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Let $R$ be a commutative, associative ring with unit, consider $R[x]$. Is it true that the following two statements are equivalent? $$ R[x] \text{ is the direct product of rings} $$

$$ R[x] \text{ has zero divisors} $$

If it's a product of rings, then it's obvious that it has zero divisors, but I can't prove the other implication.

KReiser
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NeoFanatic
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No. Let $R=k[y]/(y^2)$ for $k$ a field. Then $R[x]\cong k[x,y]/(y^2)$ which has zero divisors but is not the direct product of rings since it has a unique minimal prime $(y)$.

KReiser
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  • Okay, that's good example, but i suppose this statement is true if $R$ is field? – NeoFanatic Oct 05 '22 at 14:47
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    @NeoFanatic Sure, but that is because in that case both statements are false anyway. – Tobias Kildetoft Oct 05 '22 at 14:57
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    @NeoFanatic $R[x]$ is a direct product iff $R$ is, and $R[x]$ has zero divisors iff $R$ does, but there's a big difference between having zero-divisors and being a direct product. Direct products have disconnected spectrum while rings with zero divisors may be irreducible (hence connected) but not reduced, or reduced but not irreducible (and reducible may or may not be connected). – KReiser Oct 05 '22 at 15:19