How to find the generating element for Ideals

[xenonforlife]

I need to find the generating element a such that Ideal I in Z can be represented as I = aZ.

1) 2Z + 3Z
2) 2Z ∩ 3Z

Not getting a clue how to proceed.

 

[daon2]

Write down what they are as sets first. Then it might become clearer.

 

[xenonforlife]

I did...
2Z = {0,1}
3Z = {0,1,2}

2Z+3Z should be a set of ordered pairs {(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)}

and the set 2Z ∩ 3Z should be {0,1}

now 2Z+3Z can be constructed with the elements {(1,0),(0,1)} so does that become the generator?? This is where I am getting confused because if that is the case it wont matter whether its 2Z+3Z or say nZ+mZ, it will always be {(1,0),(0,1)}??

similarly 2Z ∩ 3Z can be constructed from the element {1} alone but that is the same case for the whole of Z, this is what is creating doubts in my mind.

 

[daon2]

2Z is the set of even integers, 3Z is the set {0, 3, 6, 9, ..., }.

You are confusing \(\displaystyle 2\mathbb{Z} = \{2k;\,\, k\in\mathbb{Z}\}\) with \(\displaystyle \mathbb{Z}/2\mathbb{Z} = \{k+2\mathbb{Z}; \,\, k\in \mathbb{Z}\}\). The second is a quotient group. In that quotient group, even integrs are equivalent to 0, odd integers are equivalent to 1.

The ideal formed by the sum of two ideals \(\displaystyle I+J = \{i + j;\,\, i\in I, j\in J\}\). In particular:

\(\displaystyle 2\mathbb{Z} + 3\mathbb{Z} =\{2n+3m;\,\, m,n \in \mathbb{Z}\}\).

Hint:

The generating element of \(\displaystyle \,\, r\mathbb{Z} + s\mathbb{Z} \,\,\) depends on \(\displaystyle r,s\).

 

[xenonforlife]

How silly of me to get confused between those two...anyhow isnt it true that we will get all the elements of Z in the ideal 2Z+3Z, so in this case the generator element should indeed be {1} or {-1} but this element is not present in any of the sets 2Z or 3Z


and in the case of the intersection of the two ideals the generator element can be {6} or {-6}. Am I getting it a little bit or am still wandering?

 

[xenonforlife]

But I guess if the question would have been 4Z+6Z then the generating element has to be {2} or {-2}...that is the generating element in this case will be the r-s (difference of r and s) for the ideal rZ+sZ and in the case of the intersection of the two ideals it has to be the LCM of r and s, am I right?

 

[daon2]

But I guess if the question would have been 4Z+6Z then the generating element has to be {2} or {-2}...that is the generating element in this case will be the r-s (difference of r and s) for the ideal rZ+sZ and in the case of the intersection of the two ideals it has to be the LCM of r and s, am I right?

The second one is correct. In the first one it is the set of all linear combinations of the integers r and s. What do you know about every linear combination of two integers? Your 4Z+6Z answer is right, it is <2>, but the reasoning is off.

 

[xenonforlife]

The second one is correct. In the first one it is the set of all linear combinations of the integers r and s. What do you know about every linear combination of two integers? Your 4Z+6Z answer is right, it is <2>, but the reasoning is off.

In the first case the linear combination of 2 and 3 will give me the whole Z space...in general I guess it will give me the space (r-s)Z...what am I missing?

 

[daon2]

In the first case the linear combination of 2 and 3 will give me the whole Z space...in general I guess it will give me the space (r-s)Z...what am I missing?

Does 2Z + 5Z give you 3Z? Consider 2(-2) + 5(1). Is that a multiple of 3?

edit: Yes, for your particular question, you would get all of Z, but if it had asked, say 3Z+9Z you would have not been correct. In fact of a|b then aZ + bZ = aZ, since bZ is a subset of aZ. Focus on the fact that they are linear combinations. See lemma 1.5 of your text.

