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I’m reading Lang’s “Algebra” (3rd edition) and came across this result: Let $\mathfrak{a}_1,...,\mathfrak{a}_n$ be ideals of $A$ and assume that $\mathfrak{a}_i+\mathfrak{a}_j=A$ for $i \neq j$. Consider the map $$f:A \to \prod_{i=1}^n A/\mathfrak{a}_i=(A/\mathfrak{a}_1)\times ... \times (A/\mathfrak{a}_n).$$ The kernel of $f$ is $\ker(f)=\displaystyle \bigcap_{i=1}^n \mathfrak{a}_i$ and since $f$ is surjective then $A/\displaystyle\bigcap \mathfrak{a}_i \to \prod A/\mathfrak{a}_i$ is an isomorphism.

Is there any way to find the inverse of $A/\displaystyle \bigcap \mathfrak{a}_i \to \prod A/\mathfrak{a}_i$ explicitly? How can I define the multiplication on $\displaystyle \prod A/\mathfrak{a}_i$?

Pedro Martinez
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Yes, the proof of surjectivity provides an inverse. This is a classical case of something that should not be buried in the proof since it may be useful for calculations.

For simplicity, I will only explain the case $n = 2$. The general case is somewhat similar (but not in a straight forward way).

More generally, let $U,V$ be subgroups of an abelian* group $A$ with $U + V = A$. Then the canonical map $$\alpha : A/(U \cap V) \to A/U \times A/V, \quad [a] \mapsto ([a],[a])$$ is an isomorphism.

Namely, the map is clearly injective. For surjectivity, let $([a_1],[a_2]) \in A/U \times A/V$. By assumption, we can write $$a_1 = u_1 + v_1, \quad a_2 = u_2 + v_2$$ for some $u_i \in U$, $v_i \in V$. Then $a := v_1 + u_2$ is an element with $$a \equiv a_1 \bmod U, \quad a \equiv a_2 \bmod V,$$ which precisely means that $[a]$ is a preimage.

Therefore,

$$\varphi : A/U \times A/V \to A/(U \cap V),\quad ([u_1 + v_1],[u_2 + v_2]) \mapsto [v_1 + u_2]$$

is the desired inverse map. It is as constructive as it can get provided you have a constructive evidence for $U + V = A$.

Conversely, assume that $\alpha$ is surjective. Then in particular every element $([a_1],0)$ has a preimage, where $a_1 \in A$. This means there will be some $a \in A$ with $a \equiv a_1 \bmod U$ and $a \equiv 0 \bmod V$. Thus, $a \in V$ and $a-a_1 \in U$. This simplifies to $a_1 \in U + V$. This shows that $U + V = A$ is equivalent to surjectivity of $\alpha$.

*It also works for normal subgroups of a group which is not necessarily abelian.

Martin Brandenburg
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  • Thank you! How do you define the product of $[u_1+v_1]$ and $[u_2+v_2]$ in this case (to prove that $\varphi(xy)=\varphi(x)\varphi(y)$)? – Pedro Martinez Oct 28 '24 at 23:55
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    These are additive groups. And there is no need to define anything as if it was not defined already. Please look at the definition of quotient groups and quotient rings. $[x] + [y] := [x+y]$. Also, inverses of group (or ring) homomorphisms are always group (or ring) homomorphisms. You don't need to prove that here again in this special case. – Martin Brandenburg Oct 29 '24 at 00:09
  • Yes, but I need to show that directly (without the fact that the inverse of homomorphisms are homomorphisms). – Pedro Martinez Oct 29 '24 at 00:43
  • And in such case, what is the multiplicative unit? I'm so confused with that. – Pedro Martinez Oct 29 '24 at 01:42