1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Ring Isomorphism Proof

  1. Feb 11, 2013 #1

    Zondrina

    User Avatar
    Homework Helper

    1. The problem statement, all variables and given/known data

    The question : http://gyazo.com/5372336302b5ef289b305172bcd16a2a

    2. Relevant equations

    First Isomorphism theorem.

    3. The attempt at a solution

    Define [itex]\phi : \mathbb{Q}[x]/<x^2-2> → Q[ \sqrt{2} ] \space | \space \phi (f(x)) = f( \sqrt{2})[/itex]

    So showing phi is a homomorphism is quite easy so I'll skip those details.

    My question lies in my argument for phi being a bijection. I COULD show it's 1-1 and onto which would mean it's an isomorphism, but I need practice with the first isomorphism theorem so I'm going to try it and hope I know what's going on.

    So, I believe I need to show phi is onto and then argue about the kernel.

    To show phi is onto, suppose (a+b√2) is in Q[√2] and f(x) is in Q[x]/<x2-2> so that f(x) = ax+b.

    Now, [itex]\phi (f(x)) = f( \sqrt{2}) = a + b \sqrt{2}[/itex]. Hence for every a+b√2 in Q[√2] there exists f(x) in Q[x]/<x2-2> such that phi(f(x)) = a+b√2.

    Hence phi is onto.

    Now consider that ker(phi) = { f(x) in Q[x]/<x2-2> | f(√2) = 0 }. Since x2-2 is in ker(phi) and x2-2 is of the smallest degree, we conclude that ker(phi) = x2-2 and hence by the first isomorphism theorem, Q[x]/<x2-2> is isomorphic to Q[√2].

    This is my first time trying to apply the theorem rather than doing it the long way of showing 1-1 and onto correspondence.

    If anyone could tell me where I may have went wrong or if it looks good, it would be much appreciated.
     
  2. jcsd
  3. Feb 11, 2013 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your domain for ##\phi## is incorrect. It should be
    $$\phi : \mathbb{Q}[x] \rightarrow Q[\sqrt{2}]$$
    The polynomial ##f(x)## is in ##\mathbb{Q}[x]##, not ##\mathbb{Q}[x]/\langle x^2 - 2\rangle##.
     
  4. Feb 11, 2013 #3

    Zondrina

    User Avatar
    Homework Helper

    Ah this makes sense to me.

    It wouldn't make sense for my f(x) to be inside a set of cosets. So supposing I switch all instances of Q[x]/<x2-2> to Q[x]. Does the argument still apply?
     
  5. Feb 11, 2013 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, your argument is fine. However, I suggest being a bit more explicit in your argument about the kernel of ##\phi##. Perhaps something along the following lines: "the kernel of ##\phi## is an ideal, and ##\mathbb{Q}[x]## is a principal ideal domain, so we know the kernel is generated by a single element of ##\mathbb{Q}[x]##, namely the minimal polynomial of ##\sqrt{2}##. The minimal polynomial of ##\sqrt{2}## is ##x^2 - 2## because..."
     
  6. Feb 11, 2013 #5

    Zondrina

    User Avatar
    Homework Helper

    because (√2)2 - 2 = 0 so that x2-2 = ker(phi) (i.e x2-2 generates ker(phi)).
     
  7. Feb 11, 2013 #6

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    To be more precise, ##x^2 - 2## satisfies the definition of the minimal polynomial for ##\sqrt{2}##: (1) it is monic, (2) ##\sqrt{2}## is a root, and (3) if ##p(x)## is a monic polynomial of smaller degree, then ##\sqrt{2}## cannot be a root of ##p(x)## because...

    (This may seem obvious, but if it's one of your first homework assignments on this material, being pedantic is never a bad idea.)
     
  8. Feb 11, 2013 #7

    Zondrina

    User Avatar
    Homework Helper

    Woah, now this was never mentioned in the book, but a quick read up on wiki has informed me of what you mean.

    (1) Is satsified because the coefficient of the highest powered term is 1.
    (2) (√2)2 - 2 = 0 so √2 is a root of x2-2.
    (3) If p(x) is a monic polynomial of smaller degree, then √2 cannot be a root of p(x) because the only polynomials of smaller degree are of degree 1 and the constant polynomial. i.e either p(x) = ax + b or p(x) = a for some constant a.

    Thus p(√2) = a√2 + b or p(√2) = a which tells us that x2-2 is indeed the minimal polynomial.
     
