Should $R$ be commutative as we claim that $c \in R$ is a root of $f \in R[x]$ iff $x-c$ divides $f$.

Question

A theorem in Algebra by Hungerford says:

Let $R$ be a commutative ring with identity and $f\in R[x]$. Then $c \in R$ is a root of $f$ iff $x-c$ divides $f$.

Proof: $f(x) = q(x)(x-c)+f(c)$ by Remainder Theorem. If $x-c$ divides $f$, $f(x) = h(x)(x-c) = f(x) = q(x)(x-c) + f(c)$ with $h \in R[x]$, whence $(h(x) - q(x))(x-c) = f(x)$. Since $R$ is commutative, $f(c) = (h(c) - q(c))(c - c) = 0$($R[x]$ is a universal objective in a specific category). Therefore $c$ is a root.

If I go this way:

Since $x-c$ divides $f$, we have $f(x) = h(x)(x-c)$. Let $h(x) = \sum_\limits{i=0}^{n-1}a_ix^i$, then $f(x) = a_{n-1}x^n + \sum_\limits{i=1}^{n-1}(a_{i-1} - a_ic)x^i - a_0c$. Therefore $f(c) = a_{n-1}c^n + \sum_\limits{i=1}^{n-1} (a_{i-1}c^i - a_ic^{i+1}) + a_0c= 0$ and $c$ is a root.

No commutativity is assumed in my proof but Algebra emphasizes the commutativity(with a sentence "Commutativity is not required for the converse.".

Am I wrong or Hungerford is somehow misleading?

You have implicitly assumed commutativity in your expansion of $f(x)$ as $h(x)(x-c)$! — CardioidAss22, Jan 23 '19 at 14:16
@Cardioid_Ass_22 Do you mean the concept of "divide" is based on commutativity? — XT Chen, Jan 23 '19 at 14:19
Perhaps my comment was phrased poorly, but sort of. Since your ring is non-commutative, there is also the possibility that $x-c$ dividing $f(x)$ means $f(x)=(x-c)h(x)$ rather than $f(x)=h(x)(x-c)$. — CardioidAss22, Jan 23 '19 at 14:35
@Cardioid_Ass_22 You are right. But it sounds tricky since all goes well if we change the definition of division or polynomial from right to left. — XT Chen, Jan 23 '19 at 14:58

Jose Brox · Accepted Answer · 2019-01-23T15:22:28.153

You are right: the factor theorem does not need commutativity. You have to be a bit careful, though: for starters, you can have left or right polynomials, depending on where you put the coefficients with respect to the variable, division can be done on the left or on the right, and substitution is not well behaved with respect to products, it is not a homomorphism anymore. That being said, the following results are true for left polynomials:

1) Division algorithm: If $f,g$ are left polynomials with invertible leading coefficient, then there exist unique right quotient $q_r$, right remainder $f_r$, left quotient $q_l$ and left remainder $f_l$ such that $$f=q_rg+f_r=gq_l+f_l$$ (the subindices are motivated by the position of the divisor).

2) Remainder theorem: If $f(X)=\sum_{i=0}^n a_iX^i$, the right and left remainders of the division of $f$ by $X-c$ are respectively $$f_r(c)=\sum_{i=0}^n a_ic^i, f_l(c)=\sum_{i=0}^n c^ia_i.$$

3) Factor theorem (aka root lemma): $f$ has $c$ as root iff there exists a left polynomial $g$ such that $f(X)=g(X)(X-a)$.

4) Transitivity of roots on the right: If $f,g$ are left polynomials, then $(fg)(a)\neq f(a)g(a)$ in general, but if $g(a)=0$ then $(fg)(a)=0$. Warning: $f(a)=0$ does not imply $(fg)(a)=0$.

5) Evaluation in products: In general, the correct evaluation for left polynomials is $(fg)(a)=[f(X)g(a)]|_{X=a}$. In the particular case that $b:=g(a)$ is invertible, you can also use $(fg)(a)=f(bab^{-1})g(a)$.

You can see some of these results in A first course in noncommutative rings by T.Y. Lam, at the start of chapter 16.

Should $R$ be commutative as we claim that $c \in R$ is a root of $f \in R[x]$ iff $x-c$ divides $f$.

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