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Denote by $\mathbb{H}[x_1,\dots,x_n]$ the ring of polynomials in $n$ variables with quaternionic coefficients, where the variables commute with each other and with the coefficients. Two polynomials $P,Q\in \mathbb{H}[x_1,\dots,x_n]$ are similar, if $P=a Q b$ for some $a,b\in \mathbb{H}$. A ring $\mathbb{K}$ is factorial, if the equality $P_1\cdot\dots\cdot P_n=Q_1\cdot\dots\cdot Q_m$, where $P_1,\dots, P_n,Q_1,\dots,Q_m\in \mathbb{K}$ are irreducible (and noninvertible) elements, imply that $n=m$ and there is a permutation $s\in S_n$ such that $P_k$ is similar to $Q_{s(k)}$ for each $k=1,\dots,n$?

By [1, Theorem 1] and [2, Theorem 2.1] it follows that $\mathbb{H}[x]$ is factorial.

Is the ring $\mathbb{H}[x,y]$ factorial?

This question is a continuation of the following ones:

When the determinant of a 2x2 polynomial matrix is a square?

Pythagorean 5-tuples

[1] Oystein Ore, Theory of non-commutative polynomials, Annals of Math. (II) 34, 1933, 480-508.

[2] Graziano Gentili and Daniele C. Struppa, On the Multiplicity of Zeroes of Polynomials with Quaternionic Coefficients, Milan J. Math. 76 (2008), 15-25, DOI 10.1007/s00032-008-0093-0.

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1 Answer 1

up vote 8 down vote accepted

Just to remove the question from the 'Unanswered' list: $(x-i)\cdot((x+i)(y+j)+1)=((y+j)(x+i)+1)\cdot(x-i)$, hence $\mathbb{H}[x,y]$ is not factorial.

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