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Suppose that $P_1$ and $P_2$ are coprime between them in $\mathbb{K}[X]$. Show that $$ \ker ( (P_1 \times P_2 )(f) )=\ker(P_1 (f)) \oplus \ker(P_2(f)).$$

I tried by starting off by showing that the right hand equality is true, that is $\mathbb{K}[X]$ is indeed equal to $\ker(P_1 (f)) \oplus \ker(P_2(f))$. My first attempt was to show that $$\ker(P_1 (f)) \cap \ker(P_2(f))= \{ 0 \} .$$

But I got nowhere. I think I could have used the Bezout theoreme, that is:

If $P_1$ and $P_2$ are coprime, then $\exists U,V \in \mathbb{K}[X]$ so that $P_1U(X) + P_2V(X)=1$

But nothing comes to my mind.

Rócherz
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John Mayne
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    This result is known as the Kernels lemma and is generalised to the case of $n$ pairwise coprime polynomials. – Bernard Jul 19 '17 at 12:08

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Bézout is the way to go: $ P_1(X)U(X) + P_2(X)V(X)=1$.

$\ker (P_1P_2)(f) = \ker P_1 (f) + \ker P_2(f)$

We have $v = (P_1 U)(f)v + (P_2V)(f)v$. If $v \in \ker (P_1P_2)(f)$, then $(P_1 U)(f)v \in \ker P_2(f)$ and $ (P_2V)(f)v \in \ker P_1(f)$.

If $u \in \ker P_1(f)$ and $ w \in \ker P_2(f)$, then $(P_1P_2)(f)(u+w)=0+0=0$.

$\ker P_1 (f) \cap \ker P_2(f)= 0$

$v = (UP_1)(f) v + (VP_2)(f)v$ and so $\ker P_1 (f) \cap \ker P_2(f)= 0$.

lhf
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  • How is defined the product of a vector by a polynomial? – Bernard Jul 19 '17 at 12:09
  • @Bernard, right! I forgot to add $(f)$. Done, thanks. But one can make sense of $Pv$ by seeing $V$ as a $K[X]$-module: $Pv = P(f)v$. – lhf Jul 19 '17 at 12:12
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    I know, but I doubt the O.P. even knows what a module is. Also a fundamental (obvious) point is that for any polynomials $P, Q$, $P(f)$ and $Q(f)$ commute. – Bernard Jul 19 '17 at 12:19