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    Ok, I've done the first part but I'm having trouble with showing that f is an isomorphism of rings??


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    (Original post by maths10101)
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    ^
    Ok, I've done the first part but I'm having trouble with showing that f is an isomorphism of rings??


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    What trouble are you having exactly? f is a ring isomorphism if it is bijective and

    f(x+ y) = f(x) + f(y) and f(xy) = f(x)f(y) for all x, y \in R.

    Which part in the proof is tripping you up?

    This is my last post before bed, so hopefully someone else will reply to you.
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    (Original post by Zacken)
    What trouble are you having exactly? f is a ring isomorphism if it is bijective and

    f(x+ y) = f(x) + f(y) and f(xy) = f(x)f(y) for all x, y \in R.

    Which part in the proof is tripping you up?

    This is my last post before bed, so hopefully someone else will reply to you.
    yeah that part is fine...but how do you show that a function of such is bijective?
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    (Original post by maths10101)
    yeah that part is fine...but how do you show that a function of such is bijective?
    Not quite sure about this - somebody jump in and correct me, but have you covered ideals and kernels? If so, that should let you prove the function is injective.
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    (Original post by Zacken)
    Not quite sure about this - somebody jump in and correct me, but have you covered ideals and kernels? If so, that should let you prove the function is injective.
    Yup, showing that the kernel is trivial is a good move.
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    (Original post by Gregorius)
    Yup, showing that the kernel is trivial is a good move.
    Thanks for the reassurance.
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    Pretty advanced for year 9...
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    (Original post by morgan8002)
    Pretty advanced for year 9...
    Crazy curriculum changes.
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    Assuming you've shown \displaystyle f is a homomorphism then take two distinct elements in your subring which is \displaystyle M_2(\mathbb{R}) with some extra conditions on the elements of the matrix.

    A general element of your subring looks like \displaystyle \begin{pmatrix}a & b\\ -b & a\end{pmatrix}.

    Consider where they are mapped to under \displaystyle f is it possible for the two elements to get mapped to the same element in \displaystyle \mathbb{C} and still be distinct? ... Conclude \displaystyle f is injective.

    I think this a more "direct approach" alternatively note that \displaystyle f is injective if and only if \displaystyle \text{ker}(f) is trivial. So you could show that the only element that gets mapped to the zero in the complex numbers under f is in fact the zero of your subring.

    Pick an arbitrary element in \displaystyle \mathbb{C} can you find a matrix in your subring that gets maps to your arbitrary complex number under \displaystyle f? ... Conclude \displaystyle f is surjective.

    Hence \displaystyle f is a homomorphism and bijective so \displaystyle f is indeed a isomorphism.
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    (Original post by poorform)
    Assuming you've shown \displaystyle f is a homomorphism then take two distinct elements in your subring which is \displaystyle M_2(\mathbb{R}) with some extra conditions on the elements of the matrix.

    A general element of your subring looks like \displaystyle \begin{pmatrix}a & b\\ -b & a\end{pmatrix}.

    Consider where they are mapped to under \displaystyle f is it possible for the two elements to get mapped to the same element in \displaystyle \mathbb{C} and still be distinct? ... Conclude \displaystyle f is injective.

    I think this a more "direct approach" alternatively note that \displaystyle f is injective if and only if \displaystyle \text{ker}(f) is trivial. So you could show that the only element that gets mapped to the zero in the complex numbers under f is in fact the zero of your subring.

    Pick an arbitrary element in \displaystyle \mathbb{C} can you find a matrix in your subring that gets maps to your arbitrary complex number under \displaystyle f? ... Conclude \displaystyle f is surjective.

    Hence \displaystyle f is a homomorphism and bijective so \displaystyle f is indeed a isomorphism.

    Thanks mate, a lot of help that has had!
 
 
 
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