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C00002 00002	%folio 797 galley 12 (C) Addison-Wesley 1978	*
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%folio 797 galley 12 (C) Addison-Wesley 1978	*
\def\\#1({\mathop{\hjust{#1}}(}\def\+#1\biglp{\mathop{\hjust{#1}}\biglp}
\ansno 8. If $u(x) = v(x)w(x)$, where $u(x)$ has
integer coefficients while $v(x)$ and $w(x)$ have rational coefficients,
there are integers $m$ and $n$ such that $m\cdot v(x)$ and $n \cdot
w(x)$ have integer coefficients. Now $u(x)$ is primitive, so
we have
$$u(x) = \pm\,\+pp\biglp m \cdot v(x)\bigrp\+pp\biglp n \cdot w(x)\bigrp.$$

\ansno 9. We can extend Algorithm E as follows:
Let $\biglp u↓1(x), u↓2(x), u↓3, u↓4(x)\bigrp$ and $\biglp v↓1(x),
v↓2(x), v↓3, v↓4(x)\bigrp$ be quadruples satisfying the relations
$u↓1(x)u(x) + u↓2(x)v(x) = u↓3u↓4(x)$, \ $v↓1(x)u(x) + v↓2(x)v(x)
= v↓3v↓4(x)$. The extended algorithm starts with $\biglp 1, 0,
\\cont(u), \\pp(u(x))\bigrp$ and $\biglp 0, 1, \\cont(v), \\pp(v(x))\bigrp$
and manipulates these quadruples in such a way as to preserve
the above conditions, where $u↓4(x)$ and $v↓4(x)$ run through the
same sequence as $u(x)$ and $v(x)$ do in Algorithm E\null. If $au↓4(x)
= q(x)v↓4(x) + br(x)$, we have $av↓3\biglp u↓1(x), u↓2(x)\bigrp
- q(x)u↓3\biglp v↓1(x), v↓2(x)\bigrp = \biglp r↓1(x), r↓2(x)\bigrp
$, where $r↓1(x)u(x) + r↓2(x)v(x) = bu↓3v↓3r(x)$, so the extended
algorithm can preserve the desired relations. If $u(x)$ and
$v(x)$ are relatively prime, the extended algorithm eventually
finds $r(x)$ of degree zero, and we obtain $U(x) = r↓2(x)$, $V(x)
= r↓1(x)$ as desired.\xskip$\biglp$In practice we would divide $r↓1(x)$,
$r↓2(x)$, and $bu↓3v↓3$ by $\gcd\biglp\\cont(r↓1),\\ cont(r↓2)\bigrp$.$\bigrp
$\xskip Conversely, if such $U(x)$ and $V(x)$ exist, then $u(x)$ and
$v(x)$ have no common prime divisors, since they are primitive
and have no common divisors of positive degree.

\ansno 10. By successively factoring polynomials
that are reducible into polynomials of smaller degree, we must
obtain a finite factorization of any polynomial into irreducibles.
The factorization of the {\sl content} is unique. To show that
there is at most one factorization of the primitive part, the
key result is to prove that if $u(x)$ is an irreducible factor
of $v(x)w(x)$, but not a unit multiple of the irreducible polynomial
$v(x)$, then $u(x)$ is a factor of $w(x)$. This can be proved
by observing that $u(x)$ is a factor of $v(x)w(x)U(x) = rw(x)
- w(x)u(x)V(x)$ by the result of exercise 9, where $r$ is a
nonzero constant.

\ansno 11. The only row names needed
would be $A↓1$, $A↓0$, $B↓4$,
$B↓3$, $B↓2$, $B↓1$, $B↓0$, $C↓1$, $C↓0$, $D↓0$. In general, let $u↓{j+2}(x)
= 0$; then the rows needed for the proof are $A↓{n↓2-n↓j}$
through $A↓0$, $B↓{n↓1-n↓j}$ through $B↓0$, $C↓{n↓2-n↓j}$ through $C↓0$, 
$D↓{n↓3-n↓j}$ through $D↓0$, etc.

\ansno 12. If $n↓k = 0$, the text's proof of (24) shows that the value of
the determinant is $\pm h↓k$, and this equals $\pm\lscr↓k↑{n↓{k-1}}/
\prod↓{1<j<k}\lscr↓{\!j}↑{\delta↓{j-1}(\delta↓j-1)}$.
If the polynomials have a factor
of positive degree, we can artificially assume that the polynomial
zero has degree zero and use the same formula with $\lscr↓k=
0$.

