# Why is commutativity needed for polynomial evaluation to be a ring homomorphism?

In this proof they prove it for fields, but they say that the proof holds for a commutative ring with unity. But why do we need it to be commutative, and why do we need a unity? Why does it not hold if we are only working with a ring?

In the last part of the proof they work with $1$, is this why we need the unity? I guess 1 here is the unity and not the number one, but would this part not work without the unity?

But I do not see where they use commutativity?

• You have the spelling wrong, cummutative refers to things that are likely to cause rain: en.wikipedia.org/wiki/Cumulonimbus_cloud Mar 30, 2015 at 22:16
• Try writing it up in the noncommutative case and be careful that nothing is assumed (such as the formula for the multiplication of polynomials) and see where it breaks down. If it does not, then you do not need the assumption.
– Eoin
Mar 30, 2015 at 22:23
• But why do we need it to be commutative..? Nobody said it needs to be commutative, they have only said commutativity is sufficient for this. If F is contained in the center of E, then evaluation will still be a homomorphism, even if E is noncommutative. Mar 30, 2015 at 22:27
• @rschwieb But then wouldn't they write that we only need it to hold in a ring with unity? They have earlier made a case to distinguish rings with unity and commutative rings. Mar 30, 2015 at 22:34
• @user119615 Does your book define ring homomorphisms as preserving identity? Then that would e why. In general, the same map between commutative rings without identity would still be a homomorphism of rings. Mar 30, 2015 at 22:58

As you've noted yourself, the last line of the proof depends on the ring having an $1$ (that it, a multiplicative identity), which is what "ring with unity" means. So even if the theorem were to be true for rings without unity, it can't possibly be so "with the identical proof.

The ring needs to be commutative, because polynomials don't work well over non-commutative rings. For example, the product of two first-degree polynomials would be $$(aX+b)(cX+d) = aXcX + bcX + aXd + bd$$ but according to the usual rule for the coefficients of a product of polynomials we ought to get $$acX^2+(bc+ad)X + bd$$ and that's only equal to the above expression if $X$ commutes with $c$ and $d$.

We could still simply declare the polynomial multiplication rule to be valid in the noncommutative case and work with purely formal polynomials, investigating where that leads us. But the cost of doing that is that exactly the theorem you're looking at fails to hold. In more primitive terms, what happens is that the product of the values of two polynomial functions wouldn't equal the value of the product polynomial for the same input, or $p(x)q(x)\ne (pq)(x)$ in general.

• The theorem itself doesn't fully make sense if the ring has no unity. Indeed, if the ring has no unity, then $x$ is not a well-defined polynomial! Mar 30, 2015 at 22:28
• Thanks, but when they defined a polynomial they always put the x at the end. In this case it is about commutativity of the elements in E and F. Mar 30, 2015 at 22:31
• @darij: Yes, there's that too. So the $\phi_\alpha(x)=\alpha$ part of the conclusion wouldn't be meaningful. Mar 30, 2015 at 22:31
• @user119615: The putting-x-at-the-end is exactly what creates the problem -- because if you multiply $(ax)\cdot(b)$ you don't get $(ab)x$ unless $xb=bx$. So you need something to make sure that $\alpha$ commutes with all possible coefficients -- and one possible something is that $E$ is commutative, though as rschwieb pointed out less than this can do in a pinch. Mar 30, 2015 at 22:35
• @user119615: Yes, exactly. Mar 30, 2015 at 23:47

Presumably in defining $F[x]$ you're assuming the indeterminate $x$ commutes with everything in $F$, so $(ax)(bx) = abx^2$. We have $\phi_\alpha(ax) = a\alpha$ and $\phi_\alpha(bx) = b\alpha$ but $\phi_\alpha(abx^2) = ab\alpha^2$. If $b$ and $\alpha$ don't commute, this may not be the same as $(a\alpha)(b\alpha)$.

• Thanks, but you mean $\alpha$ not 0?($\phi_{\alpha}$not $\phi_{0}$) Mar 30, 2015 at 22:39

Evaluating $\ a*x = x*a\$ at $\, x = e\in E\,$ shows that every coefficient $\,a\,$ must commute with every element of the target ring, i.e. the image of $F$ in $E$ is central. It turns out that this necessary condition is sufficient for the evaluation map to be a ring hom.

Remark $\:R\:$-algebras are the rings $\:A\:$ into which one can evaluate $R[x]$ for every $\:a \in A\:$. More directly, an $\:R$-algebra is just a ring $A\:$ containing a central subring $R'$ that's a ring image of $\: R\:,\:$ i.e. $\: R'\:$ is either an embedding of $\:R\:$ or $\:R/I\:$ for some ideal $\;I\in R\:.\;$ Being central is precisely the condition needed for elts in $\:R'\:$ to serve as "coefficients" in the sense that this makes polynomial rings $\: R[x]\:$ be universal $\:R\:$-algebras. Namely, the fact that the coefficients commute with all elts of $\:A\:$ is precisely what is required to make the evaluation map be a ring homomorphism $\: R[x]\to A\:,\:$ viz $\;\; r\; x = x\: r\;$ in $\:R[x]\;\Rightarrow\; r\: a = a\:r\;\;$ by evaluation $\: x\to a\in A,\:$ i.e. by definition polynomial multiplication assumes that the coefficients commute with the indeterminates, so this property must remain true at values of indeterminates if evaluation is to be a ring homomorphism.