Kernel of ring homomorphism is an ideal I am asked to show that if f is a ring homomorphism from R to R' then kernel of f is an ideal of R. 

According to definition of ideal : A non empty subset of R is an ideal for any two elements of ideal their substraction must be in that ideal and
the product of any element of R and an element of ideal must be in ideal,
I am not able to prove second condition. Please help
Let $f: R\rightarrow R'$ be a ring homomorphism. We assume that $R$ and $R'$ both have an identity. Since $0 \in\ker(f)=\{x\in R\mid f(x)=0\}, \ker(f)$ is non-empty.
Let $u,v \in \ker(f), r \in R$, then $f(ru)=f(r)f(u)=0,\ f(ur)=f(u)f(r)=0,\ f(u-v)=f(u)-f(v)=0$.
Let $\varphi: (R_1, +_1, \circ_1) \rightarrow (R_2, +_2, \circ_2)$ be a ring homomorphism.
Then the kernel of $\varphi$ is an ideal of $R_1$.
Proof:
By Kernel of Ring Homomorphism is Subring, $\ker(\varphi)$ is a subring of $R_1$.
Let $s \in \ker(\varphi)$, so $\varphi(s) = 0$ in $R_2$.
Suppose $x \in R_1$. Then:
\begin{align*}
\varphi(x \circ_1 s) &amp; = \varphi(x) \circ_2 \varphi(s) &amp; \text{Definition of Morphism Property} \\
&amp; = \varphi(x) \circ_2 0_{R_2} &amp; \text{as}\ s \in \ker(\varphi)  \\
&amp; = 0_{R_2} &amp; \text{Properties of}\ 0_{R_2} 
\end{align*}
and similarly for $\varphi(s \circ_1 x)$.
The result follows.