  9. Feb 11, 2013 #8

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Right, if ##p(x) \in \mathbb{Q}[x]## is a monic polynomial of degree 1, then it is of the form ##p(x) = x + a##. If we had ##p(\sqrt{2}) = 0## then that means ##\sqrt{2} + a = 0##, or ##\sqrt{2} = -a##, a contradiction because ##a## is rational and ##\sqrt{2}## is not.

    And the only monic polynomial of degree 0 is ##p(x) = 1##, which clearly does not satisfy ##p(\sqrt{2}) = 0##, so this tells us that ##x^2 - 2## is indeed the minimal polynomial.

    By the way, if this is all new to you, you may want to take a few minutes to convince yourself why we say "the" minimal polynomial, i.e., why is it unique? Can't there be some other monic polynomial of degree 2 with ##\sqrt{2}## as a root?
     
  10. Feb 11, 2013 #9

    Zondrina

    User Avatar
    Homework Helper

    I had a look over the proof for uniqueness and this line jumped at me :

    "there is at least one polynomial in ker(subα) that generates ker(subα). Such a polynomial will have least degree among all non-zero polynomials in ker(subα), and a(x) is taken to be the unique monic polynomial among these."

    Source : http://en.wikipedia.org/wiki/Minimal_polynomial_(field_theory)
     
  11. Feb 11, 2013 #10

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Here's a simple proof of uniqueness. If ##p(x)## and ##q(x)## both satisfy the requirements of a minimal polynomial for ##\sqrt{2}##, then ##r(x) = p(x) - q(x)## is also a polynomial satisfying ##r(\sqrt{2}) = 0##, and the degree of ##r(x)## is strictly lower than that of ##p(x)## and ##q(x)## because the leading coefficients of ##p(x)## and ##q(x)## are both 1, so the leading term cancels when you subtract them.

    If ##r(x)## is not the zero polynomial, then we can normalize it by dividing by its leading coefficient, thereby obtaining a monic polynomial satisfying ##r(\sqrt{2}) = 0##. This contradicts the minimality of ##p(x)## and ##q(x)##. We conclude that ##r(x)## must be the zero polynomial, so ##p(x) = q(x)##.
     
  12. Feb 11, 2013 #11

    Zondrina

    User Avatar
    Homework Helper

    That's a lot better and easier to follow for sure. So since p = q, the minimal polynomial is monic and unique.
     
  13. Feb 11, 2013 #12

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    By the way, we can also make a direct argument for why the kernel is equal to ##\langle x^2 - 2\rangle##, without mentioning minimal polynomials. We just use the fact that we already established, namely that if ##p(x)## is a polynomial of degree less than 2, such that ##p(\sqrt{2}) = 0##, then ##p(x)## must be the zero polynomial.

    First, note that the kernel consists precisely of those polynomials ##p(x)## for which ##p(\sqrt{2}) = 0##.

    If ##p(x)## is an element of ##\langle x^2 - 2\rangle##, that means ##p(x) = (x^2 - 2)q(x)##, where ##q(x)## is a polynomial in ##Q[x]##, and evaluating at ##x = \sqrt{2}## shows that ##p(\sqrt{2}) = 0##, so ##p(x)## is in the kernel. Therefore ##\langle x^2 - 2 \rangle \subset \textrm{ker}(\phi)##.

    Conversely, if ##p(x) \in \textrm{ker}(\phi)##, then by the division algorithm, we may write ##p(x) = q(x)(x^2 - 2) + r(x)##, where the degree of ##r(x)## is strictly less than the degree of ##x^2 - 2##. But ##r(x) = p(x) - q(x)(x^2 - 2)##, so ##r(\sqrt{2}) = 0##, which means ##r(x) \in \textrm{ker}(\phi)##. We already established that this forces ##r(x)## to be the zero polynomial. Thus ##p(x) = q(x)(x^2 - 2)##, which shows that ##p(x) \in \langle x^2 - 2 \rangle##. Thus ##\textrm{ker}(\phi) \subset \langle x^2 - 2 \rangle## and we're done.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Ring Isomorphism Proof
  1. Ring isomorphism (Replies: 3)

  2. Isomorphic rings (Replies: 0)

  3. Isomorphic rings (Replies: 1)

  4. Ring Isomorphism (Replies: 9)

  5. Rings Isomorphism. (Replies: 8)

Loading...