# 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?

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)\ \ \ \ \ \ \$$ [Remainder Theorem]

$$(2)\ \ \ f(a) = 0\,\Rightarrow\, x\!-\!a\mid f\ \ \$$ [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$$

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.

• Thank you. That's an interesting example I'll have to remember. – user124910 Apr 3 '14 at 21:25