#### Exercise 4.5.1

Let $p$ be a monic polynomial over the field $F$, and let $f$ and $g$ be relatively prime polynomials ovef $F$. Prove that the g.c.d. of $pf$ and $pg$ is $p$.

I proved this in more generality as part of Exercise 7 of the previous section. That makes me think that was not the solution they were looking for. Or maybe they want a proof here using prime factorization. Here are both proofs.

Solution 1: We use the definition of g.c.d. in terms of ideals. Since gcd$(f,g)=1$, the ideal generated by $f$ and $g$ is the same as the ideal generated by $1$. In other words $fF[x]+gF[x]=F[x]$. Thus $$pfF[x]+pgF[x]=pF[x].$$ Since $p$ is monic by assumption, it follows that $p$ satisifes the definition of gcd$(f,g)$ in $F[x]$.

Solution 2: We use the comments at the top of page 137. Factor into primes $f=f_1^{r_1}\cdots f_j^{r_j}$, $g=g_1^{s_1}\cdots g_k^{s_k}$ and $p=p_1^{t_1}\cdots p_{\ell}^{t_{\ell}}$. Since g.c.d.$(f,g)=1$ none of the $f_i$’s equal any of the $g_i$’s. Thus the common factors of $pf$ and $pg$ are exactly the $p_i$’s. In other words the prime factorizations are

$$fp=f_1^{r_1}\cdots f_j^{r_j} \cdot p_1^{t_1}\cdots p_{\ell}^{t_{\ell}}$$$$gp=g_1^{s_1}\cdots g_k^{s_k} \cdot p_1^{t_1}\cdots p_{\ell}^{t_{\ell}}.$$Since none of the $f_i$’s equal any of the $g_i$’s, it follows that g.c.d$(f,g)=p_1^{t_1}\cdots p_{\ell}^{t_{\ell}}$ which equals $p$.

#### Exercise 4.5.2

Assuming the Fundamental Theorem of Algebra prove the following. If $f$ and $g$ are polynomials over the field of complex numbers, then g.c.d$(f,g)=1$ if and only if $f$ and $g$ have no common root.

Solution: By the Fundamental Theorem of Algebra, we can factor $f$ and $g$ all the way to linear factors

$$f=(x-a_1)^{n_1}\cdots(x-a_k)^{n_k}$$$$g=(x-b_1)^{m_1}\cdots(x-b_{\ell})^{m_{\ell}}.$$The roots of $f$ are exactly $a_1,\dots,a_k$, the roots of $g$ are exactly $b_1,\dots,b_{\ell}$.If gcd$(f,g)=1$ then (by the comments at the top of page 137) $f$ and $g$ have no common factors, and therefore $a_i\not=b_j$ $\forall$ $i,j$. Thus (by Corollary 1, page 128) $f$ and $g$ have no common roots. And if $f$ and $g$ have no common roots then (by Corollary 1, page 128) none of the factors $(x-a_i)$ can equal any of the factors $(x-b_j)$. Thus gcd$(f,g)=1$.

#### Exercise 4.5.3

Let $D$ be the differentiation operator on the space of polynomials over the field of complex numbers. Let $f$ be a monic polynomial over the field of complex numbers. Prove that

$$f=(x-c_1)\cdots(x-c_k)$$ where $c_1,\dots,c_k$ are distinct complex numbers if and only if $f$ and $Df$ are relatively prime. In other words, $f$ has no repeated roots if and only if $f$ and $Df$ have no common root. (Assume the Fundamental Theorem of Algebra.)

Solution: First assume all $c_i$’s are distinct. Then by Theorem 11, page 137, we know $f$ and $Df$ are relatively prime.

Now assume $f$ and $f’$ are relatively prime. Then again by Theorem 11 we know $f$ is a product of distinct irreducibles. Since $\mathbb C$ is algebraically closed each of those irreducibles must be of the form $x-a$. It follows that $f$ must be of the form $(x-c_1)\cdots(x-c_n)$ for distinct $c_1,\dots,c_n$.

#### Exercise 4.5.4

Prove the following generalization of Taylor’s formula. Let $f$, $g$, and $h$ be polynomials over a subfield of the complex numbers, with $\deg f\leq n$. Then

$$f(g)=\sum_{k=0}^n\frac{1}{k!}f^{(k)}(h)(g-h)^k.$$(Here $f(g)$ denotes `$f$ of $g$.’)

