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Math Help - Algebra, Problems For Fun (40)

  1. #1
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    Algebra, Problems For Fun (40)

    This is the last problem of this series of problems that I've been posting for a while now:

    Let F be a field and R=\left \{\begin{pmatrix}x & 0 & y \\ 0 & x & z \\ 0 & 0 & x \end{pmatrix}: \ \ x,y,z \in F \right \}. It's easy to see that R is a commutative ring with identity element. Find all maximal ideals of R.

    Hint:
    Spoiler:
    R has only one maximal ideal.
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Clearly S=\left\{\begin{pmatrix}x & 0 & 0\\<br />
0 & x & 0\\<br />
0 & 0 & x<br />
\end{pmatrix}:x\in F\right\} is an ideal of R and the quotient ring R/S is a commutative ring with identity I_3+S.

    Let’s let A(x,y,z) denote the element \begin{pmatrix}x & 0 & y\\<br />
0 & x & z\\<br />
0 & 0 & x<br />
\end{pmatrix}+S in R/S.

    Then if x\ne0, A(x,y,z)\cdot A(x^{-1},-yx^{-2},-zx^{-2})=I_3+S. Hence R/S is a field and so S is a maximal ideal.
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  3. #3
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    Quote Originally Posted by TheAbstractionist View Post
    Clearly S=\left\{\begin{pmatrix}x & 0 & 0\\<br />
0 & x & 0\\<br />
0 & 0 & x<br />
\end{pmatrix}:x\in F\right\} is an ideal of R and the quotient ring R/S is a commutative ring with identity I_3+S.

    Letís let A(x,y,z) denote the element \begin{pmatrix}x & 0 & y\\<br />
0 & x & z\\<br />
0 & 0 & x<br />
\end{pmatrix}+S in R/S.

    Then if x\ne0, A(x,y,z)\cdot A(x^{-1},-yx^{-2},-zx^{-2})=I_3+S. Hence R/S is a field and so S is a maximal ideal.
    S contains the identity element of R. so if it was an ideal of R, it would be equal to R. but it's not! so S is not an ideal of R.

    you can also see that by checking that S is not closed under multiplication by elements of R.
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    This is the last problem of this series of problems that I've been posting for a while now:

    Let F be a field and R=\left \{\begin{pmatrix}x & 0 & y \\ 0 & x & z \\ 0 & 0 & x \end{pmatrix}: \ \ x,y,z \in F \right \}. It's easy to see that R is a commutative ring with identity element. Find all maximal ideals of R.

    Hint:
    Spoiler:
    R has only one maximal ideal.
    (I've used (x,y,z) = \begin{pmatrix}x & 0 & y \\ 0 & x & z \\ 0 & 0 & x \end{pmatrix}, with the equivalent multiplication, because it's quicker to type...)

    The subring with x=0 is clearly an ideal: (x,y,z).(a,b,c) = (xa, xb+ya, xc+za) \Rightarrow (0,y,z).(a,b,c) = (0, ya, za)=(a,b,c).(0,y,z).

    However, this ideal is also maximal. This is because if I is an ideal containing an element (x,y,z), x \neq 0, then [(x,y,z)(x^{-1},0,0)](1, -yx^{-1}, -zx^{-1}) = id and so I=R.

    Subsequently this ideal contains all other proper ideals, as if it didn't then there would be an ideal with x \neq 0, a contradiction. Thus it is the only maximal ideal.
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