Post #1815105
2026-04-07 22:22 UTC
@hallasurvivor This is related to the fact that \( Q[x] \) does not have unique factorization. When dealing with polynomial rings over commutative fields, we can prove unique factorization as a consequence of the Euclidean algorithm. And this doesn't work for skew-fields. The key point seems to be that you need to multiply things like \( ax^m bx^n = ab x^{m+n} \). This requires you to commute the indeterminate \( x \) with \( b \).
Perhaps there's something more conceptual...
Replies (1)
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@hallasurvivor@sunny.garden 2026-04-08 03:08
@francisbrb This is what I had in mind! I think the more conceptual phrasing is that R[x] is the free R-algebra on one *commutative* generator, and so the "evaluation at a" map R[x] --> R is only a ring hom when a is central. The reason for this is essentially your argument about multiplying terms.