Showing that the set define as follows is subring of its subset that a ring.

Question

Let $R$ and $S$ be a ring and two ring homomorphism $f:R \to S$ and $g:R \to S$. Show that $H = \{x \in R | f(x) = g(x)\}$ is subring of R.

My attempt:

Let $x,y \in H$. Then, $f(x)=g(x)$ and $f(y)=g(y)$. Now,

$f(x-y) = f(x)-f(y) = g(x)-g(y) \in H$

and

$f(xy) = f(x)f(y) = g(x)g(y) \in H$

Hence, $H$ is subring of R.

Am I right?

Answer 1

$H$ is non empty as $1_R\in R$ , $f(1_R)=1_S=g(1_R).$ Let $x,y \in H$. $f(x)=g(x),f(y)=g(y)\in H$, Now, $f(x)-f(y) = f(x-y)= g(x-y)\in H$ Thus closed under subtraction. and

$f(x)f(y) = f(xy) = g(xy) \in H$ ,closed under multiplication. Hence, $H$ is subring of R.

Answer 2

As $f$ and $g$ are ring homomorphisms, $f(0_R) = 0_S = f(0_R)$, so, $0_R$ is an element in $R$ for which $f$ and $g$ has the same value, in other words, $0_R \in H$.

Now, take $x$ and $y$ in $H$, that is, $x$ and $y$ are elements of $R$ such that $f(x) = g(x)$ and $f(y) = g(y)$. Now, $x-y$ is an element of $R$ such that $$f(x-y) = f(x)-f(y) = g(x)-g(y) = g(x-y),$$ that is, $x-y \in H$. Similarly we can show that $xy \in H$.