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I'm stuck on an exercice :

Let $A,B$ $\in \mathbb{Z}[X]$, where $B$ is a monic polynomial. Consider $A = BQ + R$ the Euclidean division of $A$ by $B$ in $\mathbb{C}[X]$.

1) Show that $Q$ and $R$ are in $\mathbb{Q}[X]$.

2) Show that $Q$ and $R$ are in $\mathbb{Z}[X]$.

(Before those questions, I have shown that when two polynomials $P$,$Q$ $\in$ $\mathbb{Q}[X]$, and $PQ\in \mathbb{Z}[X]$ then $P\in\mathbb{Z}[X]$ and $Q\in\mathbb{Z}[X]$)

EDIT : I have to do this without induction, and to start with the first question.

MiKiDe
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    Try applying the long division algorithm to divide $ A $ by $ B $. Do fractions ever show up? (In addition, quotients and remainders in Euclidean division are unique in $ K[X] $, where $ K $ is a field.) – Ege Erdil Feb 11 '17 at 10:27
  • How can I do that with this abstract expression? But I understand that the GCD(A,B) is in $\mathbb{Z}[X]$, but how to formalize your idea? – MiKiDe Feb 11 '17 at 10:37
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    You may simply induct on the degree of $ A $. – Ege Erdil Feb 11 '17 at 10:51

2 Answers2

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Hint:

Use induction on the degree of the dividend $A$.

The assertion is true if $\deg A<\deg B$, since $A=0\cdot B +A$.

Suppose now $\deg A=n\ge\deg B=m$, and $Q, R\in\mathbf Z[X]$ if the degree of the dividend is less than $n$. As $B$ is monic, the first step of the division algorithm consists in replacing $A$ with $A'=A-a_nX^{n-m}B$. Observe $\deg A'\le n-1$.

Bernard
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  • Yes, but it shows that $Q$ and $R$ are in $Z[X]$, but I have to show that the both polynimials are in $Q[X]$ at first. So I guess that my teacher wants me to use something else that induction, btw your idea works well! – MiKiDe Feb 11 '17 at 11:14
  • If they're in $\mathbf Z[X]$, they're in $\mathbf Z[X]$. Or replace $\mathbf Z$ with $\mathbf Q$. There's not a single word to change. – Bernard Feb 11 '17 at 11:18
  • Yes I know that. But I'm sure that my teacher wants another method, because otherwise she won't ask to prove at first on $Q[X]$ and after on $Z[X]$ .

    But I keed your method in mind!

     – MiKiDe Feb 11 '17 at 11:26
  • I don't see wht she would want. There's nothing special with $\mathbf Q$: it's just a particular case of a general result about polynomial rings over fields being Euclidean domains. – Bernard Feb 11 '17 at 12:35
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$(1)\ $ By the division algorithm in $\Bbb Q[X]\,$ we have $\, A = B\bar Q + \bar R,\,$ $\,\bar Q,\bar R\in\Bbb Q[X].\,$ By the uniqueness of the quotient and remainder in $\,\Bbb C[X]\,$ we deduce $\,Q = \bar Q\,$ and $\,R = \bar R\,$ are both in $\,\Bbb Q[X].$

$(2)\ $ If $\,Q\,$ has a nonintegral coef then we can write $\,Q = C + Q',\,$ for $\, C\in\Bbb Z[X]\,$ and $\,Q'\,$ with nonintegral lead coef. Then $\,A = B (C+Q') + R$ $\Rightarrow\, A - BC = BQ' + R.\, $ Comparing lead coefs, using $\,B$ monic, implies that the lead coef of $\,Q'$ is integral, contradiction. So all coef's of $Q$ are integral, i.e. $\,Q\in\Bbb Z[X],\,$ hence also$\,R = A-BQ\in\Bbb Z[X]$.

Bill Dubuque
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