Let R be a ring φ : R → R be a ring homomorphism. Let S := {x ∈ R | φ(x) = x}.Prove that S is a subring of R.

To prove that $S$ is a subring of $R$, we need to show that $S$ is closed under addition, multiplication, and contains the additive identity and additive inverses.

1. Additive Identity:

   • We need to show that $0 \in S$.
   • Since $\phi$ is a ring homomorphism, $\phi(0) = 0$.
   • Therefore, $0 \in S$.

2. Additive Inverses:

   • Let $x \in S$. We need to show that $-x \in S$.
   • Since $x \in S$, $\phi(x) = x$.
   • Because $\phi$ is a ring homomorphism, $\phi(-x) = -\phi(x)$.
   • Thus, $\phi(-x) = -x$, so $-x \in S$.

3. Closed under Addition:

   • Let $x, y \in S$. We need to show that $x + y \in S$.
   • Since $x, y \in S$, $\phi(x) = x$ and $\phi(y) = y$.
   • Because $\phi$ is a ring homomorphism, $\phi(x + y) = \phi(x) + \phi(y)$.
   • Thus, $\phi(x + y) = x + y$, so $x + y \in S$.

4. Closed under Multiplication:

   • Let $x, y \in S$. We need to show that $x \cdot y \in S$.
   • Since $x, y \in S$, $\phi(x) = x$ and $\phi(y) = y$.
   • Because $\phi$ is a ring homomorphism, $\phi(x \cdot y) = \phi(x) \cdot \phi(y)$.
   • Thus, $\phi(x \cdot y) = x \cdot y$, so $x \cdot y \in S$.

Since $S$ contains the additive identity, is closed under addition and multiplication, and contains additive inverses, $S$ is a subring of $R$.