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Let $R$ be a commutative ring and $f\in R[X]$ be a monic non-constant polynomial. How can one show that there exists a commutative ring $S$ so that $R$ is a subring of $S$ and $f$ can be written as a product of monic polynomials of degree $1$ in $S[X]$?

I tried to mimic a construction of a splitting field of a polynomial over a field: define $S_1=R[X]/(f)$ where $(f)\subseteq R[X]$ is the ideal generated by $f$. Then $R$ embeds in $S_1$ since the leading coefficient of $f$ is not a zero-divisor. Moreover $f$ has a root $\alpha=X+(f)$ in $S_1$.

Now if $S_1$ was a field then we could write $f=(X-\alpha)g$ for some $g\in S_1[X]$ and we could continue similarly with $g$. However I think this depends on the fact that a field is a Euclidean domain and $S_1[X]$ may not be one. This is where I'm stuck at the moment.

feebly
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    The division by $X-\alpha$ works for any commutative ring. (Moreover, the division algorithm stands for any monic polynomial instead of $X-\alpha$, and on any commutative ring.) – user26857 Aug 19 '15 at 14:47
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    Thank you. The answers found in http://math.stackexchange.com/questions/116029/why-can-we-use-the-division-algorithm-here are helpful in this context. – feebly Aug 19 '15 at 15:07

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