MHB Communitative ring, map R / ( I /\ J) -> R/I x R/J

[Fernando Revilla]

Commutative ring, map R / ( I /\ J) -> ( R/I ) x ( R/J )

I quote an unsolved question posted in MHF (November 25th, 2012) by user needhelp2.

Say that R is a commutative ring and the I and J are ideals. Show that
the map : R=(I intersection J) maps to R/I R/J given by (r + (I intersection J)) maps to (r + I; r + J) is
well defined and is an injection. Show further more that is a surjection if and
only if I + J = R.

P.S. Communicative note: Of course I meant in the title, commutative instead of communitative.

 

[Fernando Revilla]

I suppose you mean \phi:R\;/\;(I\cap J) \to (R\;/\;I)\times (R\;/\;J),\;\phi(r+I\cap J)=(r+I,r+J)

(a) \phi is well defined. Suppose r+I\cap J=r'+I\cap J, this implies r-r'\in I\cap J, that is r-r'\in I and r-r'\in J. As a consequence r+I=r'+I and r+J=r'+J or equivalently, (r+I,r+J)=(r'+I,r'+J): the image does not depend on the representants.

(b) \phi is injective. Suppose \phi(r_1+I\cap J)=\phi(r_2+I\cap J) then, (r_1+I,r_1+J)=(r_2+I,r_2+J), hence r_1-r_2\in I, r_1-r_2\in J or equivalently r_1-r_2\in I\cap J which implies r_1+I\cap J=r_2+I\cap J: \phi is injective.

(c) \phi is a surjection \Leftrightarrow\; R=I+J.

\Rightarrow) Let s\in R, as \phi is a surjection there exists r\in R such that \phi(r+I\cap J)=(0+I,s+J), that is r+I=0+I and r+J=s+J. This implies r\in I and s-r\in J, so s=r+j with r\in I and j\in J. As a consequence I+J\subset R\subset I+J, or equivalently R=I+J.

\Leftarrow) (Left as an exercise for the reader). :)

 

[Deveno]

suppose $R = I+J$. then for any $r \in R$ we have $r = x+y$. for some $x \in I, y \in J$.

let $(r + I,r'+J)$ be any element of $R/I \times R/J$.

writing $r = x + y, r' = x' + y'$ we have:

$r+I = (x+y)+I = (y+x)+I = y+I + x+I = y+ I + I = y + I$ and:

$r'+J = (x'+y')+J = x'+J + y'+J = x'+J + J = x' + J$

let $s = x'+y$. then

$\phi(s+(I\cap J)) = \phi((x'+y)+(I\cap J)) = ((x'+y)+I,(x'+y)+J)$

$= ((x'+I)+(y+I),(x'+J)+(y+J)) = (I+(y+I),(x'+J)+J)= (y+I,x'+J) = (r+I,r'+I)$

so $\phi$ is surjective.

oh snap! this is the chinese remainder theorem in disguise, isn't it?

 

[Fernando Revilla]

oh snap! this is the chinese remainder theorem in disguise, isn't it?

A very interesting question. :)

 

[Deveno]

If $R = \Bbb Z$ then the condition $I + J = R$ is equivalent to:

$(a) + (b) = (1)$ that is, a and b are co-prime: gcd(a,b) = 1 (using tacitly the fact that $\Bbb Z$ is a principal ideal domain, which follows from the fact that it is euclidean).

In this case, $I\cap J = (a) \cap (b) = (\text{lcm}(a,b)) = \left(\frac{ab}{\gcd(a,b)}\right) = (ab)$

We can thus conclude that if gcd(a,b) = 1:

$\Bbb Z/(ab) \cong \Bbb Z/(a) \times \Bbb Z/(b)$ a more familiar form of the CRT.