of the three
standard forms.
\\28\\Section 4\\
Mention the fact that for distinct points~$P,Q,R$ on a Weierstrass equation, we
have $P+Q+R={\cal O}$ if and only if $P,Q,R$ are colinear. More generally,
include an exercise to prove that if $P,Q,R$ are distinct points on any elliptic
curve, then $P+Q+R={\cal O}*{\cal O}$ if and only if $P,Q,R$ are colinear.
\\33\\Line $-2$ of Exercise 1.8\\
$5-adic$ \SB $5$-adic. (The ``adic'' should not be italicized.)
\\34\\Exercise 1.11(c)(iii)\\
Use ``$P*({\cal O}*(Q*R))=R*({\cal O}*(P*Q))$'', since that matches better with
$(P+Q)+R=P+(Q+R)$.
\\36\\Exercise 1.18\\
Use $Q_1,Q_2,\ldots,Q_7$ for the names of the points in this exercise,
since this curve is considered on page~31, where the point~$(2,5)$ is
called~$P_2$.
\\37\\Chapter I Exercises\\
Add a new exercise to show that $xy^2+\cdots$ is smooth if and only if
$y^2+\cdots$ is smooth. (See attached note from Tate.)
\\39\\Line $-6$\\
``of $2P$ and equal'' \SB ``of $2P$ equal''
\\42\\Line $-1$\\
``order ${1\over2}m$'' should be ``order $m$''.
\\48\\Last two lines\\
``so $r(x)$ and $s(x)$ are integers'' makes it sound like the polynomials
are constant. Change to ``so $r(x)$ and $s(x)$ take on integer values when
evaluated at the integer $x$.''
\\51\\Figure 2.6\\
The figure in the $st$-plane is really not accurate. In general there can be
more than one value of $s$ for a given value of $t$. See the attached sheets
for a corrected version. This means that the proof is incomplete, we need to
consider vertical lines (i.e., $\alpha=\infty$). Figure 2.7 should also be
corrected to match the new version of Figure 2.6.
\\52\\Paragraph 3\\
If $P_1\ne P_2$ and $t_1=t_2$, it is not true that $P_1=-P_2$. This argument
needs to be rewritten.
\\67\\Line 2\\
``contant'' \SB ``constant''
\\77\\$3^{\rm rd}$ Displayed Equation\\
``$\skew3\bar{\bar C}:y^2=x^2+\bar{\bar a}x^2+\bar{\bar b}x$'' \SB
``$\skew3\bar{\bar C}:y^2=x^3+\bar{\bar a}x^2+\bar{\bar b}x$''.
\\78\\$2^{\rm nd}$ Displayed Equation\\
``$\bar u={1\over2}c_1\omega_1+c_2\omega_2=c_1\bar\omega_1+c_2\bar\omega_2$'' \SB
``$\bar u=c_1\omega_1+c_2\omega_2=2c_1\bar\omega_1+c_2\bar\omega_2$.''
\\81\\Line $-8$\\
``$(\bar\lambda x+\bar\nu)^2=f(x)$'' \SB
``$(\bar\lambda x+\bar\nu)^2=\bar f(x)$'', or else write it out in full as
``$(\bar\lambda x+\bar\nu)^2=x^3+\bar a x^2+\bar b x$''.
\\87\\$2^{\rm nd}$ Displayed Equation\\
``$\pm(\hbox{rational number})^2$'' \SB ``$\pm(\hbox{integer})^2$''
\\94--96\\Examples 1 and 3\\
If we only check the allowable~$b_1$'s modulo squares, then we have to
allow $M,e,N$ to have some common factors. The point is that
every~$(x,y)\in\Gamma$ leads to a factorization~$b=b_1b_2$ and to a solution of
$N^2=b_1M^4+aM^2e^2+b_2e^4$ with $M,e,N$ pairwise relatively prime, but~$(x,y)$
need not lead to a square-free~$b_1$. Basically, if we replace~$b_1$ by its
square-free part, then we have to allow $M,e,N$ to have common factors dividing
the square part we canceled.
