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\centerline{Mathematics 351 -- Abstract Algebra 1}
\centerline{Solutions for Problem Set 3}
\centerline{September 22, 2007}
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\noindent
{\it Section 4.3}
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\noindent
5.  $\Rightarrow$:  If $f$ and $g$ are associates, then $f = ug$, where
$u$ is a unit in $F[x]$.  By Corollary 4.9, this says $f = ug$ where
$u \ne 0$ is a constant polynomial.  The equation $f = ug$ shows $g | f$
by definition.  Since $u^{-1}$ exists in $F$, we also have $u^{-1}f = g$,
which shows $f | g$.
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\noindent
$\Leftarrow$:  Assume $f | g$ and $g | f$. Then we have equations $g = fk$
and $f = g\ell$ for some $k, \ell \in F[x]$.  Substituting from the
second equation into the first and using the associative law for 
multiplication in $F[x]$, we have $g = (k\ell)g$.  By considering
degrees, we see that $\deg(k\ell) = 0$, so both $k$ and $\ell$ must
be nonzero constants.  Hence $f,g$ are associates in $F[x]$.
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\noindent
6.  If $x^2 + 1 = (ax + b)(cx + d)$ for $a,b,c,d\in \mathbb{Q}$, then by equating
coefficients on both sides, $ac = 1$, $bd = 1$ and $ad + bc = 0$.  This
Shows $d = \frac{1}{b}$ and $c=\frac{1}{a}$, so $\frac{a}{b} + \frac{b}{a} = 0$.
Multiplying both sides of the last equation by $ab$, $a^2 + b^2 = 0$, which
is only possible if $a = b = 0$, which is absurd since $ac = bd = 1$.
This contradiction shows $x^2 + 1$ must be
irreducible in $\mathbb{Q}[x]$.
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\noindent
14.  The two ways are $x^2 + x = x(x+1)$ and $x^2 + x = (x+3)(x+4)$ (note
$x^2 + 7x + 12 \equiv x^2 + x\! \bmod 6$).  {\it Comment:}  This problem
shows that unique factorization for polynomials (as in Theorem 4.12) can
fail in $R[x]$ when $R$ is not a field!
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\noindent
16.  $\Rightarrow$:  Assume that $p$ is irreducible in $F[x]$ and $p$ does
not divide $g$.  We must show that $(p,g) = 1$.  Suppose not. Then 
There is some polynomial $d$ of positive degree with $d|p$ and $d|g$.
Since $p$ is irreducible, the first divisibility statement can only be
true if $d$ is an associate of $p$.  But then $p|g$ as well since $p$ and $d$
differ only by a nonzero constant multiple.  This is a contradiction to 
the assumption that $d$ does not divide $g$.  Hence $(p,g) = 1$.
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\noindent
$\Leftarrow$:  Assume that $p$ is a polynomial such that for {\it every}
$g \in F[x]$, either $p|g$ or $(p,g) = 1$.  We want to show that $p$
must be irreducible.  Suppose not.  Then $p = hk$ for two polynomials
$h,k$ of positive degrees.  Apply the hypothesis with $g = h$.
we get that either $p|h$ or $(p,h) = 1$.  But both of these lead
to contradictions.  First, $p$ cannot divide $h$, since $p = hk$
and $\deg(k) \ge 1$.  Second, $(p,h) = 1$ is also impossible since
$h$ is a common divisor of $h$ and $p$ of positive degree.
Therefore, $p$ satisfying the given condition must be irreducible.  
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\noindent
20.  The hypothesis is that $p,q$ are both irreducible, but $p \ne cq$ for
any nonzero constant $c \in F$.  By Exercise 16 above, since $p$ is 
irreducible, either $p|q$ or $(p,q) = 1$.  However, $p|q$ is not 
possible since $q$ irreducible says that the only factors of $q$ are
units and associates of $q$.  But we are given that $p$ and $q$ are not 
associates.  Hence $p$ and $q$ are relatively prime.
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\noindent
21.  (a) The reducible monic polynomials of degree $2$ in $\mathbb{Z}_p[x]$ come
in two ``flavors:''  $(x + a)(x + b)$ with $a,b \in F$ distinct, and
$(x + a)^2$ for $a \in F$.  There are $p$ of the second type because
$a$ can be any one of the $p$ elements of $\mathbb{Z}_p$.  There are
$\frac{p(p-1)}{2}$ of the second type because:
\begin{itemize}
\item $a$ can be any one of the $p$ elements $\mathbb{Z}_p$, 
\item $b$ can be any one of the $p - 1$ elements of $\mathbb{Z}_p$ different from $a$, but
\item every such polynomial is obtained in exactly two different ways by this construction.
\end{itemize}
The total number is $p + \frac{p(p-1)}{2} = \frac{p(p+1)}{2}$.
