Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 11 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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.