\\98\\Line 15\\
Change ``$N^2=68M^4-e^4$'' to ``$N^2=17M^4-4e^4$''. (Although it is true that
both equations have non-trivial~$p$-adic solutions for all~$p$, the first
equation doesn't actually have a solution modulo~4 if we require~$N$ and~$e$ to
be relatively prime.)
\\98\\Line $-2$\\
``has rank 15'' \SB ``has rank at least 15''. (Update to give current record,
which is now at least 19.)
\\99\\Line $-3$\\
The second coordinate should be $\displaystyle -{\nu^3\over y_1y_2}$
instead of $\displaystyle {\nu^3\over y_1y_2}$. Exercise 3.10 on Page
105 also needs to be changed.
\\100\\Top of Page and Theorem\\
``We observed in Chapter I that there are two possible pictures for the
singularity $S$'' is not correct, there are three possibilities. The theorem
should be restated to include the third case $y^2=x^2(x-1)$. Further, over
$\QQ$, the structure in general is more complicated. This is explained in
Exercise~3.15, so possibly just mention that there exists the third case and
refer the reader to exercise~15 for more details.
\\100\\$3^{\rm rd}$ Displayed Equation\\
``$\displaystyle (x,y)\longmapsto{x\over y}$'' \SB
``$\displaystyle (x,y)\longmapsto{y\over x}$''
\\105\\Exercise 3.7(c)\\
The first condition in the table should be
``$\displaystyle{\ZZ\over4\ZZ}$, if $D=4d^4$ for some $d$,''.
\\105\\Exercise 3.11\\
``1 if $P={\cal O}$'' \SB ``0 if $P={\cal O}$''.
\\107\\Line $-6$\\
``an element of $\FF_p$.)'' \SB ``an element of $\FF_p$).''
\\109\\Line $-1$ of Paragraph 3\\
``non-residues.)'' \SB ``non-residues).''
\\117\\Line 6\\
``$\beta_1\beta_2\beta_3=3k-2$'' \SB
``$\beta_1\beta_2\beta_3=(3k-2)p$''. (The $p$ was omitted on the RHS.)
\\126\\Line 2\\
``1, 000, 000'' \SB ``$1,000,000$''. (Close up space after the commas by
using math mode.)
\\132\\Pollard's Algorithm, Step 4\\
Replace ``Calculate $D=\gcd(a^k-1,n)$'' with something like
``Calculate $b\equiv a^k-1\pmod n$, and then $D=\gcd(b,n)$''.
\\132\\Pollard's Algorithm, Step 4\\
Change the last line to ``If $D=n$, either go back to Step~2 and choose
another~$a$, or go back to Step~1 and take a smaller~$k$.'' The reason for
the change is the (unlikely) possibility that every~$p$ dividing~$n$ has
the property that~$p-1$ divides~$k$.
\\135\\Equation for $\lambda$ in Center of Page\\
The equation given for $\lambda$ is actually the formula for $x(2Q)$.