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\noindent
(b)  Every monic polynomial of degree 2 is either reducible or irreducible
(these are mutually exclusive properties).  There are $p^2$ total monic
polynomials $f = x^2 + \alpha x + \beta$ of degree 2 in $\mathbb{Z}_p[x]$, 
since $\alpha$ can be any one of the $p$ elements of $\mathbb{Z}_p$ and 
for each $\alpha$, $\beta$ can be any one of the $p$ elements of $\mathbb{Z}_p$.
Hence the number of monic irreducibles of degree $2$ is 
\[ 
p^2 - \frac{p(p+1)}{2} = \frac{p(p-1)}{2}.
\]

\noindent
22. (a) The quick way to see this is to note that for all $a \in \mathbb{Z}_3$, 
since $\mathbb{Z}_3[x]$ is a commutative ring:
\[
(x + a)^3 = x^3 + 3 a x^2 + 3 a^2 x + a^3 \equiv x^3 + a \!\bmod 3.
\]
This shows $x^3 + a$ is reducible.
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\noindent
(b)  Similar to (a):
\[
(x + a)^5 = x^5 + 5 a x^4 + 10 a^2 x^3 + 10 a^3 x^2 + 5 a^4 x +  a^5 \equiv x^5 + a \!\bmod 5.
\]
(where we used the fact $a^5 \equiv a \!\bmod 5$ for all $a$, easy to check, and 
mentioned in class).
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\noindent
23.  (a)  Since $x^2 + 2$ is quadratic, using some results (Corollary 4.18 in 
particular) from the next section,
it suffices to show that it has no
roots in $\mathbb{Z}_5$.  This is clear since $0^2 + 2 \equiv 2$, 
$1^2 + 2 \equiv 3$, $2^2 + 2 \equiv 1$, $3^2 + 2 \equiv 1$, $4^2 + 1 \equiv 3$.  
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\noindent
(b)  $x^4 - 4 = (x^2 + 2)(x^2 - 2)$.  Both factors can be seen to be irreducible.
The first one is by part (a), the other one is by a similar argument.
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\noindent
{\it Section 4.4}
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\noindent
4.  (a)  $x - 2$ is a factor of the polynomial if and only if $2$ is 
a root, or $16 - 40 + 20 + 6 + k = 0$, which says $k = -2$.  
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\noindent
(b)  $x + 1$ is a factor if and only if $-1 \equiv 4$ is a root in $\mathbb{Z}_5$.
This says  $1 - 2 - 3 + k + 1 = 0$, or $k = -3 \equiv 2 \!\bmod 5$.
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\noindent
5. By Theorem 4.15, $x - 1_F$ divides $f(x) = a_nx^n + \cdots a_1 x + a_0$
if and only if $1_F$ is a root of $f$.  This is equivalent to 
\[
0_F = f(1_F) = a_n + a_{n-1} + \cdots + a_1 + a_0,
\]
which is what we wanted to show.
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\noindent
8.  (e)  Since this is a cubic polynomial, we can test for irreducibility
by seeing if there are roots in $\mathbb{Z}_{11}$.  With $x = 4$, we have
$4^3 - 9 = 64 - 9 = 55 \equiv 0 \!\bmod 11$.  Hence $x - 4$ is a factor
of $x^3 - 9$ in $\mathbb{Z}_{11}$, and this polynomial is reducible.
(f)  If we substitute $x = 1$ we get $1^4 + 1^2 + 1 \equiv 0\!\bmod 3$, so 
$1$ is a root and this polynomial is reducible too.
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\noindent
9.  (See Exercise 21 from Section 4.3 above!) There are precisely $\frac{3(3-1)}{2} = 3$
monic irreducibles of degree 2 in $\mathbb{Z}_3[x]$:
\[
x^2 + 1, x^2 + x + 2, x^2 + 2x + 2.
\]
The systematic ways to find them are either
\begin{itemize}
\item to list all monic polys of degree 2, then
cross off the ones that have roots in $\mathbb{Z}_3$, or
\item to make all of the reducible monic polys of degree 2
in $\mathbb{Z}_5$ as in Exercise 21 from Section 4.3, multiply them out,
and take all the polynomials not obtained that way.
\end{itemize}
Similarly, in $\mathbb{Z}_5[x]$,
we get $\frac{5(5-1)}{2} = 10$ such polynomials:
\begin{eqnarray*}
&&x^2 + 2, x^2 + 3, x^2 + 2x + 4, x^2 + x + 1, x^2 + 3x + 3, \\
&&x^2 + 3x + 4, x^2 + 4x + 1, x^2 + 4x + 2, x^2 + x + 2, x^2 + 2x + 3.
