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Math Help - Jacobson Radical

  1. #1
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    Jacobson Radical

    Let k be a field. Let A be the k-subspace of M_3(k) spanned by

    1=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix},  \alpha =\begin{pmatrix}0&1&0\\1&0&0\\0&0&-1\end{pmatrix},\beta =\begin{pmatrix}0&0&0\\0&0&0\\1&1&0\end{pmatrix},

    which can be shown to actually be a k-algebra.

    Determine the Jacobson radical of A, which is the intersection of all maximal left ideals.

    I know at least \beta (and hence the left ideal it generates) is in the Jacobson radical, because it is nilpotent (it is known that the Jacobson radical contains all nilpotent elements). I don't know where to proceed from here, though...
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  2. #2
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    Re: Jacobson Radical

    Quote Originally Posted by topspin1617 View Post
    Let k be a field. Let A be the k-subspace of M_3(k) spanned by

    1=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix},  \alpha =\begin{pmatrix}0&1&0\\1&0&0\\0&0&-1\end{pmatrix},\beta =\begin{pmatrix}0&0&0\\0&0&0\\1&1&0\end{pmatrix},

    which can be shown to actually be a k-algebra.

    Determine the Jacobson radical of A, which is the intersection of all maximal left ideals.

    I know at least \beta (and hence the left ideal it generates) is in the Jacobson radical, because it is nilpotent (it is known that the Jacobson radical contains all nilpotent elements). I don't know where to proceed from here, though...
    first of all, this claim that "the Jacobson radical contains all nilpotent elements" is true in commutative rings not in all rings. for example J(M_2(\mathbb{C}))=(0) but obviously M_2(\mathbb{C}) has non-zero nilpotent elements.

    what is true in general is that the Jacobson radical contains all nilpotent ideals.

    to answer your question, just use this fact that a \in J(A) if and only if 1-xa is invertible for all x \in A. you'll eventually need to consider two cases: \text{char}(k) \neq 2 and \text{char}(k)=2.
    Last edited by NonCommAlg; August 11th 2011 at 02:09 AM.
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