Let $E$ be a $K$-vector space, and $u$ an $E$-linear function.

Let $P$ be a polynomial verifying $P(u) = 0$ and $P \neq 0$, and $A$, $B$ two polynomials verifying $P = AB$ and $A \wedge B = 1$.

Using an arithmetic theorem, prove that $E = Im A(u) + Im B(u)$. The only theorem I think of is Bézout ($AU + BV = 1$ where $U$ and $V$ are polynomials) but I don't see how to use it here.

  • 1
    $\begingroup$ Minor notation detail: the real and imaginary part functions are usually used with parentheses: e.g., $\Re(i) = 0$. $\endgroup$ Apr 15, 2017 at 21:57
  • $\begingroup$ I think you want $\ker $ instead of $\Im$ to conclude $\ker P(u) = \ker A(u) \oplus \ker B(u)$. $\endgroup$
    – lhf
    Apr 15, 2017 at 22:57
  • $\begingroup$ yup syntax problem, didn't see it, and it's actually the image of A(u) and B(u), and as Im A(u) is included in Ker B(u) and same thing with B and A reversed I can conclude that E = Im A(u) + Im B(u) = Ker B(u) + Ker A(u) $\endgroup$
    – nougatine
    Apr 16, 2017 at 12:46

1 Answer 1


image version

$E = \operatorname{im} A(u) + \operatorname{im} B(u)$

Let $\alpha \in E$. Bézout gives $AU + BV = 1$. Then $\alpha = (AU)(u)(\alpha)+(BV)(u)(\alpha)$. Notice that $(AU)(u)(\alpha) \in \operatorname{im} A(u)$ and $(BV)(u)(\alpha) \in \operatorname{im} B(u)$.

This answers your question. But actually we have:

$E = \operatorname{im} A(u) \oplus \operatorname{im} B(u)$

Indeed, let $\gamma \in \operatorname{im} A(u) \cap \operatorname{im} B(u)$. Write $\gamma=A(u)(\alpha)$ and $\gamma=B(u)(\beta)$. Then $B(u)(\gamma)=(BA)(u)(\alpha)=P(u)(\alpha)=0$. Similarly, $A(u)(\gamma)=0$. Therefore, $\gamma \in \ker A(u) \cap \ker B(u)=0$, as proved below.

kernel version

More generally:

If $P=AB$ with $\gcd(A,B)=1$, then $\ker P(u) = \ker A(u) \oplus \ker B(u)$

Let's prove that in steps.

$\ker P(u) \subseteq \ker A(u) + \ker B(u)$

Let $\alpha \in \ker P(u)$. Then $0=P(u)(\alpha)=A(u)B(u)(\alpha)=B(u)A(u)(\alpha)$. Bézout gives $AU + BV = 1$. Then $\alpha = (AU)(u)(\alpha)+(BV)(u)(\alpha)$. Notice that $(AU)(u)(\alpha) \in \ker B(u)$ and $(BV)(u)(\alpha) \in \ker A(u)$.

$\ker P(u) \supseteq \ker A(u) + \ker B(u)$

Let $\alpha \in \ker A(u)$ and $\beta \in \ker B(u)$. Then $P(u)(\alpha + \beta)=BA(u)(\alpha)+AB(u)(\beta)=0$.

$\ker A(u) \cap \ker B(u) = 0$

Let $\alpha \in \ker A(u) \cap \ker B(u)$. Then $\alpha = (AU)(u)(\alpha)+(BV)(u)(\alpha)=(UA)(u)(\alpha)+(VB)(u)(\alpha)=0$.

Applying this to the problem, we have $E=\ker P= \ker A(u) \oplus \ker B(u)$.


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