0

Let $R$ a commutative ring and let $f(x) \in R[x]$.

(i) Prove that if $(x-a)^2|f(x)$, then $(x-a)|f'(x)$.

(ii)Prove that if $(x-a)|f(x)$ and $(x-a)|f'(x)$, then $(x-a)^2|f(x)$.

Any help in order to prove the last statements? I have been trying to prove these straight forward but I cannot see how to attack this exercises, thanks.

Hagen von Eitzen
  • 374,180
Cos
  • 1,799
  • Did you test it for $f(x)=(x-a)^2$? – Dietrich Burde Feb 26 '19 at 19:10
  • Testing the way you propose will help in order to prove (ii), right? @DietrichBurde – Cos Feb 26 '19 at 19:16
  • Im stucked proving (ii), now. If I have $(x-a)g(x)=f(x)$ and $(x-a)(h(x)=f'(x)$,how i see $(x-a)^2r(x)=f(x)$. I´ve trying developing derivative product rule as the hint below but im going in circles @DietrichBurde – Cos Feb 26 '19 at 19:33

2 Answers2

1

Write $f(x)=(x-a)^kg(x)$ and apply the product rule.

Hagen von Eitzen
  • 374,180
  • Thanks! This was really helpful in order to prove (i) but still dont know how to end the proof for (ii) @HagenVonEitzen – Cos Feb 26 '19 at 19:19
0

Hint for $(ii)$

Since $(x - \alpha) | f(x)$ one has $f(x)=g(x)(x-\alpha)$ and $f'(x) = g'(x)(x-\alpha) + g(x)$.

On the other side $f'(x) = (x- \alpha) \cdot h(x)$ simillary.

Hence: $g'(x)(x-\alpha) + g(x) = (x-\alpha) \cdot h(x)$ $\iff (x - \alpha)\left(g'(x) + \frac{g(x)}{(x-\alpha)}\right) = (x - \alpha)h(x)$

Can you continue from here?

Sqyuli
  • 600