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Math Help - Subring Elements

  1. #1
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    Subring Elements

    Let R be a ring and a an element of R. Let S = \{ x \in R: ax = 0_R\}. S is a subring of R.

    Let R = M_2(\mathbb{Z}_5) (the ring of 22 matrices with entries from the field \mathbb{Z}_5) and

    a= \begin{pmatrix}4 & 2 \\ 3 & 4 \end{pmatrix}. Determine the elements of the subring S defined previously. (I) How many elements are in S? (II) Show that S is not an ideal of R.

    Attempt:

    (I) the elements of S are 2x2 matrices x such that ax=0_R.

    let x=\begin{pmatrix}a & b \\ c & d \end{pmatrix}

    ax=\begin{pmatrix}4 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix}a & b \\ c & d \end{pmatrix} = \begin{pmatrix}4a+2c & 4b+2d \\ 3a+4c & 3b+4d \end{pmatrix}= \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}

    If I solve the homogeneous system
    4a+2c=0
    4b+2d=0
    3a+4c=0
    3b+4d=0

    I get a=b=c=d=0.

    So does this mean the only element in the subring S is the zero matrix?

    (II) I will use the "ideal test":

    Since the zero matrix belongs to S, S \neq \emptyset.

    Then S \triangleleft R \iff (x-y) \in S and rS, Sr \subseteq S \forall r \in R. (x,y are in S).

    The question says show that S is NOT an ideal. But the problem is that S passes the ideal test since (x-y) is equal to the zeo matrix and therefore a(x-y)=0_R \in S. The same is true with multipication. So, is something wrong with the question or did I make a mistake?!
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  2. #2
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    Quote Originally Posted by demode View Post
    Let R be a ring and a an element of R. Let S = \{ x \in R: ax = 0_R\}. S is a subring of R.


    This may depend on the definition. Many authors require that if R is unitary then in order to be considered a

    candidate to be a subring a subset must contain the unit of R. In this case, though, it

    must be not so since then a = 0.


    Let R = M_2(\mathbb{Z}_5) (the ring of 22 matrices with entries from the field \mathbb{Z}_5) and

    a= \begin{pmatrix}4 & 2 \\ 3 & 4 \end{pmatrix}. Determine the elements of the subring S defined previously. (I) How many elements are in S? (II) Show that S is not an ideal of R.

    Attempt:

    (I) the elements of S are 2x2 matrices x such that ax=0_R.

    let x=\begin{pmatrix}a & b \\ c & d \end{pmatrix}

    ax=\begin{pmatrix}4 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix}a & b \\ c & d \end{pmatrix} = \begin{pmatrix}4a+2c & 4b+2d \\ 3a+4c & 3b+4d \end{pmatrix}= \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}

    If I solve the homogeneous system
    4a+2c=0
    4b+2d=0
    3a+4c=0
    3b+4d=0

    I get a=b=c=d=0.


    Even without checking this cannot be right since the matrix x is singular, and indeed:

    3a+4c=2(5a+2c)\,,\,\,3b+4d=2(4b+2d)\Longrightarrow c=-2a=3a\,,\,\,d=-2b=3b\Longrightarrow

    any matrix of the form \begin{pmatrix}a&b\\3a&3b\end{pmatrix} belongs to S


    So does this mean the only element in the subring S is the zero matrix?

    (II) I will use the "ideal test":

    Since the zero matrix belongs to S, S \neq \emptyset.

    Then S \triangleleft R \iff (x-y) \in S and rS, Sr \subseteq S \forall r \in R. (x,y are in S).

    The question says show that S is NOT an ideal. But the problem is that S passes the ideal test since (x-y) is equal to the zeo matrix and therefore a(x-y)=0_R \in S. The same is true with multipication.


    No, it's not. Try to find a counterexample to x\in S\,,\,r\in R\Longrightarrow rx\in S now that you know the general

    form of elements in S

    Tonio


    So, is something wrong with the question or did I make a mistake?!
    .
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  3. #3
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    So if the elements are of the form \begin{pmatrix}a&b\\3a&3b\end{pmatrix}, then a \in \{0,1,2,3,4\}, b \in \{0,1,2,3,4\} in \mathbb{Z}_5. 5x5=25 possible combinations of a and b. Therefore there are 25 elements in S?

    Try to find a counterexample to x\in S\,,\,r\in R\Longrightarrow rx\in S now that you know the general

    form of elements in S
    Let a=b=1 then x= \begin{pmatrix}1&1\\3&3\end{pmatrix}, and r= \begin{pmatrix}1&1\\1&1\end{pmatrix}.

    rx= \begin{pmatrix}4&4\\4&4\end{pmatrix}, and

    a(rx)= \begin{pmatrix}4&2\\3&4\end{pmatrix} \begin{pmatrix}4&4\\4&4\end{pmatrix} = \begin{pmatrix}4&4\\3&3\end{pmatrix} \neq R_0

    Therefore rx \notin S. Is this okay?
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  4. #4
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    Quote Originally Posted by demode View Post
    So if the elements are of the form \begin{pmatrix}a&b\\3a&3b\end{pmatrix}, then a \in \{0,1,2,3,4\}, b \in \{0,1,2,3,4\} in \mathbb{Z}_5. 5x5=25 possible combinations of a and b. Therefore there are 25 elements in S?



    Let a=b=1 then x= \begin{pmatrix}1&1\\3&3\end{pmatrix}, and r= \begin{pmatrix}1&1\\1&1\end{pmatrix}.

    rx= \begin{pmatrix}4&4\\4&4\end{pmatrix}, and

    a(rx)= \begin{pmatrix}4&2\\3&4\end{pmatrix} \begin{pmatrix}4&4\\4&4\end{pmatrix} = \begin{pmatrix}4&4\\3&3\end{pmatrix} \neq R_0

    Therefore rx \notin S. Is this okay?


    Yes and yes, but it is S, not R_o.

    Tonio
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    Quote Originally Posted by tonio View Post
    Yes and yes, but it is S, not R_o.

    Tonio
    Why is that? S is defined as x in R such that ax=0_R. Since rx is in R, but a(rx) is not equal to 0_R, it follows that rx doesn't belong to S. Then what's wrong with what I did?
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  6. #6
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    Quote Originally Posted by demode View Post
    Why is that? S is defined as x in R such that ax=0_R. Since rx is in R, but a(rx) is not equal to 0_R, it follows that rx doesn't belong to S. Then what's wrong with what I did?

    I think I do. You wrote a(rx)\neq R_0 , instead of a(rx)\neq 0_R

    Tonio
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