# Polynomial remainder theorem for polynomials over ring

Polynomials over field are Euclidean domain. So for any $a$

$f(x)=(x-a)Q(x) + R(a)$ where $R(a) = f(a)$.

So we have a corollary: $a$ is a root of polynomial $f(x) \iff (x-a)|f(x)$.

Is this true for polynomials over rings? For example let $\mathbb F[x_1, x_2, ..., x_n]$ be a polynomial of several variables. We can also see it as a ring of polynomials over ring: $\mathbb F[x_2, ..., x_n][x_1]$. If we keep $deg(f)$ as a Euclidean norm is this true that remainder norm is less than norm of divider?

Thanks!

• What is the precise statement you are asking about for polynomials over rings? That they are a Euclidean domain? Or that $a$ is a root of $f(x)$ iff $x-a$ divides $f(x)$? Or something else? Jun 4, 2017 at 23:24
• @EricWofsey Is Polynomial remainder theorem true for polynomials over ring. Jun 4, 2017 at 23:27
• What is the "polynomial remainder theorem"? Jun 4, 2017 at 23:27
• @EricWofsey Highlighted statement Jun 4, 2017 at 23:29

The polynomial division algorithm works over any ring as long as you are dividing by a monic polynomial. Indeed, if you want to divide a polynomial $f(x)=ax^n+\dots$ by a polynomial $g(x)=x^m+\dots$, all you need to do is multiply $g(x)$ by $ax^{n-m}$ to cancel the leading term of $f(x)$, and then you can repeat this to lower the degree until the degree is less than $n$.
In particular, for any ring $R$, the division algorithm works for $g(x)=x-a$ if $a\in R$ to give that there exists $Q(x)\in R[x]$ and $r\in R$ such that $$f(x)=(x-a)Q(x)+r.$$ If $R$ is commutative, you can then evaluate both sides at $x=a$ to find that $r=f(a)$, thus giving your desired statement.
• Sir, whst if $R$ is non commutative? Is the corollary in the question still holds? How? Please reply Sep 10, 2021 at 3:45
• I think the factor theorem fails if $R$ is non commutative ring. For example: In case of ring of polynomials over Hamilton's quaternions. Please reply Sep 10, 2021 at 4:00