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#### Krizalid

##### Active member

- Feb 9, 2012

- 118

- Thread starter Krizalid
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- Feb 9, 2012

- 118

- Jun 26, 2012

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The second is the easiest: since $0\in J$ then $\phi (ker (\phi))=0\in J$. The third is not complicated: think of the homomorphism $\phi (x)=diag(x,x)$, where $diag(x,x)$ is the 2x2 diagonal matrix with x's on the diagonal. It can't be an ideal.

The first: Let $J^{-1}:=\phi^{-1}(J)$. If $a\in J^{-1}$ and $r\in R$. Then $\phi(ra)=\phi(r) \phi(a) \in J$, since J is an ideal. So $ra\in J^{-1}$.