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    Stuck on iv.

    a, b and c can be three possible numbers so is the answer simply 3^3?

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    (Original post by cooldudeman)
    Stuck on iv.

    a, b and c can be three possible numbers so is the answer simply 3^3?

    Posted from TSR Mobile
    It is. You've shown that every member of the quotient is of the form [a+bx+cx^2]+I; conversely, it's obvious that every polynomial of the form a+bx+c x^2 induces a different member [a+bx+cx^2]+I in the quotient, so we have a bijection taking (a,b,c) \mapsto [a+bx+c x^2]+I.

    There are, as you say, 27 elements of the left-hand set, so there must be 27 of the right-hand set.

    The question has constructed a field extension \mathbb{F}_3(\theta), where \theta^3 + \theta + 1 = 0.
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    (Original post by Smaug123)
    It is. You've shown that every member of the quotient is of the form [a+bx+cx^2]+I; conversely, it's obvious that every polynomial of the form a+bx+c x^2 induces a different member [a+bx+cx^2]+I in the quotient, so we have a bijection taking (a,b,c) \mapsto [a+bx+c x^2]+I.

    There are, as you say, 27 elements of the left-hand set, so there must be 27 of the right-hand set.

    The question has constructed a field extension \mathbb{F}_3(\theta), where \theta^3 + \theta + 1 = 0.
    Thanks so much. Would you be able to help me on c too.

    It doesn't specifically say what I is so I'm a little confused on that. I'm trying to show that I is a subset of (a) and vise versa.

    I also am considering the definition of an ideal too. But how do I know that a natural number exists in I...

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    (Original post by cooldudeman)
    Thanks so much. Would you be able to help me on c too.

    It doesn't specifically say what I is so I'm a little confused on that. I'm trying to show that I is a subset of (a) and vise versa.

    I also am considering the definition of an ideal too. But how do I know that a natural number exists in I...

    Posted from TSR Mobile
    You know I is an ideal of Z, so I contains integers as it is obvious it is a subset of Z. Suppose m is an integer in I. Then if m is greater than or equal to 0, you have m is a natural. If not, then you know I is a subgroup under addition so you must have an additive inverse of m in I as well, and this is -m. If m < 0, then -m > 0 and so in both cases you have a natural number in I.

    As for the actual question, it can't tell you explicitly what I is because you're trying to prove a statement about the general structure of ideals in Z. Specialising defeats the point of the question. I'd encourage you to keep on thinking about it, because for this question any hint is a big one, I think. You have that (a) is a subset of I for any a in I, so a good strategy is try and pick an a that is special somehow, so that you can prove I is a subset of (a).

    Spoilers for more specific hint.

    Spoiler:
    Show
    Instead, pick the lowest natural in I (if it's not immediately obvious that there is one, think about the structure of the naturals), and see what information you can glean about the ideal generated by it.
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    (Original post by cooldudeman)
    Thanks so much. Would you be able to help me on c too.

    It doesn't specifically say what I is so I'm a little confused on that. I'm trying to show that I is a subset of (a) and vise versa.

    I also am considering the definition of an ideal too. But how do I know that a natural number exists in I...

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    I is an ideal of \mathbb{Z}, and therefore it can only contain integers.

    You're going to have trouble showing that I \subset \langle a \rangle if you don't know what a is. Suppose I gave you the ideal I = \langle 6, 15 \rangle \subset \mathbb{Z}. The question is asking you to show that there is some a such that I = \langle a \rangle. Can you tell me what a is in this case?
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    (Original post by Smaug123)
    I is an ideal of \mathbb{Z}, and therefore it can only contain integers.

    You're going to have trouble showing that I \subset \langle a \rangle if you don't know what a is. Suppose I gave you the ideal I = \langle 6, 15 \rangle \subset \mathbb{Z}. The question is asking you to show that there is some a such that I = \langle a \rangle. Can you tell me what a is in this case?
    Im not understanding what the comma is all about. <6, 15> what does the comma mean in this? I have never seen notation like this before...
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    (Original post by cooldudeman)
    Im not understanding what the comma is all about. <6, 15> what does the comma mean in this? I have never seen notation like this before...
    Just like <a> is the ideal generated by a, so elements of the form na where n is in Z, <a, b> is the ideal generated by a and b so elements of the form na + mb for n and m in Z.

    So <6, 15> is elements of the form 6n + 15m for n and m in Z.
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    (Original post by Smaug123)
    I is an ideal of \mathbb{Z}, and therefore it can only contain integers.

    You're going to have trouble showing that I \subset \langle a \rangle if you don't know what a is. Suppose I gave you the ideal I = \langle 6, 15 \rangle \subset \mathbb{Z}. The question is asking you to show that there is some a such that I = \langle a \rangle. Can you tell me what a is in this case?
    (Original post by SParm)
    Just like <a> is the ideal generated by a, so elements of the form na where n is in Z, <a, b> is the ideal generated by a and b so elements of the form na + mb for n and m in Z.

