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Let $K$ be a field, and consider the ring $R=\{f\in K[x]\mid f'(1)=f''(1)=0\}$. Show that $R$ is not a UFD (Unique Factorization Domain).

My thoughts: I can show that elements such as $(x-1)^3$ and $(x-1)^4$ are irreducible in $R$. Can this be used to show $R$ is not a UFD? I am not sure the best route to take. Should we exhibit an element with a non-unique factorization into irreducibles, or should we find two elements who do not have a GCD? Another thing we may be able to do, is consider a quotient of $R$ by an irreducible polynomial and show it has zero-divisors (hence it is not a domain, so the polynomial we choose isn't prime, but every irreducible polynomial in a UFD must be prime).

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  • $\begingroup$ An element with non-unique factorisation is easiest, methinks. $\endgroup$ May 11, 2020 at 15:27
  • $\begingroup$ IF you knew $z^3$ and $z^4$ were irreducible, then you would have no trouble coming up with inequivalent factorizations of $z^{12}$. $\endgroup$
    – rschwieb
    May 11, 2020 at 15:29

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Silly me. Following the advice of @rschwieb, we have the non-unique factorizations $$ (x-1)^{12}=(x-1)^3(x-1)^3(x-1)^3(x-1)^3 $$ and $$ (x-1)^{12}=(x-1)^4(x-1)^4(x-1)^4 $$ into irreducibles. Therefore $R$ is not a UFD. Just to add this bit of detail: $(x-1)^3$ and $(x-1)^4$ are irreducible in $R$ since if they weren't, they'd either have a linear or quadratic factor. But any linear or quadratic factor will either have a non-zero constant first or second derivative, violating the definition of $R$.

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