Interesting problem with ideals and function

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Discussion Overview

The discussion revolves around a problem involving ideals in the context of ring theory. Participants explore the properties of a specific function defined from a ring \( A \) to the Cartesian product \( A \times \mathbb{Z} \), and whether the image of an ideal under this function remains an ideal in the product ring. The scope includes theoretical aspects of ring homomorphisms and ideal properties.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that the function \( f \) is an injective ring homomorphism, arguing that if \( f(a) = f(b) \), then \( a = b \).
  • Another participant confirms that the image of \( f \) is isomorphic to \( A \times \{0\} \), suggesting that the second coordinate does not contribute to the structure of \( f(A) \).
  • Participants check the axioms of an ideal, asserting that the subgroup property and absorption properties hold for \( J \times \{0\} \) in \( A \times \mathbb{Z} \).
  • One participant questions the interest level of the problem, implying that the properties being discussed may be trivial.
  • Corrections and clarifications are made regarding the problem statement, with participants reiterating the triviality of the subgroup property and absorption properties.

Areas of Agreement / Disagreement

While some participants agree on the triviality of certain properties being discussed, there is no consensus on the overall interest or significance of the problem itself. The discussion remains somewhat unresolved regarding the perceived complexity or novelty of the problem.

Contextual Notes

Participants have noted that the subgroup property and absorption properties are straightforward, but there may be underlying assumptions or definitions that are not fully articulated. The discussion does not resolve whether the problem is interesting or significant.

Krizalid1
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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 $A\times Z.$
 
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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).
 
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?
 
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I've corrected the problem and added things I should've posted before.
Sorry for the inconvenients.
 
Krizalid said:
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.$\qquad \blacksquare$
 

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