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I need to prove that the ring of dual numbers over a field $K$, $K[\epsilon]$, defined by $a+b\epsilon$, with $a, b \in K$ has exactly three ideals. I already proved that it is isomorphic to $K[X]/(X^2)$.

I'm wondering if I need to use the correspondence theorem to prove this, or if there is another (perhaps quicker) way to do this? Thanks!

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    $\begingroup$ In a way or another one, you have to use that ideal of the dual numbers are ideals of $k[X]$ containing $(X^2)$. You have $(X^2), (X)$ and the whole ring if you allow it. $\endgroup$ – Nicolas Hemelsoet Mar 1 '18 at 20:27
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The ideals in a quotient $R/I$ correspond to the ideals of $R$ that contain $I$. So you are counting ideals in $K[X]$ that contain $(X^2)$. Since $K[X]$ is a principal ideal domain, every ideal is generated by a (wlog. monic) polynomial $f$ and $(f)\supset (X^2)$ mean that $f$ divides $X^2$. The monic polynomials that divide $X^2$ are precisely $1$, $X$ and $X^2$; thus there are three ideals.

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Since $(a + b\epsilon)(a^{-1} - ba^{-2}\epsilon) = 1,$ we see that for all nonzero $a$, $a + b\epsilon$ is a unit. So the ring of dual numbers over $k$ has a unique maximal ideal $(\epsilon)$ and the ring is local. Now we have three ideals, including the zero ideal and the whole ring.

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  • $\begingroup$ Does this prove that there are no more than 3? $\endgroup$ – The Coding Wombat Feb 27 at 21:13

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