Exercise 4.3.1

Question

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In (4.9) we represented elements of $F(\alpha)$ uniquely using remainders on division by the minimal polynomial of $\alpha$. In the exercise you will adapt the proof of Proposition 4.3.4 to the case of quotient rings. Suppose that $f \in F[x]$ has degree $n>0$. Prove that every coset on $F[x]/\langle f angle$ can be written as
$$a_0+a_1x+\cdots+a_{n-1} x^{n-1} + \langle f angle,$$
where $a_0,a_1,\ldots,a_{n-1} \in F$ are unique.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Let $f \in F[x],\deg(f) = n>0$, and $y \in F[x]/\langle f angle$. There exists $g \in F[x]$ such that $y = g +\langle f angle$.

The division of $g$ by $f$ gives 
$$g = qf+r,\  \deg(r)  < \deg(f)=n.$$
Thus $g-r =qf \in \langle f angle$, and consequently $y=g+\langle f angle = r+ \langle f angle$.

As $\deg(r)<n$, $r = a_0+a_1x+\cdots+a_{n-1}x^{n-1},\ a_0,a_1,\ldots,a_{n-1} \in F$.

Every $y \in F[x]/\langle f angle$ can be written as

$$y = a_0+a_1x+\cdots+a_{n-1}x^{n-1} + \langle f angle, \ a_0,a_1,\ldots,a_{n-1} \in F.$$

{\bigskip}

Unicity:

Suppose that $y \in g+\langle f angle$ is written as
\begin{align*}
y &= a_0+a_1x+\cdots+a_{n-1}x^{n-1} + \langle f angle\
&=b_0+b_1x+\cdots+b_{n-1}x^{n-1} + \langle f angle\
(a_i, b_i \in F, i = 0,\ldots,n-1).
\end{align*}
Then there exist two polynomials $a,b \in \langle f angle$ such that $$p = \sum\limits_{k=0}^{n-1}{a_kx^k} + a = \sum\limits_{k=0}^{n-1}{b_kx^k} + b.$$
Let $r = \sum\limits_{k=0}^{n-1}{a_kx^k}, s = \sum\limits_{k=0}^{n-1}{b_kx^k}$.
By definition of $\langle f angle$,  there exists $q_1,q_2\in F[x]$ such that $$p = q_1f + r=q_2f+s, \qquad \deg(r)<n, \deg(s)<n.$$ 
The unicity of the remainder in the division of $p$ by $f$ shows that  $r=s$, so $a_i=b_i, i=0,\ldots,n-1$.

{\bigskip}

Conclusion: Every element in  $F[x]/\langle f angle$ is written as
$$a_0+a_1x+\cdots+a_{n-1}x^{n-1} + \langle f angle, \qquad a_0,a_1,\ldots,a_{n-1} \in F.$$
where $a_0,a_1,\ldots, a_{n-1}$ are unique.
\end{proof}

Exercise 4.3.1

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Exercise 4.3.1

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