# Ring homomorphism homework including ideals and surjectivity.

$R$ is a ring and $I$ and $J$ are ideals of $r$. Show that the ring homomorphism $h:R \rightarrow R/I \times R/J, r \mapsto (r+I,r+J)$ is surjective iff $I+J=R$

give a description of the kernel of $h$ in terms of the ideals $I$ and $J$

Have some ideas about this that its basically saying that the elements of R get sent to the cartesian product +I and +J hence if it is surjective then r+I and r+J must hit all points in the set R in which case I+J=. But I don't feel this is very rigorous or if it is infact true. I also feel like it may be due to one of the isomorphism theorems.

Not really sure how to go about the kernel part.

1. If $h$ is surjective, then $(0+I,1+J)=h(r)$.
2. If $I+J=R$, can you write $(x+I,y+J)=(r+I,r+J)$, for some $r$?
Assume $I+J=R$. Let arbitrary $(a+I,b+J)\in R/I\times R/J$ be given. As $I+J=R$, there exist $i\in I$, $j\in J$ with $a-b=i+j$. Let $r=a-i=b+j$. Then $h(r)=(r+I,r+J)=(a+I,b+J)$. We conclude that $h$ is surjective.
Now assume that $h$ is surjective. let $a\in R$ be given. We want to find $i\in I$, $j\in J$ with $a=i+j$. By assumption, $h(r)=(a+I,0+J)$ for some $r$. This implies $r+J=J$, i.e. $r\in J$, and $a+I=r+I$, i.e. $a-r\in I$. Let $j=r$ and $i=a-r$.