Exercise 9.1.4

Question

\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}}

Let $R$ be an integral domain, and let $f,g\in R[x]$, where $f
e 0$. If $K$ is the field of fractions of $R$, then we can divide $g$ by $f$ in $K[x]$ using the division algorithm of Theorem A.1.14. This gives $g = qf+r$, though $q,r \in K[x]$ need not lie in $R[x]$.
\be
\item[(a)] Show that dividing $x^2$ by $2x+1$ in $\Q[x]$ gives $x^2 = q\cdot (2x+1) + r$, where $q,r \in \Q[x]$ are not in $\Z[x]$, even though $x^2$ and $2x+1$ lie in $\Z[x]$.
\item[(b)] Show that if $f$ is monic, then the division algorithm gives $g=qf+r$, where $q,r \in R[x]$. Hence the division algorithm works over $R$ provides we divide by monic polynomials. 
\ee

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}
\begin{enumerate}
\item[(a)]
$x^2 = (\frac{1}{2}x - \frac{1}{4})(2x+1)+\frac{1}{4}$.
The quotient $q(x) = \frac{1}{2}x - \frac{1}{4}$ is not in $\Z[x]$.

\item[(b)]
Let $f = x^m + b_{m-1}x^{m-1}+ \cdots+b_0$ be a fixed monic polynomial in  $R[x]$.

We show by induction on the degree $n$ the proposition

$P(n) : \forall g \in R[x], \deg(g) = n \Rightarrow \exists (q,r) \in R^2, \ g = qf+r,\  \deg(r) < \deg(f)$

(with the convention $\deg(0) = -\infty$).

We suppose that $P(k)$ is true for all $k<n$, and we prove $P(n)$. Let $g$ be any polynomial in $R[x]$.

$\bullet$ If $\deg(g) < m = \deg(f)$, then the pair $(q,r) = (0,g)$ is an answer.

$\bullet$ Suppose that $\deg(g) \geq m$.  Write $g = a_n x^n+ a_{n-1}x^{n-1}+ \cdots +a_0$, with $\deg(g) = n \geq m$ and $a_i \in R, i=0,\ldots,n$.

The polynomial $g_1 = g - a_n x^{n-m} f \in R[x]$ satisfies $\deg(g_1) < n$. We can then apply to it the induction hypothesis:

$g_1 =  q_1 f+r, q_1 \in R[x],\ r \in R[x],\ \deg(r) < \deg(f)$.

Then $g = (a_n x^{n-m} + q_1) f+ r$.

If we write $q = a_n x^{n-m} + q_1$, then $q \in \mathbb{R}[x]$ and $g = fq+r,\ (q,r) \in\mathbb{R}[x]^2, \deg(r) < \deg(f)$. The pair $(q,r)$ is an answer, and the induction is done.

In particular, if $g,f \in \mathbb{Z}[x]$ , and $g = f q$, the unicity of the Euclidean division in $\mathbb{Q}[x]$ and the preceding result shows that $q \in \mathbb{Z}[x]$.

\end{enumerate}
\end{proof}

Exercise 9.1.4

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Exercise 9.1.4

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