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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