MHB Interesting problem with ideals and function

[Krizalid1]

Hi guys, it's been a while! Here's an interesting problem.

Let $A$ be a ring and $Z$ be the ring of $\mathbb Z.$ Consider the cartesian product $A\times Z.$
Define $A\times Z$ the product $(a,n)\cdot(b,m)=(ma+nb+ab,nm).$
Let $f:A\longmapsto A\times Z$ be defined by $f(a)=(a,0)$ for all $a$ in the ring $A.$ Prove that if $J$ is an ideal of $A,$ its image below the function $f$ is an ideal of the ring $A\times Z.$

 

[Deveno]

f is clearly an injective ring homomorphism since if f(a) = f(b), (a,0) = (b,0), and by the definition of equality in the cartesian set product, a = b. It is trivial to check that the image of f is A x {0}, which is ring-isomorphic to A.

In other words, the 0 in the 2nd coordinate in Z x A is "just along for the ride", it doesn't contribute any structure to f(A):

(a,0) + (b,0) = (a+b,0+0) = (a+b,0)

(a,0)*(b,0) = (ab,0*0) = (ab,0).

 

[I like Serena]

Checking the axioms of an ideal:

1. $(J \times \{0\}, +)$ is a sub group of $(A \times \mathbb Z, +)$.
2. $\forall (j,0) \in J \times \{0\}, \forall (r,z) \in A \times \mathbb Z: (j,0) \cdot(r,z) \in J \times \{0\}$.
3. $\forall (j,0) \in J \times \{0\}, \forall (r,z) \in A \times \mathbb Z: (r,z) \cdot(j,0) \in J \times \{0\}$.

These are all trivially true.

Erm... how is it interesting?

 

[Krizalid1]

I've corrected the problem and added things I should've posted before.
Sorry for the inconvenients.

 

[I like Serena]

I've corrected the problem and added things I should've posted before.
Sorry for the inconvenients.

Okay... the subgroup property with respect to + is still trivial.
For right absorption we get:

$$\forall (j,0) \in J \times \{0\}, \forall (a,z) \in J \times \mathbb Z: \\
\qquad (j,0) \cdot (a,z) = (zj+0a+ja, 0z) = (j',0) \in J \times \{0\}$$
This is true, because:

• $zj$ is a summation of elements in $J$, which is also an element of $J$,
• $0a=0$,
• $ja \in J$ because $J$ is an ideal of $A$.

Left absorption is similar.$\qquad \blacksquare$