The following sounds so natural, I'm surprised I have never asked it before:

Question 1.Let $R$ be a commutative ring. Let $P \in R\left[X\right]$ be a polynomial. Can we find a commutative ring $S$ that contains $R$ as a subring, and a bunch of elements $a_1, a_2, \ldots, a_m, b_1, b_2, \ldots, b_m \in S$ such that $P = \left(a_1 X + b_1\right) \left(a_2 X + b_2\right) \cdots \left(a_m X + b_m\right)$ in $S\left[X\right]$ ?

In other words, can every polynomial over a commutative ring be factored as a product of polynomials of degree $\leq 1$ over a possibly larger commutative ring?

The answer to this question is known to be positive if $P$ is monic (here, it suffices to adjoin the roots of $P$ to $R$ one by one, using the standard $R\left[X\right] / \left(P\right)$ construction). Thus, by mirroring, the answer is also positive if the constant term of $P$ is $1$.

I suspect that the answer in the general case is negative. However, perhaps above two positive cases can be combined under the following general roof:

Question 2.Assume that the content of $P$ (that is, the ideal of $R$ generated by the coefficients of $P$) is the whole $R$. Is the answer to Question 1 positive?

An example for Question 2 would be the polynomial $2X^2 + X + 2$ over $R = \mathbb{Z}/8$. Over $R = \mathbb{Z} / 4$, the same polynomial factors as $\left(2X+1\right)\left(X+2\right)$ even without extending the ring.

An example for Question 1 that seems particularly suited to disprove it might be the polynomial $\alpha X^2 + \beta X + \alpha$ over $R = \mathbb{Q}\left[\alpha, \beta\right] / \left(\alpha, \beta\right)^2$. But it's one thing to come up with the polynomial, another to actually prove that it does not factor into linear factors...

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