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- Jun 22, 2012

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Is the following argument - working from definitions - correct

Does (a) + (b) = (a,b)?

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By definition (Dummit and Foote page 251) [TEX] (a, b) = \{r_1a + r_2b \ | \ r_1, r_2 \in R \} [/TEX]

[Note (a, B) includes the terms [TEX] r_1a [/TEX] and [TEX] r_2b [/TEX] since [TEX] r_1 [/TEX] or [TEX] r_2 [/TEX] can equal 0.]

Also by definition we have [TEX] (a) = \{r_1a \ | \ r_1 \in R \} [/TEX] and [TEX] (b) = \{r_2b \ | \ r_2 \in R \} [/TEX]

Now if by '+' we mean the "addition" (union or putting together) of sets then we have

[TEX] (a) + (b) = \{r_1a, r_2b \ | \ r_1, r_2 \in R \} [/TEX]

so we are missing the 'addition' terms [TEX] r_1a + r_2b [/TEX] of (a, B).

But if we take (as we probably should) the '+' to mean the sum of ideals - then the definition is (Dummit and Foote page 247) for ideals X and Y in R

[TEX] X + Y = \{x + y \ | \ x \in X, y \in Y \} [/TEX]

Working, then, with this definition we have

[TEX](a) + (b) = \{ r_1a + r_2b \ | \ r_1, r_2 \in R \} [/TEX] and this is the same as the definition of (a, b) so (a) + (b) = (a, b) from the definitions.

**Is this correct?**If the above is correct then seemingly for an ideal generated by the set [TEX] A = \{ a_1, a_2, ... ... a_n \} [/TEX] we have that

[TEX] (a_1, a_2, ... ... a_n) = (a_1) + (a_2) + ... ... (a_n) [/TEX]

Is this correct?

Just another vaguely connected question.

Given a ring R consisting of the elements [TEX] \{a_1, a_2, ... ... a_n \} [/TEX]

do there always (necessarily?) exist ideals [TEX] A_1 = (a_1) , A_2 = (a_2), ... ... A_n = (a_n) [/TEX]

Help with confirming (or otherwise) my reasoning & clarifying the above issues would be much appreciated.

**Note that if you agree with my reasoning above then a brief confirmation would be very helpful.**

Peter

[Also posted on MHF]