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  1. #1 Algebra 
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    Now that I am without a professor, I may have some algebra questions that someone hopefully can help me with.

    Here's the one that has me stuck right now. The book says "Let and . Show that these two rings are not ring-isomorphic.

    But aren't they isomorphic by the map ?


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  3. #2 Re: Algebra 
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    Quote Originally Posted by AlexP
    Now that I am without a professor, I may have some algebra questions that someone hopefully can help me with.

    Here's the one that has me stuck right now. The book says "Let and . Show that these two rings are not ring-isomorphic.

    But aren't they isomorphic by the map ?




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  4. #3  
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    Oh geez...dumb mistake. I did those calculations and somehow managed not to see that. Thanks. I will keep working on it.
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  5. #4  
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    When showing that two fields are not isomorphic, is it sufficient to show that the two fields without zero are not isomorphic as multiplicative groups? It makes sense to me but I'd like to verify it. If so then I've shown that the fields of real and complex numbers are not isomorphic...if not then I'm still stuck.
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  6. #5  
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    Quote Originally Posted by AlexP
    When showing that two fields are not isomorphic, is it sufficient to show that the two fields without zero are not isomorphic as multiplicative groups?
    That would do it.
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  7. #6  
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    Thanks. It certainly made sense, the book just never mentioned doing it that way. So then has an element of order 4 () but does not, so they are not isomorphic since isomorphisms preserve orders. Is there another way though?
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  8. #7  
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    Quote Originally Posted by AlexP
    Thanks. It certainly made sense, the book just never mentioned doing it that way. So then has an element of order 4 () but does not, so they are not isomorphic since isomorphisms preserve orders. Is there another way though?
    They will all boildown to the fact that has no .
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  9. #8  
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    ok. What struck me right away was the fact that the complex numbers are two-dimensional, but I had no way to formalize that notion.

    Back to showing that is not isomorphic to ...

    Since homomorphisms preserve subrings and both and contain as a subring, the isomorphism must send to , but then it must be the identity which is not possible since .

    Reasons why I doubt the truth of this: 1) Who says it must send to ? 2) Even if it does, does the isomorphism acting identically on a subring imply that it acts identically on the whole field?

    But this is the lines along which I'm thinking... I have not been able to come up with anything else because the two fields very much 'act' the same.

    EDIT: If we look at the subring of integers instead, then it does take to , but that still doesn't imply that the isomorphism is identical on the whole field because not every element is an integer multiple of 1.
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  10. #9  
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    The task is to construct a field of order 25.

    works because it yields , right? It's clearly order 25, and is irreducible (it doesn't have a root in ) and thus is maximal so clearly is a field. What I'm wondering is just this: if there is more than one irreducible quadratic, does it make a difference which one we mod by? I don't think it does but I'd like confirmation.

    This is in the chapter on factorization of polynomials, so this is the way it was intended to be done, in case there are any others.
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  11. #10  
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    The book gives the following two theorems:

    The 'Factor Theorem':
    "Let be a field, , and . Then is a zero of if and only if is a factor of ."

    Reducibility Test for Degrees 2 and 3:
    "Let be a field. If and or , then is reducible over if and only if has a zero in ."

    My question is...why must a polynomial be degree 2 or 3 in order to apply this reducibility theorem? If is a polynomial of any degree and has a root in then we can factor out of it, and doesn't this make it reducible? Neither nor divided by will be a unit (making it irreducible) because is an integral domain, so and the only units in are and (if applicable) .
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  12. #11  
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    Quote Originally Posted by AlexP
    The book gives the following two theorems:

    The 'Factor Theorem':
    "Let be a field, , and . Then is a zero of if and only if is a factor of ."

    Reducibility Test for Degrees 2 and 3:
    "Let be a field. If and or , then is reducible over if and only if has a zero in ."

    My question is...why must a polynomial be degree 2 or 3 in order to apply this reducibility theorem? If is a polynomial of any degree and has a root in then we can factor out of it, and doesn't this make it reducible? Neither nor divided by will be a unit (making it irreducible) because is an integral domain, so and the only units in are and (if applicable) .
    is not irreducible, but has no zeros over the real numbers.
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  13. #12  
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    Got it... Another question...

    In an example in the book, there is a polynomial with rational coefficients which he multiplies by the least common multiple of the denominators to make into a polynomial with integer coefficients. He then proceeds to show that this new polynomial is irreducible over , and I understand that this implies that it is irreducible over , but why are we allowed to multiply by a constant first? It seemed the answer would be somewhere in the proof for 'reducible over implies reducible over ' but I can't figue it out.
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  14. #13  
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    Quote Originally Posted by AlexP
    Got it... Another question...

