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My problem:

Let $c(f)$ be a content of a polynomial over $\mathbb{Z}$ and $a \in \mathbb{Z}$. Show that $c(a \cdot f)= a \cdot c(f)$.

The content of a polynomial over $\mathbb{Z}$ is the gcd (greatest common divisor) of its coefficients.

My attempt:

Let $f(x)=a_0+a_1x+a_2x+\ldots +a_nx^n$. We have:

$$\begin{align*}a \cdot f(x) &= a(a_0+a_1x+a_2x+\ldots +a_nx^n)\\ &=a\cdot [c(f) \cdot \tilde{f}], \end{align*}$$

where $\tilde{f}$ is a primitive part of $f$.

I don't know how to continue from this. I will be grateful for any help.

Luana
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1 Answers1

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Let $\lbrace c_0, ..., c_n \rbrace$ be the coefficients of $f$. Then the coefficients of $af$ are $\lbrace ac_0, ..., ac_n \rbrace$.

Let's prove that $$\gcd(ac_0, ..., ac_n) = a \gcd(c_0, ..., c_n)$$

First, for each $k \in \lbrace 0, ..., n \rbrace$, $\gcd(c_0, ..., c_n)|c_k$, and $a|a$, so $a\gcd(c_0, ..., c_n)|ac_k$, and because this is true for all $k$, you deduce that $$a\gcd(c_0, ..., c_n)|\gcd(ac_0, ..., ac_n) \quad \quad\quad \quad (1)$$

Conversely, for each $k \in \lbrace 0, ..., n \rbrace$, $\gcd(ac_0, ..., ac_n)|ac_k$, so $\frac{1}{a}\gcd(ac_0, ..., ac_n)|c_k$, and because this is true for all $k$, you deduce that $$\frac{1}{a}\gcd(ac_0, ..., ac_n)|\gcd(c_0, ..., c_n), \quad \text{i.e.} \quad \gcd(ac_0, ..., ac_n)|a\gcd(c_0, ..., c_n)\quad \quad\quad \quad (2)$$

So you get with $(1)$ and $(2)$ that $$\gcd(ac_0, ..., ac_n)=a\gcd(c_0, ..., c_n)$$

which is exactly the same as saying that $$c(af)=ac(f)$$

TheSilverDoe
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