Say we have an ideal $I$ for $R$ and we have $R[[x]]$ as the ring of formal power series. We want to show that $R[[x]]/(I,x) \cong R/I$.
Would this proof be similar to that of showing $R[x]/I[x] \cong (R/I)[x]$ for the ring of polynomials?
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- $\begingroup$ Surely. Remember that when you factor out by $x$, you’re essentially deleting all monomials in which $x$ appears; equivalently, you’re evaluating $x$ to $0$. $\endgroup$Lubin– Lubin2018-04-18 22:31:35 +00:00Commented Apr 18, 2018 at 22:31
- $\begingroup$ @Lubin: I'm not sure it's a good idea to encourage thinking about "evaluation" in a formal power series ring. However the ideal generated by $x$ in $R[[x]]$ includes all formal power series with zero constant term and that has much the same effect as what you say about deleting monomials. $\endgroup$Rob Arthan– Rob Arthan2018-04-18 22:34:28 +00:00Commented Apr 18, 2018 at 22:34
- $\begingroup$ Ah, @RobArthan, but I evaluate formal power series all the time! You just have to make sure that the element of the complete $R$-algebra that you choose has the property that its powers go to zero. The zero element of $R$ will always satisfy this property, or, if $R=\Bbb Z_p$, then you may choose any element of $p\Bbb Z_p$. $\endgroup$Lubin– Lubin2018-04-18 23:12:58 +00:00Commented Apr 18, 2018 at 23:12
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1 Answer
$\begingroup$ $\endgroup$
The map $e\colon R[[x]]\to R$ defined by $f=\sum_{k\ge0}a_kx^k\mapsto a_0$ is a (surjective) ring homomorphism, with kernel $xR[[x]]$ (the power series with $0$ constant term).
Consider now the projection $\pi\colon R\to R/I$. What is the kernel of $\pi\circ e$? Now apply the homomorphism theorem.