Replace it with
$\displaystyle \lambda={f'(x)\over2y}={3x^2+2ax+b\over2y}\pmod{n}.$
\\136\\Computation of $kP$ at Bottom\\
The third line should be $1104P = (1372980126, 736595454)$, and all of
the points after this are incorrect. The corrected version of this table is
as follows:
$$\eqalign{
2^ 4 P &= 16 P = ( 385062894 , 618628731 )\cr
(2^ 4 + 2^ 6 )P &= 80 P = ( 831572269 , 1524749605 )\cr
(2^ 4 + 2^ 6 + 2^{10} )P &= 1104 P = ( 1372980126 , 736595454 )\cr
(2^ 4 + 2^ 6 + 2^{10}+ 2^{12} )P &= 5200 P = ( 1247661424 , 958124008 )\cr
\hbox{(previous partial sum)} + 2^{13} P
&= 13392 P = ( 1548582473 , 1559853215 )\cr
\hbox{(previous partial sum)}+ 2^{14} P
&= 29776 P = ( 201510394 , 7154559 )\cr
\hbox{(previous partial sum)} + 2^{15} P
&= 62544 P = ( 629067322 , 264081696 )\cr
\hbox{(previous partial sum)} + 2^{17} P
&= 193616 P = ( 844665131 , 537510825 )\cr
\hbox{(previous partial sum)} + 2^{19} P
&= 717904 P = ( 886345533 , 342856598 )\cr
\hbox{(previous partial sum)} + 2^{20} P
&= 1766480 P = ( 370579416 , 1254954111 )\cr
\hbox{(previous partial sum)} + 2^{21} P
&= 3863632 P = ( 77302130 , 514483068 )\cr
\hbox{(previous partial sum)} + 2^{23} P
&= 12252240 P = ( 1225303014 , 142796033 )\cr
}
$$
\\137\\Table at Top of Page\\
The table heading should be ``$2^iP \pmod{1715761513}$''. (The modulus
listed, $246082373$, is the modulus used in the Pollard example earlier
in the section.) The entries in the table are correct.
\\137\\Line $3$\\
The value of ``$kP$'' is incorrect. This line should read
$$
kP=12252240 ( 2 , 1 ) \equiv ( 1225303014, 142796033 ) \pmod{1715761513}.
$$
\\137\\Line $-5$ ff\\
The book asserts that no factor is found with~$P=(2,1)$ and~$1\le
b\le253$, but a factor is found with~$b=254$. Mossinghoff did not find
a factor with~$b=254$, but did find a factor with~$b=42$. (Guth found
a factor using $P=(17,1)$, $b=4$, $c=-4980$.) For Mossinghoff's version one
gets the table
$$\eqalign{
2^ 4 P &= 16 P = ( 1126060215 , 1502149623 )\cr
(2^ 4 + 2^ 6 )P &= 80 P = ( 1711657470 , 477996011 )\cr
(2^ 4 + 2^ 6 + 2^{10} )P &= 1104 P = ( 234439070 , 38804882 )\cr
(2^ 4 + 2^ 6 + 2^{10}+ 2^{12} )P &= 5200 P = ( 1158684598 , 1064974943 )\cr
\hbox{(previous partial sum)} + 2^{13} P
&= 13392 P = ( 487240237 , 1393430236 )\cr
\hbox{(previous partial sum)}+ 2^{14} P
&= 29776 P = ( 1236999455 , 390791552 )\cr
\hbox{(previous partial sum)} + 2^{15} P
&= 62544 P = ( 1695955849 , 1498221355 )\cr
\hbox{(previous partial sum)} + 2^{17} P
&= 193616 P = ( 1616297325 , 461346409 )\cr
\hbox{(previous partial sum)} + 2^{19} P
&= 717904 P = ( 373023881 , 1510113896 )\cr
\hbox{(previous partial sum)} + 2^{20} P
&= 1766480 P = ( 1211273029 , 1248862167 )\cr
\hbox{(previous partial sum)} + 2^{21} P
&= 3863632 P = ( 1115004543 , 1676196055 )\cr
}
$$
Now the material on the bottom of page 137 (starting at line $-5$) and the top
half of page 138 can be replaced with:
\par
we find that we are able to compute $kP~(\bmod~n)$ for all $b=3,4,5,\ldots,41$.