Solution: Since we are working over the base field $\mathbb C$ by Theorem 3 page 126 there is a natural isomorphism between the algebra $\mathbb C[x]$ and the algebra of polynomial functions from $\mathbb C$ into $\mathbb C$. Taylor’s Theorem is therefore also a theorem about polynomial functions from $\mathbb C$ to $\mathbb C$. So we may plug any complex numbers we want in for $x$ and $c$ in the statement of Taylor’s Theorem and the equality remains valid. In particular we can subsitute $g$ in for $x$ and $h$ in for $c$ to obtain the desired formula. We then simply translate over to the algebra $\mathbb C[x]$ via the isomorphism in Theorem 3 to obtain the corresponding result in $\mathbb C[x]$.

For the remaining exercises, we shall need the following definition. If $f$, $g$, and $p$ are polynomials over the field $F$ with $p\not=0$, we say $f$ **is congruent to $g$ modulo** $p$ if $(f-g)$ is divisible by $p$. If $f$ is congruent to $g$ modulo $p$, we write $$f\equiv g\pmod p.$$

#### Exercise 4.5.5

Prove for any non-zero polynomial $p$, that congruence modulo $p$ is an equivalence relation.

(a) It is reflexive: $f\equiv f\pmod p$.

(b) It is symmetric: if $f\equiv g\pmod p$, then $g\equiv f\pmod p$.

(c) It is transitive: if $f\equiv g\pmod p$ and $g\equiv h\pmod p$, then $f\equiv h\pmod p$.

Solution:

(a). In this case $p$ must divide $f-f$ which equals zero. But everything divides zero. Just take $d=0$ and then $p=d(f-f)$. Thus $f\equiv f\pmod p$ is always true.

(b). Assume $f\equiv g\pmod p$. Then $p$ divides $f-g$. Thus $\exists$ $d$ s.t. $f-g=pd$. Let $d’=-d$. Then $g-f=pd’$ thus $p$ divides $g-f$. Thus $g\equiv f\pmod p$.

(c). Assume $f\equiv g\pmod p$ and $g\equiv h\pmod p$. Then $\exists$ $d$ and $d’$ such that $f-g=pd$ and $g-h=pd’$. Adding these two equations gives $f-h=pd+pd’$. Let $d”=d+d’$. Then $f-h=pd”$. Thus $f\equiv h\pmod p$.

#### Exercise 4.5.6

Suppose $f\equiv g\pmod p$ and $f_1\equiv g_1\pmod p$.

(a) Prove that $f+f_1\equiv g+g_1\pmod p$.

(b) Prove that $ff_1\equiv gg_1\pmod p$.

Solution:

(a) By assumption there are polynomials $d$ and $d’$ such that $f-g=pd$ and $f_1-g_1=pd’$. Adding these two equations gives $$f+f_1-g-g_1=pd+pd’.$$ Or equivalently $$(f+f_1)-(g+g_1)=p(d+d’).$$ Thus $f+f_1\equiv g+g_1\pmod p$.

(b) By assumption there are polynomials $d$ and $d’$ such that $f-g=pd$ and $f_1-g_1=pd’$. Now \begin{align*}ff_1-gg_1&=ff_1-g_1f+g_1f-gg_1\\&=f(f_1-g_1)+g_1(f-g)\\&=fpd’+g_1pd=p(fd’+g_1d).\end{align*} Thus $p$ divides $ff_1-gg_1$. Thus $ff_1\equiv gg_1\pmod p$.

#### Exercsie 4.5.7

Use Exercise 6 to prove the following. If $f$, $g$, and $p$ are polynomials over the field $F$ and $p\not=0$, and if $f\equiv g\pmod p$, then $h(f)\equiv h(g)\pmod p$.

Solution: It follows from Exercise 6 part (b) that $f\equiv g\pmod p$ $\Rightarrow$ $f^2\equiv g^2\pmod p$ (since $f\equiv f\pmod p$ and $g\equiv g\pmod p$). By induction $f^n\equiv g^n\pmod p$ for all $n=1,2,3,\dots$. Let $c\in F$. Then letting $f_1=g_1=c$ we get $cf\equiv cg\pmod p$ for any $c\in F$. Thus if $h(x)=cx^n$ we have shown the result is true. Thus the result is true for all monomials. Now we can obtain the result on the sum of monomials using part (a) of Exercise 6. Since the general $h$ is a sum of monomials, the general result follows.

Exercise 4.5.8

If $p$ is an irreducible polynomial and $fg\equiv 0\pmod p$, prove that either $f\equiv 0\pmod p$ or $g\equiv 0\pmod p$. Give an example which shows that this is false if $p$ is not irreducible.

Solution: $fg\equiv0\pmod p$ implies $p$ divides $fg$. By the Corollary to Theorem 8, page 135, it follows that $p$ divides $f$ of $p$ divides $g$. Thus $f\equiv0\pmod p$ or $g\equiv0\pmod p$.

From http://greggrant.org