    So <6, 15> is elements of the form 6n + 15m for n and m in Z.
    Hi very very sorry that I didnt reply. Been very busy. I had my exam today and the same question came up but m is and integer and not a natural number.

    This is whay I did. Please can someone check it. if this was out of 7 marks, how many do you think I would get. Really not sure if what I did made sense...

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    (Original post by cooldudeman)
    Hi very very sorry that I didnt reply. Been very busy. I had my exam today and the same question came up but m is and integer and not a natural number.

    This is whay I did. Please can someone check it. if this was out of 7 marks, how many do you think I would get. Really not sure if what I did made sense...

    Posted from TSR Mobile
    Sorry to say, you seem to have proven the converse.

    You were asked to show:

    IF I is an ideal THEN it has a certain form.

    You've assumed it has a certain form, and shown (didn't check the details as I'm not well up even on basic ring theory) that it is an ideal.
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    (Original post by cooldudeman)
    Hi very very sorry that I didnt reply. Been very busy. I had my exam today and the same question came up but m is and integer and not a natural number.
    \langle -m \rangle = \langle m \rangle, of course, so WLOG m is greater than 0.

    This is whay I did. Please can someone check it. if this was out of 7 marks, how many do you think I would get. Really not sure if what I did made sense...
    To be honest, it doesn't really. You need to show that I is principal: that there exists m such that I = \langle m \rangle.

    For 1), you've written "maximal" instead of "minimal". Nonzero ideals in \mathbb{Z} don't have maximal elements.
    It's all a bit unclear for me, to be honest, and I'm not really sure how you're trying to prove it. Your final line doesn't make sense anyway: "I = \langle m \rangle for all m \in \mathbb{Z}" is not true of any ideal.

    I think what you have written is more convincing as a proof that sets of the form \langle m \rangle are ideals of \mathbb{Z}. I see some of the right answer in there, but also some bits that aren't right.

    To answer the question as stated in a "model answer" kind of way (bearing in mind that solutions anyone comes up with on the spot are usually much less well-structured than this):

    Spoiler:
    Show
    Let m be the least positive element of I. We show that I = \langle m \rangle.

    Firstly: I \subset \langle m \rangle. Let a \in I, and we show that a \in \langle m \rangle.

    By the division algorithm, any a \in I may be written as a = qm + r, with r &lt; m. But since m, a \in I, must have a-qm \in I, so r \in I.

    But m was the least positive element of I, and r is nonnegative and less than m, so r must be zero. That is, m \vert a, so a \in \langle m \rangle.

    Conversely, any element a m \in \langle m \rangle is an element of I because it is a multiple of m \in I.
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    (Original post by Smaug123)
    \langle -m \rangle = \langle m \rangle, of course, so WLOG m is greater than 0.


    To be honest, it doesn't really. You need to show that I is principal: that there exists m such that I = \langle m \rangle.

    For 1), you've written "maximal" instead of "minimal". Nonzero ideals in \mathbb{Z} don't have maximal elements.
    It's all a bit unclear for me, to be honest, and I'm not really sure how you're trying to prove it. Your final line doesn't make sense anyway: "I = \langle m \rangle for all m \in \mathbb{Z}" is not true of any ideal.

    I think what you have written is more convincing as a proof that sets of the form \langle m \rangle are ideals of \mathbb{Z}. I see some of the right answer in there, but also some bits that aren't right.

    To answer the question as stated in a "model answer" kind of way (bearing in mind that solutions anyone comes up with on the spot are usually much less well-structured than this):

    Spoiler:
    Show
    Let m be the least positive element of I. We show that I = \langle m \rangle.

    Firstly: I \subset \langle m \rangle. Let a \in I, and we show that a \in \langle m \rangle.

    By the division algorithm, any a \in I may be written as a = qm + r, with r &lt; m. But since m, a \in I, must have a-qm \in I, so r \in I.

    But m was the least positive element of I, and r is nonnegative and less than m, so r must be zero. That is, m \vert a, so a \in \langle m \rangle.

    Conversely, any element a m \in \langle m \rangle is an element of I because it is a multiple of m \in I.
    OK I see. Thanks for the model answer.

    Do you think Id even get one mark lol...

    Thank god this was the ONLY thing I struggled on.
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    (Original post by cooldudeman)
    OK I see. Thanks for the model answer.

    Do you think Id even get one mark lol...

    Thank god this was the ONLY thing I struggled on.
    No idea how they mark these things - some of your lines (your Chunk 3) would appear in a proof, so you might get something for that.
 
 
 
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