{\sl Notes:} The value $R(u, v)$ of Sylvester's determinant
is called the {\sl resultant} of $u$ and $v$, and the quantity
$(-1)↑{\hjust{\:e deg}(u)(\hjust{\:e deg}(u)-1)/2}\lscr(u)↑{-1}R(u, u↑\prime )$ is
called the {\sl discriminant} of $u$, where $u↑\prime$ is the derivative of $u$. 
If $u(x) = a(x - α↓1)
\ldotsm (x - α↓m)$ and $v(x) = b(x - β↓1) \ldotsm (x - β↓n)$,
we have $R(u, v) = a↑nv(α↓1) \ldotsm v(α↓m) = (-1)↑{mn}b↑mu(β↓1)
\ldotsm u(β↓n) = a↑nb↑m \prod↓{1≤i≤m,1≤j≤n}(α↓i - β↓j)$. It follows
that the polynomials of degree $mn$ in $y$ defined as the respective
resultants with $v(x)$ of $u(y - x)$, $u(y + x)$, $x↑mu(y/x)$, and $u(yx)$
have as respective roots the sums $α↓i + β↓j$, differences
$α↓i - β↓j$, products $α↓iβ↓j$, and quotients $α↓i/β↓j$ $\biglp$when
$v(0) ≠ 0\bigrp $. This idea has been used by R. G. K.
Loos to construct algorithms for arithmetic on algebraic numbers
[{\sl SIAM J. Computing} ({\sl c}. 1979), to appear].

If we replace each row $A↓i$ in Sylvester's matrix by
$$(b↓0A↓i + b↓1A↓{i+1} +\cdots + b↓{n↓2-1-i}A↓{n↓2-1})
- (a↓0B↓i + a↓1B↓{i+1} + \cdots + a↓{n↓2-1-i}B↓{n↓2-1}),$$
and then delete rows $B↓{n↓2-1}$
through $B↓0$ and the last $n↓2$ columns, we obtain an $n↓1
\times n↓1$ determinant for the resultant instead of the original
$(n↓1 + n↓2) \times (n↓1 + n↓2)$ determinant. In some cases
the resultant can be evaluated efficiently by means of this
determinant; see {\sl CACM \bf 12} (1969), 23--30, 302--303.

J. T. Schwartz has shown that resultants and Sturm sequences for
polynomials of degree $n$ can be evaluated with $O\biglp n(\log n)↑2\bigrp$
operations as $n→∞$.\xskip [``Probabilistic algorithms for verification of
polynomial identities,'' to appear.]

\ansno 13. One can show by induction on $j$ that the values of $\biglp
u↓{j+1}(x),g↓{j+1},h↓j\bigrp$ are replaced respectively by $\biglp\lscr↑{1+p↓j}w(x)
u↓j(x),\lscr↑{2+p↓j}g↓j,\lscr↑{p↓j}h↓j\bigrp$ for $j≥2$, where $p↓j=n↓1+n↓2-2n↓j$.
$\biglp$In spite of this growth, the bound (26) remains valid.$\bigrp$

\ansno 14. Let $p$ be a prime of the domain, and let $j,k$ be
maximum such that $p↑k\rslash v↓n = \lscr(v)$, $p↑j\rslash v↓{n-1}$.
Let $P = p↑k$. By Algorithm R we may write $q(x) = a↓0 + Pa↓1x
+\cdots + P↑sa↓sx↑s$, where $s = m - n ≥ 2$. Let
us look at the coefficients of $x↑{n+1}$, $x↑n$, and $x↑{n-1}$
in $v(x)q(x)$, namely $Pa↓1v↓n + P↑2a↓2v↓{n-1} +\cdotss$, $a↓0v↓n
+ Pa↓1v↓{n-1} +\cdotss$, and $a↓0v↓{n-1} + Pa↓1v↓{n-2} +\cdotss$,
each of which is a multiple of $P↑3$. We conclude from the first
that $p↑j\rslash a↓1$, from the second that $p↑{\min(k,2j)}\rslash
a↓0$, then from the third that $P\rslash a↓0$. Hence $P\rslash
r(x)$.\xskip $\biglp$If $m$ were only $n + 1$, the best we could prove would
be that $p↑{\lceil k/2\rceil }$ divides $r(x)$; e.g., consider
$u(x) = x↑3 + 1$, $v(x) = 4x↑2 + 2x + 1$, $r(x) = 18$. On the other hand, an
argument based on determinants of matrices like (21) and (22) can be used to
show that \def\\{\hjust{\:e deg}}$\lscr(r)↑{\\(v)-\\(r)-1}r(x)$ is always a
multiple of $\lscr(v)↑{(\\(u)-\\(v))(\\(v)-\\(r)-1)}$.$\bigrp$