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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!

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

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

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  • $\begingroup$ 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? $\endgroup$ – Robert Lewis Mar 29 at 1:52
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    $\begingroup$ @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. $\endgroup$ – Chris Custer Mar 29 at 2:04
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    $\begingroup$ @RobertLewis The first equality comes from the assumption that it's a homomorphism, and the second comes from the definition of involution. $\endgroup$ – WhatsUp Mar 29 at 2:09
  • $\begingroup$ OK I see that now. $\endgroup$ – Robert Lewis Mar 29 at 2:09
  • $\begingroup$ Why does the ring have to be commutative? $\endgroup$ – user551155 Mar 29 at 2:14
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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.

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