Is the Polynomial Factor Theorem true over all commutative rings? I'm reading Fraliegh's text on abstract algebra, and his statement for the factor theorem is, "An element $a\in F$ ($F$ is a field) is a zero of $f(x)\in F[x]\iff$ $(x-a)$ is factor of $f(x)$ in $F[x]$." The proof is straightforward, and seems to be the most common one, but I don't see where exactly you need to use the fact that $F$ is a field; it seems like this result would hold for any ring $R$. Is this true? 
 A: Yes, your intuition correct: the Polynomial Factor Theorem works over any commutative ring since we can always divide (with remainder) by a polynomial that is monic i.e. lead coef $=1$ (or any unit = invertible element). Ditto for the equivalent Polynomial Remainder Theorem - see below.
Theorem $ $ TFAE for a polynomial $\,f\in R[x],\,$ and $\,a\in R\,$ a commutative ring.
$(0)\ \ \ f = (x\!-\!a)q + r\ $ for some $\,q\in R[x],\ r\in R\ \ \ $ [Monic Linear Division Algorithm]
$(1)\ \ \ f\bmod x\!-\!a = f(a)\ \ \ \ \ \ \ $ [Polynomial Remainder Theorem]
$(2)\ \ \ f(a) = 0\,\Rightarrow\, x\!-\!a\mid f\ \ \ $ [Polynomial Factor Theorem]
Proof $\ (0\Rightarrow 1)\ \ \ f = (x\!-\!a)q + r\,\overset{\large x\,=\,a}\Longrightarrow\, r=f(a)\,\Rightarrow\,f\bmod x\!-\!a = r = f(a) $
$(1\Rightarrow 2)\ \ \ f\bmod x\!-\!a = f(a) = 0\,\Rightarrow\, x\!-\!a\mid f$
$(2\Rightarrow 0)\ \ \ g := f-f(a)\,$ has $\,g(a) = 0\ $ so $\ g = f-f(a) = (x\!-\!a)q$
A: To add to Bill's comment, I want to point out that the factor theorem can fail quite dramatically when the ring $R$ is not commutative: a quadratic polynomial can have uncountably many roots. I'll talk specifically about an example using the quaternion ring $\mathbb H$. Elements of $\mathbb H$ have the form $a + bi + cj + dk$, where $a,b,c$, and $d$ are real numbers and $i$, $j$, and $k$ satisfy the quaternion multiplication relations (i.e. $ij=-ji=k$, $jk=-kj=i$, and $ki=-ik=j$); you add elements componentwise and multiply by distributing. The ring $\mathbb H$ is a division ring, meaning that every nonzero element of $\mathbb H$ is a unit.
Now consider the polynomial $f(x)=x^2+1\in\mathbb H[x]$. Here are three different ways to factor $f$:
\begin{align}
f(x) &= (x+i)(x-i) \\
&= (x+j)(x-j) \\
&= (x+k)(x-k).
\end{align}
In fact, $f$ has uncountably many roots: for instance, $(\cos t)\cdot i+(\sin t)\cdot j$ is a root of $f$ for any $t\in\mathbb R$. See this stack exchange post for more information.
