Prove that the group $(A,+, ◦) $ is a non-commutative ring

Question

• $A × A → A, (f, g) → f + g$, where $(f + g)(x) = f(x) + g(x)$ for all $x ∈ K$

• $A × A → A, (f, g) → f ◦ g$ where $(f ◦ g)(x) = f(g(x))$ for all $x ∈ K$

Show that $(A,+,◦)$ is a non commutative ring. So they make $f(x) = x$ the neutral element. Why?

Also i introduced $f, g, h ∈ A$ and $x ∈ K$ so $((f + g) ◦ h)(x) = (f + g)(h(x)) = f(h(x)) + g(h(x)) = (f ◦ h)(x) + (g ◦ h)(x) = (f ◦ h + g ◦ h)(x)$. So $(f + g) ◦ h = f ◦ h + g ◦ h$

And $(f ◦ (g + h))(x) = f(g(x) + h(x)) = f(g(x)) + f(h(x)) = (f ◦ g)(x) + (f ◦ h)(x)$ So $f ◦ (g + h) = f ◦ g + f ◦ h$

This shows that its non commutative? Because this is apparently associative? I don't get how this is associative. Nor commutative. The two compositions are not equal to each other in a different order. Help!

Answer 1

The neutral element (or, as I would call it, the multiplicative identity) is an element $1_A \in A$ such that for any $f \in A$, we have $f \circ 1_A = 1_A \circ f = f$. Verify that this is true for $1_A(x) = x$.

It seems that you've correctly shown that $A$ satisfies the distributivity axiom for both right and left multiplication.

In order to prove that $A$ is associative, you need to show (or otherwise establish) that $f \circ (g \circ h) = (f \circ g) \circ h$.

In order to prove that $A$ is not commutative, you need to find an counter-example to commutativity. Namely, you need a function $f$ and a function $g$, both in $A$, satisfying $f \circ g \neq g \circ f$.

In particular, we can take $f(x) = x + 1_K$, and $g(x) = x^n$. For any field $K$, there exists some $n$ so that $$ (x+1)^n \neq x^n + 1 $$

Answer 2

For any abelian group $K$ the set $A$ of endo-functions on $K$ forms a ring via the definitions given in the question. As long as $K$ has at least two distinct elements $a$ and $b$, the constant functions $x \mapsto a$ and $x \mapsto b$ do not commute with respect to function composition (the multiplication on $A$), so $A$ is a commutative ring only when $K$ is the trivial group.