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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.
 
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Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I suppose you mean [tex]\phi:R\;/\;(I\cap J) \to (R\;/\;I)\times (R\;/\;J),\;\phi(r+I\cap J)=(r+I,r+J)[/tex]

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

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

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

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

[tex]\Leftarrow)[/tex] (Left as an exercise for the reader). :)
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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?
 
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Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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.