Kernels are ideals
R,\ S : ring.
\phi : R \rightarrow S be ring homomorphism.
Let Ker(\phi) = \{r \in R \mid \phi(r) = 0 \}.
Since that \phi(a - b) = \phi(a) - \phi(b) = 0_{R} + 0_{R} = 0_{R}.
Hence, a - b \in Ker(\phi), \forall a,\ b \in Ker(\phi)
Also, by as follows,
\phi(rx) = \phi(r) \phi(x) = \phi(r) \cdot 0_{R} = 0_{R},\ \ \forall r \in R,\ \forall x \in Ker(\phi)
\phi(xr) = \phi(x) \phi(r) = 0_{R} \cdot \phi(r) = 0_{R},\ \ \forall r \in R,\ \forall x \in Ker(\phi)
rx,\ xr \in Ker(\phi),\ \ \forall r \in R,\ \forall x \in Ker(\phi) is trivial.
\therefore \text{By ideal test, } Ker(\phi)\text{ is ideal of }R
\phi : R \rightarrow S be ring homomorphism.
Let Ker(\phi) = \{r \in R \mid \phi(r) = 0 \}.
Since that \phi(a - b) = \phi(a) - \phi(b) = 0_{R} + 0_{R} = 0_{R}.
Hence, a - b \in Ker(\phi), \forall a,\ b \in Ker(\phi)
Also, by as follows,
\phi(rx) = \phi(r) \phi(x) = \phi(r) \cdot 0_{R} = 0_{R},\ \ \forall r \in R,\ \forall x \in Ker(\phi)
\phi(xr) = \phi(x) \phi(r) = 0_{R} \cdot \phi(r) = 0_{R},\ \ \forall r \in R,\ \forall x \in Ker(\phi)
rx,\ xr \in Ker(\phi),\ \ \forall r \in R,\ \forall x \in Ker(\phi) is trivial.
\therefore \text{By ideal test, } Ker(\phi)\text{ is ideal of }R
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