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The Eilenberg–Niven theorem generalizes the fundamental theorem of algebra for quaternionic polynomials,¹ and this theorem was further generalized to also encompass octonionic polynomials.²

Does similar theorem holds for the sedenion algebra?

References:

[1] Eilenberg, Samuel; Niven, Ivan, The “fundamental theorem of algebra” for quaternions, Bull. Am. Math. Soc. 50, 246–248 (1944). ZBL0063.01228.

[2] Jou, Yuh-Lin, The ‘fundamental theorem of algebra’ for Cayley numbers, Sci. Record, Acad. Sinica 3, 29–33 (1950). ZBL0039.26701.

Edit: This question is indeed answered in the negative (see the comment of user49822). Of course the sedenion algebra has zero divisors, f.i. $a = e_1+e_{10}$. But each $x \in \mathbb{S} \setminus \{0\}$ has an inverse $1/x = \bar{x}/\|x\|^2$ with $x \cdot 1/x = 1/x \cdot x = 1$. In particular $1/a = -a/2$.

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    $\begingroup$ Let $a$ be a zero divisor in sedenions, then $ax-1$ has no roots, so the obvious generalization doesn't hold. $\endgroup$
    – Wojowu
    Commented Sep 19, 2021 at 20:26
  • $\begingroup$ @Wojowu Thanks! I didn't recall that sedenions had zero divisors... 🤦🏻‍♂️ $\endgroup$
    – Tadashi
    Commented Sep 20, 2021 at 22:13
  • $\begingroup$ @Tadashi: The comment from Wojowu doesn't seem to be quite correct. Let $a = e_1+e_{10}$. Then $a$ is a zero divisor, but with $x = -a/2$ we get $ax-1=0$. Of course there is no obvious generalization. $\endgroup$
    – Dieter Kadelka
    Commented Feb 4 at 10:51
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    $\begingroup$ Wojowu's comment is easily fixed: if $a$ is a zero divisor, then the linear map $x\mapsto ax$ has a nontrivial kernel, hence is not surjective, which means that there is some $b$ for which $ax-b$ has no roots $\endgroup$
    – user49822
    Commented Feb 4 at 16:17
  • $\begingroup$ @user49822 Thank you. Of course your argument is simple and correct. $\endgroup$
    – Dieter Kadelka
    Commented Feb 4 at 16:34

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