Principal Ideals and Bezout Domains - (a,b)

[Math Amateur]

Just a further (very basic!) question:

Is the following argument - working from definitions - correct

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

---------------------------------------------------------------------------------

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

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

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

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

$$(a) + (b) = \{r_1a, r_2b \ | \ r_1, r_2 \in R \}$$

so we are missing the 'addition' terms $$r_1a + r_2b$$ 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

$$X + Y = \{x + y \ | \ x \in X, y \in Y \}$$

Working, then, with this definition we have

$$(a) + (b) = \{ r_1a + r_2b \ | \ r_1, r_2 \in R \}$$ 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 $$A = \{ a_1, a_2, ... ... a_n \}$$ we have that

$$(a_1, a_2, ... ... a_n) = (a_1) + (a_2) + ... ... (a_n)$$

Is this correct?


Just another vaguely connected question.

Given a ring R consisting of the elements $$\{a_1, a_2, ... ... a_n \}$$

do there always (necessarily?) exist ideals $$A_1 = (a_1) , A_2 = (a_2), ... ... A_n = (a_n)$$

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]

 

[TheBigBadBen]

A lot to look at here, I'll try to break it down and hope I don't miss anything.

By definition (Dummit and Foote page 251) (a,b)={r1a+r2b | r1,r2∈R}

[Note (a, B) includes the terms r1a and r2b since r1 or r2 can equal 0.]

Also by definition we have (a)={r1a | r1∈R} and (b)={r2b | r2∈R}

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

(a)+(b)={r1a,r2b | r1,r2∈R}

so we are missing the 'addition' terms r1a+r2b 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

X+Y={x+y | x∈X,y∈Y}

Working, then, with this definition we have

(a)+(b)={r1a+r2b | r1,r2∈R} and this is the same as the definition of (a, b) so (a) + (b) = (a, b) from the definitions.

I have no idea what you mean by this bit:

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

(a)+(b)={r1a,r2b | r1,r2∈R}

so we are missing the 'addition' terms r1a+r2b of (a, B).

But outside of that, your reasoning seems correct here. I guess that means that your second definition of set addition is the correct one. I assume that our discussion is limited then to commutative rings; otherwise, you would have to extend the definition of the ideal to include right-multiplication.

You can proceed to extend that definition (for finitely many elements) by induction, using the fact that the sum of any two ideals is an ideal.

Any element in a ring can be used to generate an ideal, but that ideal need not be proper. In fact, I'm fairly sure that the ideal generated by a will be R itself (not proper) iff a has a multiplicative inverse in R.