0

Let $A$ be a commutative ring. $ \phi :A \to A/I\times A/J$ be a map given by $x \to (x+ I, x+J)$.
(i) Show that $\phi$ is 1-1 iff $I\bigcap J=0$.
(ii) If $\phi$ is surjective then prove that $I+J=A$.

(i) I have done the part that if $I\bigcap J=0$ then $\phi$ is 1-1. Assuming the fact that $\phi $ is 1-1 i have proved that 0 belongs to $I\bigcap J$. But I am unable to prove that this is the only element in $ I \bigcap J$.

(ii) I need help in this part. I considered I+J but I am not sure how should I approach it.

Please help

user26857
  • 52,094

1 Answers1

1

I think by $A\times I$ you mean the residue ring $A/I$. In this case, you should notice that the sequence $$0\rightarrow I\cap J\xrightarrow{f} A\xrightarrow{\phi} A/I\times A/J\xrightarrow{h} A/I+J\rightarrow 0,$$ where $f(x)=x, \phi(x)=(x+I,x+J)$ and $h(x+I,y+J)=x-y+(I+J)$ is exact, that is, $f$ is injective, $im(f)=\ker(\phi), im(\phi)=\ker(h)$ and $h$ is surjective.

Indeed, it is trivial to see that $f$ is injective, $h$ is surjective, $im(f)\subseteq\ker(\phi)$ and $im(\phi)\subseteq\ker(h)$. On the other hand, given $x\in\ker(\phi)$, we have $(x+I,x+J)=0$, i.e., $x\in I$ and $x\in J$. In other words, $x\in I\cap J=im(f)$. Finally, if $(x+I,y+J)\in\ker(h)$, then $x-y\in I+J$. Let $i\in I$ and $j\in J$ be two elements such that $x-y=i+j$ and take $z=x-i=y+j$. Hence $\phi(z)=(z+I,z+J)=(x-i+I,y+j+J)=(x+I,y+J)$.

From the exactness of the sequence above we conclude that $\phi$ is injective if and only if $I\cap J=im(f)=0$ and that $\phi$ is surjective if and only if $A/I+J=im(h)=(A/I\times A/J)/\ker(h)=(A/I\times A/J)/im(\phi)=0$, that is, $A=I+J$.

Rafael
  • 3,789
  • Honda isn't there a way without using the exactness? The intructor hasn't taught exactness till this point? I think it will make the problem hard because first I will have to prove exactness then stated result. –  Oct 15 '21 at 11:45
  • @user319735 you can do all the math without talking about exact sequences. At the end of the day, what you're doing is proving the exactness of the sequence above. – Rafael Oct 15 '21 at 21:06
  • I am not able to prove $ker (h)=im(\phi)$. image $\phi$ =(x+I, x+J) and ker(h)=x-y +I+J =0 . which implies that x-y $\in $ I+J . But how are they equal? –  Jan 11 '22 at 19:11
  • $h\phi(x)=h(x+I,x+J)=(x-x)+(I+J)=0$, i.e., $im(\phi)\subseteq\ker(h)$. On the other hand, if $h(x+I,y+J)=0$, then $x-y\in I+J$. Let $i\in I, j\in J$ such that $x-y=i+j$ and take $z=x-i=y+j$. Thus $\phi(z)=(z+I,z+J)=(x-i+I,y+j+J)=(x+I,y+J)$. Therefore $\ker(h)\subseteq im(\phi)$. – Rafael Jan 11 '22 at 23:11