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Let $R ≠ 0$ be a principal ideal domain that is not a field. Let $0 ≠ x \in R.$

Show that if $x = yz$ for $y, z \in R \backslash R^\times$ with $\text{gcd}(y, z) = 1$, then there exists an idempotent $e \in R/x$ with $e ≠ 0, 1$ such that $e \cdot R/x \cong R/y$ and $(1 - e) \cdot R/x \cong R/z.$

I think I almost have this solved myself; the Chinese remainder theorem tells us that $R/x \cong R/y \oplus R/z$. In general, I also know that we have $R/x \cong e \cdot R/x \oplus (1 - e) \cdot R/x$ for an idempotent. But struggle in showing that indeed $e \cdot R/x \cong R/y$ and that $e$ can be non trivial.

J. W. Tanner
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Minerva
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1 Answers1

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Yes you have already done all the work here. That is you have established an isomorphism $$\theta\colon R/x\to R/y \times R/z,$$mapping $$a+xR\mapsto (a+yR,a+zR).$$

Let $e+xR=\theta^{-1}(1+yR,0+Rz)\in R/x$.

Then we have $$\theta(e+xR)=(1+yR,0+zR),\quad \theta((1-e)+xR)=(0+yR,1+zR)$$

Thus $\theta$ identifies: $$eR/x\cong (1+yR,0+zR)(R/y\times R/z)\cong R/y,$$ $$(1-e)R/x\cong (0+yR,1+zR)(R/y\times R/z)\cong R/z.$$

Finally note that we cannot have $e=0,1$ as then $1+yR=0+yR$ or $1+zR=0+zR$ in which case either $y$ or $z$ is a unit.

tkf
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