\end{eqnarray*}
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\noindent
12.  If $c_n a^n + \cdots + c_1 a + c_0 = 0_F$, then multiply both sides
by $(a^{-1})^n$ to obtain $c_n + c_{n-1} a + \cdots + c_1 a^{n-1} + c_0 a^n$.
This shows $a^{-1}$ is a root of the ``reversed'' polynomial
$c_n + c_{n-1} x + \cdots + c_0 x^n$.
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\noindent
13.  (a)  If $f,g$ are associates in $F[x]$, then $g = uf$ for some 
$u \ne 0 \in F$.  Then for all $a \in F$, $g(a) = u f(a)$.  If $a$ is a 
root of $f$, then $f(a) = 0_F$, so $g(a) = u \cdot 0_F = 0_F$ as well.  
Conversely, if $a$ is a root of $g$, then $g(a) = 0_F$.  Hence $0_F = u f(a)$, so 
$f(a) = u^{-1} 0_F = 0_F$, and $a$ is a root of $f$.
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\noindent
(b)  The answer is no, since for instance $g$ could equal $f$ times a polynomial
in $F[x]$ that has no roots in $F$.  An explicit counterexample is 
$f(x) = x - 1$ and $g(x) = (x - 1)(x^2 + 1)$ in $\mathbb{Q}[x]$.  These
have the same roots in $F$ (namely $x = 1$), but $f,g$ are not associates.
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\noindent
16.  Consider the polynomial $h = f - g$.  By the given information 
$\deg(h) = \deg(f - g) \le n$.  But $h(c_i) = f(c_i) - g(c_i) = 0$
for $i = 0,\ldots, n$, so $h$ has $n + 1$ roots in $F$.  Therefore, by
Corollary 4.16, $h = 0$ (the zero polynomial), and $f = g$.  
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\noindent
18.  Let $f = c_n x^n + \cdots + c_1 x c_0\in \mathbb{Q}[x]$.
Apply $\varphi$ to both sides of the equation $0 = f(r)$:
\begin{eqnarray*}
0 = \varphi(f(r)) & = & \varphi(c_n r^n + \cdots + c_1 r + c_0)\\
& = & \varphi(c_n r^n) + \cdots + \varphi(c_1 r) +\varphi(c_0)\  \text{(sum property of homomorphisms)}\\
& = & \varphi(c_n) \varphi(r)^n + \cdots + \varphi(c_1)\varphi(r) +\varphi(c_0)\  \text{(mult. property of homomorphisms)}\\
&=& c_n \varphi(r)^n + \cdots + c_1 \varphi(r) + c_0\ (\text{since}\ c_i \in \mathbb{Q}) 
\end{eqnarray*}
This shows that $\varphi(r)$ is also a root of $f$.  
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\noindent
19.  (a)  $\Leftarrow$:  If $a$ is a multiple root of $f$, then 
$f(x) = (x - a)^k q(x)$ for some $k \ge 2$, and some $q \in \mathbb{R}[x]$.
Hence by the product rule for derivatives 
\[
f'(x) = k(x - a)^{k-1} q(x) + (x - a)^k q'(x).
\]
Since $k \ge 2$, $k - 1\ge 1$, so $f'(a) = 0$ and $a$ is also a root of $f'$.
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\noindent
$\Rightarrow$:  Conversely, assuming $a$ is a root of $f$ and $f'$,
write $f(x) = (x - a)^k q(x)$ where $q(a) \ne 0$ (i.e. $k$ is as large as possible). 
Then by the equation above, 
\[
0 = f'(a) = k(a - a)^{k-1} q(a) + (a - a)^k q'(a) = k(a - a)^{k-1}q(a).
\]
Since $q(a) \ne 0$, $k - 1 \ge 1$, and hence $k \ge 2$.  This shows $a$
is a multiple root of $f$.
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\noindent
(b)  We show the contrapositive:  If $f$ has a multiple root, then
$(f,f') \ne 1$.  But this is clear from the above, since if $k \ge 2$, 
then $(x - a)|f$ and $(x - a)|f'$.
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\noindent
24.  Consider the mapping
\begin{eqnarray*}
\varphi_a : F[x] & \to & F\\
               f & \mapsto & f(a).
\end{eqnarray*}
Then $\varphi_a$ is a homomorphism of rings since for all $f,g \in F[x]$,
\begin{eqnarray*}
\varphi_a(f + g) &=& (f+g)(a) = f(a) + g(a) = \varphi_a(f) + \varphi_a(g)\\
\varphi_a(f \cdot g) &=& (f\cdot g)(a) = f(a) \cdot g(a) = \varphi_a(f) \cdot \varphi_a(g).
\end{eqnarray*}
It is surjective (onto), since for all $b \in F$ there are $f \in F[x]$ such 
that $\varphi_a(f) = f(a) = b$.  For a very simple example, take the 
constant polynomial $f = b$ (!)

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