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This is Question 19.8 of $Abstract\ Algebra:\ A\ First\ Course$ by Dan Saracino.

Let $\mathbb{F}$ be a fieald, $f(X) \in \mathbb{F}[X]$. If $a$ is a root of $f(X)$, then we can write $f(X)=(X-a)^mg(X)$, where $m \ge 1$ and $X-a$ does not divide $g(X)$. Show $m$ is uniquely determined by $f$ and $a$. That is if $f(X)=(X-a)^rh(X)$ and $X-a$ does not divide $h(X)$, then $m=r$.

Without loss of generality, I consider $r \ge m$. Then, $$(X-a)^m((X-a)^{r-m}h(X)-g(X))=0.$$ So $$\forall X\ne a,\ \ (X-a)^{r-m}h(X)=g(X).\ \ \ (1)$$

Now, one may say if $r>m$, then $X-a$ divides $g(X)$, which is a contradiction. Hence $r =m$. However, I think this argument does not work since (1) only works for $X \ne a$. Indeed, (1) does not imply $g(a)= 0$.

What should I do now to prove the statement?

Bill Dubuque
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khashayar
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    $X$ is never $a$: In the world of polynomials, $X$ is an "indeterminate" , a new symbol. You can cancel $X-a$ because $X-a$ is a nonzero polynomial (of degree $1$) and is therefore not a zero-divisor. –  Feb 25 '23 at 01:43

1 Answers1

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In first equation

$$(X-a)^m((X-a)^{r-m}h(X)-g(X))=0.$$

is as element of polynomial ring $\mathbb{F}[X]$.

Now $\mathbb{F}[X]$ is integral domain and $X-a\neq 0$ as element of $\mathbb{F}[X]$, we get equation

$$ (X-a)^{r-m}h(X)=g(X).$$

as element of polynomial ring $\mathbb{F}[X]$.

There is difference the following tow condition for $p(X)\in \mathbb{F}[X]$.

(1) For all $a\in \mathbb{F}$, $p (a)=0$.

(2)$p(X)= 0\in \mathbb{F}[X]$.

In fact, for example $\mathbb{F}=\mathbb{F}_2$ and $p(X)=X^2+X$, then this satisfy (1) and not (2).

Yos
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  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Feb 25 '23 at 02:07
  • Although my question has a duplicate, this answer gives me a better intuition in comparison to the answers to the duplicate question. This is because this answer exactly talks about my mistake. – khashayar Feb 25 '23 at 02:12