Let \(F\) be a field, and let \(R\) be the subring of \(F[X]\) of polynomials with \(X\) coefficient equal to \(0\). Prove that \(R\) is not a UFD.
Solution
The element \(X^6\) has the two inequivalent factorizations \[ X^6=(X^2)(X^2)(X^2)=(X^3)(X^3) \] into irreducibles of \(R\).
1. The ring \(R\) is a domain whose units are exactly \(F^\times\).
One has \(R=F+X^2F[X]\). This set contains \(0,1\) and is closed under subtraction. For \(f,g\in R\), the coefficient of \(X\) in \(fg\) is \(f_0g_1+f_1g_0=0\), so it is closed under multiplication as well. It is a subring of the domain \(F[X]\) and hence a domain.
If \(uv=1\) with \(u,v\in R\), degree additivity in \(F[X]\) forces \(\deg u=\deg v=0\), so both are nonzero constants. Conversely, every nonzero constant has its inverse in \(F\subset R\). Thus a nonzero nonunit in \(R\) has degree at least two: degree one is forbidden by the defining coefficient condition.
2. The displayed factorizations are into nonassociate irreducibles.
1. Their product would then have degreeat least four, a contradiction. Both monomialsare therefore irreducible in \(R\).
They are not associates: multiplication by a unit, which is a nonzero constant, cannot change degree. Consequently the two factorizations of \(X^6\) have different irreducible factors and different lengths; they cannot agree up to units and order. This violates unique factorization, so \(R\) is not a UFD.