\par
However, when we try~$b=42$, and $c=-91$, the addition law breaks
down and we find a factor of~$n$. What happens
is the following. We have no trouble making a table of~$2^iP~(\bmod~n)$
for~$0\le i\le23$, just as above. Then we start adding up the points in the
table to compute~$kP~(\bmod~n)$. At the penultimate step we find
$$\eqalign{
(2^4+2^6+2^{10}+\cdots+2^{20}+2^{21})P
&= 3863632 P \cr
&\equiv (1115004543 , 1676196055) \pmod n.\cr
}
$$
Next, we read off from the (omitted) table
$$
2^{23}P\equiv( 1267572925 , 848156341 )\pmod n.
$$
So to get~$kP$ we need to add these two points,
$$
(1115004543 , 1676196055)+( 1267572925 , 848156341 )\pmod n.
$$
To do this we have to take
the difference of their~$x$ coordinates and find the
inverse modulo~$n$. But when we try to do this, we
discover that the inverse does not exist because
$$
\gcd(1115004543-1267572925,n)
=\gcd(-152568382,1715761513)=26927.
$$
\par
So the attempt to compute $12252240 ( 2 , 1 )$ on the curve
$$
y^2 = x^3 + 42 x - 91
\pmod{1715761513}
$$
fails, but it leads to the factorization
$$
n=1715761513 = 26927 \times 63719 .
$$
One easily checks that each of these factors is prime, so this gives the full
factorization of~$n$.
\\144\\Exercise 4.17(a)\\
The $r_i$ remainders may be negative, so the condition on $r_{i+1}$ needs
absolute value signs: $-{1\over2}|r_i|10^{-6}/q^{2.955}$ or
else the bound on~$y$ should be $|y|\le10^{1317}\cdot|c|^{2000/9}$.
\\181\\Line 13\\
``smallest subfield of $\CC$ contain all of'' \SB
``smallest subfield of $\CC$ containing all of''
\\203\\Lines 8,9\\
``contains no non-empty set'' \SB
``contains no non-empty open set''
\\203\\Line $-5$\\
``because $f$ is a homomorphism'' is not strictly true, it's only true
locally. Say instead ``from given property of~$f$''.
\\204\\Line 10\\
Replace $\CC\over L$ by either $\CC/L$ or $\displaystyle{\CC\over L}$.
\\204\\Line 2 of Paragraph 2\\
``if $L$ is an integer'' \SB ``if $c$ is an integer''
\\214\\Exercise 6.4\\
``$2y\psi_{2n}=\psi_n(\psi_{n+1}\psi_{n-1}^2-\cdots$'' \SB
``$2y\psi_{2n}=\psi_n(\psi_{n+2}\psi_{n-1}^2-\cdots$''.
\par\noindent
``$4y\omega_n=\psi_{n+1}\psi_{n-1}^2-\psi_{n-2}\psi_{n+1}^2$'' \SB
``$4y\omega_n=\psi_{n+2}\psi_{n-1}^2-\psi_{n-2}\psi_{n+1}^2$''.
\\219\\Exercise 6.21(b)\\
``$z\longmapsto\bigl(4\wp(z),4\wp'(z)\bigr)$'' \SB
``$z\longmapsto\bigl(4\gamma^2\wp(z),4\gamma^3\wp'(z)\bigr)$''
\\225\\Line 6 of Paragraph 3\\
``in the the projective plane'', remove a ``the''.
\\237\\$-12$\\
There is a bad line break in the middle of $I(C_1\cap C_2,P)=1$.
\\240\\Lines 3--6\\
This ``exercise'' is difficult. Warn the reader that it is difficult, and refer
them to exercise A.17 in the case that the~8 points are distinct.
\\253\\Line 12\\
``some coefficient of $\tilde F$ is not'' \SB
``some coefficient of $F$ is not''
\\256\\Exercise A.10(a) and A.10(b)\\
``tranformation'' \SB ``transformation'' (2 times).
\\1--$\infty$\\Entire book\\
It has been strongly suggested that we write $G/H$ for quotient groups, rather
than $\displaystyle{G\over H}$.
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&Joseph H. Silverman\cr
&Mathematics Department, Box 1917\cr
&Brown University\cr
&Providence, RI 02912\ \ U.S.A\cr
&(401)\thinspace 863-1132\cr
&$\langle$jhs@gauss.math.brown.edu$\rangle$\cr
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