# Determine under what circumstances an involution on R is an automorphism

Let R be a ring. An involution on R is a function α : R → R such that, for all $$r_i ∈ R$$, we have $$α(r_1 + r_2) = α(r_1) + α(r_2), α(r_1r_2) = α(r_2)α(r_1) \,and \,α(α(r_1)) = r_1$$.

I'm not real sure how to proceed here. I know an involution is a function that is its own inverse and I know an automorphism is an isomorphism from a set to itself. So I need to show $$\alpha$$ is one-to-one and onto. Possibly something similar to this:

take $$r_1,r_2 \in R$$ and $$r_1=r_2$$, then $$f(r_1)=f(r_2) \Rightarrow f(f(r_1))=f(f(r_2)) \Rightarrow r_1=r_2 \Rightarrow R$$ is one-to-one.

Also, $$\alpha(\alpha(r_1))=r_1$$ so f is onto. Any help would be greatly appreciated!

• It looks like it's always an automorphism, if it takes $1$ to $1$. Just a tip, when showing one-one, don't assume $r_1=r_2$. – Chris Custer Mar 29 at 1:13
• Why is it always an automorphism? – user551155 Mar 29 at 1:19
• You showed it was a bijective homomorphism. – Chris Custer Mar 29 at 1:25
• OK, I think I'm with ya. Thanks! – user551155 Mar 29 at 1:31
• Guess I missed the commutative part. My ring theory could be better. – Chris Custer Mar 29 at 1:55

Let $$R$$ be a unital ring; we do not assume $$R$$ is commutative.

A an automorphism of $$R$$ is an isomorphism 'twixt $$R$$ and itself; an involution, as stated in the text of the question is a function

$$\alpha:R \to R \tag 1$$

such that

$$\forall x, y \in R, \; \alpha(x + y) = \alpha(x) + \alpha(y), \tag 2$$

and

$$\forall x, y \in R, \; \alpha(xy) = \alpha(y) \alpha(x), \tag 3$$

and

$$\alpha^2 = \Bbb I_B, \tag 4$$

where $$\Bbb I_R$$ is the identity map on $$R$$:

$$\Bbb I_R(r) = r, \; \forall r \in R. \tag 5$$

We observe that involutions are surjective, since

$$\forall x \in R, \; x = \Bbb I_R x = \alpha^2(x) = \alpha(\alpha(x)); \tag 6$$

thus, $$x$$ is always the image of $$\alpha(x)$$ under $$\alpha$$, making $$\alpha$$ an onto map.

Such $$\alpha$$ are also injective: if

$$\alpha(x) = \alpha(y), \tag 7$$

then

$$x = \Bbb I_R(x) = \alpha^2(x) = \alpha(\alpha(x))$$ $$= \alpha(\alpha(y)) = \alpha^2(y) = \Bbb I_R(y) = y. \tag 8$$

Now suppose $$R$$ is commutative; then

$$\forall x, y \in R \; \alpha(xy) = \alpha(y) \alpha(x) = \alpha(x)\alpha(y), \tag 9$$

which shows that $$\alpha$$ meets the definition of a homomorphism; if, on the other hand, we replace he resrtiction that $$R$$ be commutative with the assumption that

$$\forall x, y \in R, \; \alpha(xy) = \alpha(x) \alpha(y); \tag{10}$$

since we are given that

$$\forall x, y \in R, \; \alpha(xy) = \alpha(y)\alpha(x), \tag{11}$$

we infer that

$$\forall x, y \in R, \; \alpha(x) \alpha(y) = \alpha(y)\alpha(x); \tag{12}$$

now since from the above $$\alpha$$ is surjective,

$$\forall r, s \in R, \; \exists x, y \in R, \; r = \alpha(x), s = \alpha(y); \tag{13}$$

thus,

$$rs = \alpha(x) \alpha(y) = \alpha(y)\alpha(x) = sr, \tag{14}$$

and $$R$$ must be a commutative ring.

• So the answer to the question is when R is a commutative ring? – user551155 Mar 29 at 16:30
• @user551155: yes, but $R$ may have other related properties as well; still thinking about that, Cheers! – Robert Lewis Mar 29 at 16:58
• Doesn't $\alpha(r_1r_2)=\alpha(r_2)\alpha(r_1)$ imply commutivity? – user551155 Mar 29 at 19:57

An automorphism of a ring is a homomorphism to itself which is bijective.

Since your $$\alpha$$ is already bijective, it is an automorphism if and only if it is a homomorphism.

Let me write down the conditions for being a homomorphism.

$$α(r_1 + r_2) = α(r_1) + α(r_2), α(r_1r_2) = α(r_1)α(r_2).$$ (Here I don't assume that the ring has a unit element, but it doesn't affect the argument.)

Do you notice the difference from the conditions of being involution?

The only extra condition is $$α(r_1r_2) = α(r_1)α(r_2)$$. Thus if $$\alpha$$ is an involution and also an automorphism, then we must have $$α(r_1)α(r_2)=α(r_1r_2) = α(r_2)α(r_1)$$.

This being true for all $$r_1,r_2$$, we may substitute $$r_1=α(x)$$ and $$r_2=α(y)$$ for arbitrary $$x,y$$, and get $$xy=yx$$.

Therefore we have shown that, if there is an involution which is an automorphism, then the ring must be commutative.

Conversely, if the ring is commutative, then for any involution $$\alpha$$, we have $$α(r_1r_2) = α(r_2)α(r_1)=α(r_1)α(r_2)$$, which implies that it's an automorphism.

• I don't see why we must have $α(r_1)α(r_2)=α(r_1r_2) = α(r_2)α(r_1)$; can you prove/explain/say more about this? – Robert Lewis Mar 29 at 1:52
• @Robert I looked it up and, apparently in ring theory, involutions are generally taken to mean antihomomorphisms. Take a second look at the OP's definition. – Chris Custer Mar 29 at 2:04
• @RobertLewis The first equality comes from the assumption that it's a homomorphism, and the second comes from the definition of involution. – WhatsUp Mar 29 at 2:09
• OK I see that now. – Robert Lewis Mar 29 at 2:09
• Why does the ring have to be commutative? – user551155 Mar 29 at 2:14

Since an involution is an antihomomorphism, and an automorphism is a homomorphism, we need both. That's equivalent to $$\mathcal R$$ being commutative.

You already showed bijectiveness.