    In an example in the book, there is a polynomial with rational coefficients which he multiplies by the least common multiple of the denominators to make into a polynomial with integer coefficients. He then proceeds to show that this new polynomial is irreducible over , and I understand that this implies that it is irreducible over , but why are we allowed to multiply by a constant first? It seemed the answer would be somewhere in the proof for 'reducible over implies reducible over ' but I can't figue it out.
    Factorization of polynomials is only unique up to a non-zero multiple.
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  15. #14  
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    Going back to my question about the reducibility test for degrees 2 and 3, it IS true in the backwards direction for higher degrees, isn't it? Reducible does not necessarily imply that a polynomial has a zero over F, but having a zero over F does imply reducible since we can obviously factor out x-a.
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  16. #15  
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    Quote Originally Posted by AlexP
    Going back to my question about the reducibility test for degrees 2 and 3, it IS true in the backwards direction for higher degrees, isn't it? Reducible does not necessarily imply that a polynomial has a zero over F, but having a zero over F does imply reducible since we can obviously factor out x-a.
    Right.

    In degree 2 or 3 a reducible polynomial must have a degree 1 factor, which implies a root.
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  17. #16  
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    Ah, yes. That definitely makes it clear, thank you.

    I'm presently trying to show that is reducible over . It's proving not to be as straightforward as I thought it would be. Should it be relatively easy to factor or is another way necessary? I tried multiplying two general quadratic polynomials and linear and cubic polynomials and then solving for the necessary coefficients in the product, but it didn't work when I substituted the values in. Seems like I must have messed it up somehow but I don't know how.

    Afterthought: Going by what you've told me, I can safely assume that it's the product of two quadratics irreducible over , right?
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  18. #17  
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    Quote Originally Posted by AlexP
    Ah, yes. That definitely makes it clear, thank you.

    I'm presently trying to show that is reducible over . It's proving not to be as straightforward as I thought it would be. Should it be relatively easy to factor or is another way necessary? I tried multiplying two general quadratic polynomials and linear and cubic polynomials and then solving for the necessary coefficients in the product, but it didn't work when I substituted the values in. Seems like I must have messed it up somehow but I don't know how.

    Afterthought: Going by what you've told me, I can safely assume that it's the product of two quadratics irreducible over , right?
    Yes, it will be the product of two irreducible quadratics.

    Those quadratics will be reducible over the complex numbers. So one way to find them is to do what you tried and look for coefficients of two general quadratics, but another way is to work with the four complexs roots. If you have forgotten DeMoivre's theorem you should look it up and refresh your memory.

    You need to go through the analysis yourself, but so you can check your work I will give you the irreducble factors. They are

    and
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  19. #18  
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    I am stuck. In what way do I use the complex roots (which I may not have found correctly) to arrive at that factorization? I'd like to understand, but at the same time I'm not going to beat this problem to death.
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  20. #19  
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    Does this work?

    Show that an integral domain with the property that every strictly descending chain of ideals is finite in length must be a field.

    Suppose is an integral domain that is not a field. Then there exists such that is not a unit. We then have that is a strictly descending chain of ideals of infinite length. Therefore if every strictly descending chain of ideals in an integral domain is finite in length then we must have that it is a field.
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  21. #20  
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    Quote Originally Posted by AlexP
    I am stuck. In what way do I use the complex roots (which I may not have found correctly) to arrive at that factorization? I'd like to understand, but at the same time I'm not going to beat this problem to death.
    Just find the roots over the complex numbers and construct two polynomials with real coefficients from them.
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  22. #21  
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    Quote Originally Posted by AlexP
    Does this work?

    Show that an integral domain with the property that every strictly descending chain of ideals is finite in length must be a field.

    Suppose is an integral domain that is not a field. Then there exists such that is not a unit. We then have that is a strictly descending chain of ideals of infinite length. Therefore if every strictly descending chain of ideals in an integral domain is finite in length then we must have that it is a field.
    That works so long as you understand why (a^2) is a proper sub-ideal of (a).
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  23. #22  
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    I was going by the fact that if isn't a unit, won't be either, and in general if and isn't a unit then .
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  24. #23  
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    I really wish I didn't have to ask this, but anyway...

    I am failing to see how is a field... When we multiply two elements together, don't we get a number of the form ?

    Help, anyone? Guitarist?
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    Well, I am not sure where you difficulty lies, so let me ranble a bit.

    is undeniably a subset of the Real field so we would expect that any element can be expressed as the (possibly) trivial product of any two elements. This means, as you rightly say, all elements in takes the form

    (since .)

    To show it is a sub-field, we need to show it satisfies the field axioms (obviously!)

    Addition and subtraction i.e. additive inverse should cause no problem. Multiplication of the form should present no real difficulty once you see that, say which is rational, and that the term can similarly be piecewise resolved to Oops I edited

    Multiplicative inverse is less easy. We will have that if we can show that

    . But this is just simultaneously to solve for ,which is always possible provided the are not all zero.

    Does this help? Have I misunderstood your problem? Dunno; as I said, I was just rambling
    Last edited by Guitarist; August 16th, 2011 at 09:19 AM.
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  26. #25  
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    Quote Originally Posted by Guitarist View Post
    is undeniably a subset of the Real field so we would expect that any element can be expressed as the (possibly) trivial product of any two elements. This means, as you rightly say, all elements in takes the form

    (since .)
    But I did not say that... I said . This is (I thought) by definition when we adjoin some number to a field.

    Why do you say that elements of are of the form ?
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