 

[xenonforlife]

Does 2Z + 5Z give you 3Z? Consider 2(-2) + 5(1). Is that a multiple of 3?

edit: Yes, for your particular question, you would get all of Z, but if it had asked, say 3Z+9Z you would have not been correct. In fact of a|b then aZ + bZ = aZ, since bZ is a subset of aZ. Focus on the fact that they are linear combinations. See lemma 1.5 of your text.

Which lemma are you talking about? yes but I get your point...I was wrong in saying that it will be (r-s)Z but I still dont get how I conclude a relationship b/w r and s knowing that yes the resultant ideal is a linear combination of all the elements in rZ and sZ...

 

[daon2]

Which lemma are you talking about? yes but I get your point...I was wrong in saying that it will be (r-s)Z but I still dont get how I conclude a relationship b/w r and s knowing that yes the resultant ideal is a linear combination of all the elements in rZ and sZ...

edit: nevermind, I was confused with another poster!

There exist x,y in Z such that (a,b) = ax+by

 

[xenonforlife]

edit: nevermind, I was confused with another poster!

There exist x,y in Z such that (a,b) = ax+by

Still not getting it I guess either group theory is not my cup of tea or I'm getting dumber and dumber...

 

[daon2]

Still not getting it I guess either group theory is not my cup of tea or I'm getting dumber and dumber...

You aren't remembering something you probably saw in a previous class.

d = (a,b), the greatest common divisor of a and b, divides every linear combination. It is an important result, don't forget it. It is not hard to show now that dZ = aZ + bZ...

 

[xenonforlife]

You aren't remembering something you probably saw in a previous class.

d = (a,b), the greatest common divisor of a and b, divides every linear combination. It is an important result, don't forget it. It is not hard to show now that dZ = aZ + bZ...

Ohkkkk I never knew of such a property with respect to linear combinations, thank you for the note, I will definitely remember it from now onwards.

 

[daon2]

Ohkkkk I never knew of such a property with respect to linear combinations, thank you for the note, I will definitely remember it from now onwards.

Alright, but what you should also know that the GCD is the smallest positive linear combination. It is not usually defined that way, but it is true (Maybe the first chapter in your book has the proof. I know Gallian and Dummit and Foote do.). The special case of being "Relatively prime" means that the GCD is 1, and so there is a linear combination which adds to 1. If an ideal contains 1, it is equal to Z (or the whole ring).

 

[xenonforlife]

Alright, but what you should also know that the GCD is the smallest positive linear combination. It is not usually defined that way, but it is true (Maybe the first chapter in your book has the proof. I know Gallian and Dummit and Foote do.). The special case of being "Relatively prime" means that the GCD is 1, and so there is a linear combination which adds to 1. If an ideal contains 1, it is equal to Z (or the whole ring).

Thats very helpful indeed, i definitely need to know such useful concepts, could you please suggest a good reference book for this topic Groups, Rings and Fields. I do not need to study this in depth as this is a single chapter on my Advanced Engineering Maths course but atleast want to grab and absorb the basic concepts of the same, so do you suggest I should try to study it from Gallian or Dummit and Foote??

 

[daon2]

Thats very helpful indeed, i definitely need to know such useful concepts, could you please suggest a good reference book for this topic Groups, Rings and Fields. I do not need to study this in depth as this is a single chapter on my Advanced Engineering Maths course but atleast want to grab and absorb the basic concepts of the same, so do you suggest I should try to study it from Gallian or Dummit and Foote??

My personal favorite is Joseph Galiian's book for basic concepts. Dummit and Foote's is also a wonderful book and covers much, much more material. If you want a reference-type book (but still presented at a very readable level) the latter is probably a better option.

 

[xenonforlife]

My personal favorite is Joseph Galiian's book for basic concepts. Dummit and Foote's is also a wonderful book and covers much, much more material. If you want a reference-type book (but still presented at a very readable level) the latter is probably a better option.

Thank you so much for the help