# Factoring a polynomial into linear factors by ring extension

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...

• Do you have any example where the "just do it" solution does not work? To take your example at the end, let $S = (\mathbb{Z}/8 \mathbb{Z})[a_1, a_2, b_1, b_2] / \langle a_1 a_2-2, a_1 b_2+a_2 b_1-1, b_1 b_2 - 2 \rangle$. Does $\mathbb{Z}/8 \mathbb{Z}$ not inject into $S$? Commented Aug 25, 2022 at 9:46
• For the particular example of $2x^2+x+2$, this factors over $\mathbb{Z}/8 \mathbb{Z}$ as $(2x+5)(x+2)$. More generally, $-15$ is a square in the $2$-adics; let $\sqrt{-15}$ denote the $2$-adic square root of $-15$ which is $1 \bmod 4$. Then the factorization $2x^2+x+2 = (x-\tfrac{-1+\sqrt{-15}}{4})(2x-\tfrac{-1-\sqrt{-15}}{2})$ gives a factorization in $\mathbb{Z}/2^k \mathbb{Z}$ for any $k$. Commented Aug 25, 2022 at 9:57
• @darijgrinberg You can make Timothy Chow's claim rigorous by introducing a scale parameter: work in $R[t]/(\alpha-t\beta)$. Commented Aug 26, 2022 at 6:02
• @TimothyChow: The $K\left[a\right]$-algebra homomorphism $f:K\left[x,y,z,a\right] / \left(xa,ya,za\right) \to K\left[x_1,y_1,x_2,y_2,a\right] / \left(x_1a, x_2a, y_1a, y_2a\right)$ that sends $x,y,z$ to $x_1x_2, x_1y_2+x_2y_1, y_1y_2$, respectively (thus sending $xt^2 + yt + z \mapsto \left(x_1t+y_1\right)\left(x_2t+y_2\right)$), is injective. This is best seen using a multigrading that assigns degree $\left(0,2\right)$ to each of $x,y,z$, degree $\left(1,0\right)$ to $a$, and degree $\left(0,1\right)$ to each of $x_1,x_2,y_1,y_2$. Then, $f$ is graded, and both source ... Commented Aug 30, 2022 at 13:03
• ... and target have all their degree-$\left(i,j\right)$ components vanish when $i$ and $j$ are both positive. Commented Aug 30, 2022 at 13:04

By translation, this is also true if there is $$t\in R$$ such that $$P(t)\in R^\times$$. (In fact, in these cases we can take $$S$$ finite free, in particular faithfully flat, over $$R$$).
Now, in general, let $$t$$ be an indeterminate and consider $$R_1:=R[t, P(t)^{-1}].$$ By construction, we have $$P(t)\in R_1^\times$$ so $$P$$ can be factored over some finite free overring $$S$$ of $$R_1$$.
Geometrically, $$\mathrm{Spec}(R_1)\subset \mathbb{A}^1_R$$ is the open complement of the hypersurface defined by $$P(t)$$. Thus, to say that $$P$$ has content ideal $$c(P)=R$$ just means that $$\mathrm{Spec}(R_1)$$ is surjective (hence faithfully flat) over $$\mathrm{Spec}(R)$$. Consequently, $$R\to R_1\to S$$ is then faithfully flat, in particular injective. Note that it can be injective in other cases, but I suspect that faithful flatness is the really useful condition.
• The first paragraph is brilliant. I cannot follow the second paragraph, as I don't know the alg/geo dictionary you are tacitly using. That said, it appears that all that needs to be actually proved is that $P\left(t\right)$ is a non-zero-divisor in $R\left[t\right]$, but that follows from McCoy's theorem. Thus, we can even loosen our assumption "the content of $P$ is $R$" to "the content of $P$ is a faithful $R$-submodule of $R$". Nice!! Commented Aug 25, 2022 at 13:48
• My argument for faithful flatness can be phrased as follows: for each maximal ideal $m$ of $R$, we have $R_1/mR_1\cong (R/m)[t,\overline{P}(t)^{-1}]$ where $\overline{P}=P\bmod m[t]$. If $P$ is primitive, $\overline{P}$ is nonzero, so $R_1/mR_1\neq0$. Since $R\to R_1$ is obviously flat, it is faithfully flat. Commented Aug 25, 2022 at 15:54
• About $P$ being a nonzerodivisor: this is indeed sufficient to get injectivity, but not faithful flatness in general. This is the sort of thing I was thinking about at the end of my answer. Commented Aug 25, 2022 at 15:58
• @darijgrinberg if you only care about injectivity of the composite $A\to A[t]\to A[t,1/f(t)]$, can't you circumvent McCoy's theorem? It seems sufficient to say that for $a\in A$ the fraction $\tfrac a1$ is zero in the localization iff its annihilated by a power of $f$, recovering your condition that the content of $f$ be a faithful ideal. Commented Feb 25, 2023 at 17:45