# Set of Fixed Elements of a Ring

Let $$R$$ be a commutative ring with 1, and let $$h: R \to R$$ be a ring automorphism of $$R$$. Set $$S:= \{x\in R: h(x) = x\}$$. That is, $$S$$ is the set of all elements of $$R$$ fixed by the automorphism $$H$$.

(a) Show that $$S$$ is a subring of $$R$$.

(b) Show that $$1 \in S$$.

(c) Now suppose that $$h^2$$ is the identity map and that $$r \in R$$. Prove that there is a monic polynomial $$p \in S\lbrack x \rbrack$$ of degree 2 with coefficients in $$S$$ such that $$p(r) = 0$$.

I have worked out a solution for (a) and (b), but I am unsure where to go for (c).

(a) Since $$h$$ is a ring homomorphism, $$h(0) = 0$$, so $$0 \in S$$ and $$S \neq \varnothing$$. Now, let $$a,b \in S$$. Now, \begin{align*} h(a-b) = h(a)-h(b) = a-b, \end{align*} so $$a-b \in S$$. Finally, let $$x,y \in S$$. Now, \begin{align*} h(xy) = h(x)h(y) = xy, \end{align*} so $$xy\in S$$, thus $$S$$ is a subring of $$R$$.

(b) Let $$a \in S$$, then, \begin{align*} h(a)h(1) = h(a\cdot1) = h(a), \end{align*} so $$h(1)$$ is the same as $$1 \in R$$. In other words, $$h(1) = 1$$, so $$1 \in S$$.

I would appreciate any feedback on these two parts. Part (c) has stumped me for some time, and I'm not quite sure were to go. I am not quite seeing how it is connected to the previous part either.

(b) is weird, since it follows from resp. is included in (a). For a subring, you also need to have $$1 \in S$$ by definition. Otherwise, you would need to call it a subrng, for example.

Your proof of $$1 \in S$$ is not correct. You get $$1 \in S$$ because, by definition, a ring/group/lattice/whatever homomorphism preserves the whole structure, in particular $$h(1)=1$$ for any ring homomorphism.

About the "$$S \neq \emptyset$$" in your proof, please have a look at my comment here. It also explains in more detail why (b) is included in (a), among other things.

For (c), notice that $$r + h(r) \in S$$ and $$r \cdot h(r) \in S$$. Hence, the polynomial $$p := X^2 - (r + h(r)) \cdot X + (r \cdot h(r))$$ is contained in $$S[X]$$. Now prove $$p(r)=0$$.

The motivation behind this polynomial comes from Galois theory, in case you wonder where it comes from. But of course Galois theory is not necessary to verify that it has the desired properties.

• People arguing that the notion of "ring" is ambiguous since some authors prefer to have them non-unital and thereby perpetuating this unnecessary confusion and making literally every text on ring theory harder to read: please don't comment under this post and just read math.mit.edu/~poonen/papers/ring.pdf thanks. Dec 28, 2023 at 3:33
• That is weird. That was my first instinct for (b) as well, but this problem came from a past qualifying exam that I am studying with. I assumed that it couldn't be so easy. I wanted to use that argument to make (a) easier too... In regards to (c), I'm assuming that your previous hint is a useful way to turn use any arbitrary $r \in R$ to construct an element in $S$ which is probably how we would want to construct the coefficients in $p$. Wouldn't something like $p(x) = (x-h(x))(x+h(x))$ work? it is equal to $(x^2 - h^2(x))$ which is 0 for all $r \in R$, but maybe i'm misunderstanding. Dec 28, 2023 at 6:15
• perhaps this isn't quite correct since $h(x)$ is not necessarily in $S$. Dec 28, 2023 at 7:26
• This is not working at all. It's not even a polynomial and not using r. Please use the hint that I gave you. Dec 28, 2023 at 12:13
• sorry I am not making progress. could you give another hint? Dec 28, 2023 at 20:26