Results 1 to 9 of 9
Like Tree3Thanks
  • 1 Post By HallsofIvy
  • 1 Post By HallsofIvy
  • 1 Post By SlipEternal

Thread: Help wth rings and subrings

  1. #1
    Junior Member
    Joined
    Nov 2013
    From
    Hong Kong
    Posts
    38

    Help wth rings and subrings

    Show that the matrice $R=\begin{pmatrix}
    a & b \\
    0 & a \\
    \end{pmatrix}$ for $a,b \in \mathbb{R}$ is a subring of the ring of matrices $M_2(\mathbb{R})$, also find a ring homomorphism such that $\Phi:R \rightarrow \mathbb{R}$ that is on to.

    Its a problem in my book and I cant seem to solve it as I am quite new to rings at this point in time. Can anyone help me solve this? It will help me a great deal in trying to understand things further in my textbook.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,462
    Thanks
    2893

    Re: Help wth rings and subrings

    A "ring" is an algebraic structure that has two operations. We can think of them as "addition" and "multiplication".

    Looking only at the "addition", the set of objects is a "commutative group". That is, if x and y are two members of the set then x+ y= y+ x, there exist an additive identity, "0", such that
    x+ 0= x, and, for each x, there exist an additive inverse, y. such that x+ y= 0.

    We only require that the "multiplication" be "associative"- x(yz)= (xy)z. There is NO requirement that the multiplication be commutative or that there be a "multiplicative identity" or that every element have a "multiplicative identity".

    Here, since you are talking about a subset of "all 2 by 2 matrices", the operations are the usual ones for matrices:
    \begin{pmatrix}a & b \\ 0 & a\end{pmatrix}+ \begin{pmatrix}x & y \\ 0 & x\end{pmatrix}= \begin{pmatrix}a+ x & b+ y \\ 0 & a+ x\end{pmatrix}
    The fact that this operation is associative, has an additive identity, and has additive inverses follows from the fact that we "add matrices" by adding the numbers of corresponding entries in the matrices and that addition of numbers satisfies them.

    \begin{pmatrix}a & b \\ 0 & a \end{pmatrix}\begin{pmatrix}x & y \\ 0 & x\end{pmatrix}= \begin{pmatrix}ax & ay+ bx \\ 0 & ax\end{pmatrix}.

    Showing that multiplication is associative is a bit tedious but
    \left(\begin{pmatrix}a & b \\ 0 & a \end{pmatrix}\begin{pmatrix}x & y \\ 0 & x\end{pmatrix}\right)\begin{pmatrix}p & q \\ 0 & p\end{pmatrix}= \begin{pmatrix}ax & ay+ bx \\ 0 & ax\end{pmatrix}\begin{pmatrix}p & q \\ 0 & p\end{pmatrix}
    = \begin{pmatrix}axp & axq+ ayp+ bxp) \\ 0 & axp \end{pmatrix}.

    Now calculate \begin{pmatrix}a & b \\ 0 &  \end{pmatrix}\left(\begin{pmatrix} x & y \\ 0 & x\end{pmatrix}\begin{pmatrix}p & q \\ 0 & p\end{pmatrix}\right)
    and show that you get the same thing.
    Thanks from gaussrelatz
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2013
    From
    Hong Kong
    Posts
    38

    Re: Help wth rings and subrings

    Quote Originally Posted by HallsofIvy View Post
    A "ring" is an algebraic structure that has two operations. We can think of them as "addition" and "multiplication".

    Looking only at the "addition", the set of objects is a "commutative group". That is, if x and y are two members of the set then x+ y= y+ x, there exist an additive identity, "0", such that
    x+ 0= x, and, for each x, there exist an additive inverse, y. such that x+ y= 0.

    We only require that the "multiplication" be "associative"- x(yz)= (xy)z. There is NO requirement that the multiplication be commutative or that there be a "multiplicative identity" or that every element have a "multiplicative identity".

    Here, since you are talking about a subset of "all 2 by 2 matrices", the operations are the usual ones for matrices:
    \begin{pmatrix}a & b \\ 0 & a\end{pmatrix}+ \begin{pmatrix}x & y \\ 0 & x\end{pmatrix}= \begin{pmatrix}a+ x & b+ y \\ 0 & a+ x\end{pmatrix}
    The fact that this operation is associative, has an additive identity, and has additive inverses follows from the fact that we "add matrices" by adding the numbers of corresponding entries in the matrices and that addition of numbers satisfies them.

    \begin{pmatrix}a & b \\ 0 & a \end{pmatrix}\begin{pmatrix}x & y \\ 0 & x\end{pmatrix}= \begin{pmatrix}ax & ay+ bx \\ 0 & ax\end{pmatrix}.

    Showing that multiplication is associative is a bit tedious but
    \left(\begin{pmatrix}a & b \\ 0 & a \end{pmatrix}\begin{pmatrix}x & y \\ 0 & x\end{pmatrix}\right)\begin{pmatrix}p & q \\ 0 & p\end{pmatrix}= \begin{pmatrix}ax & ay+ bx \\ 0 & ax\end{pmatrix}\begin{pmatrix}p & q \\ 0 & p\end{pmatrix}
    = \begin{pmatrix}axp & axq+ ayp+ bxp) \\ 0 & axp \end{pmatrix}.

    Now calculate \begin{pmatrix}a & b \\ 0 &  \end{pmatrix}\left(\begin{pmatrix} x & y \\ 0 & x\end{pmatrix}\begin{pmatrix}p & q \\ 0 & p\end{pmatrix}\right)
    and show that you get the same thing.
    Thank you that was very detailed so I could follow it ! So I see this is how you do it somewhat similar to showing a some subgroup of a group.
    Then how would I now find the homomorphism which satisfies the condition in the problem?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,462
    Thanks
    2893

    Re: Help wth rings and subrings

    The function mapping \begin{pmatrix}a & b \\ 0 & a\end{pmatrix} to a is certainly "onto". Now you need to show that it is a homomorphism.
    Thanks from gaussrelatz
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2013
    From
    Hong Kong
    Posts
    38

    Re: Help wth rings and subrings

    So essentially the given mapping in the problem is the actual mapping itself? So we dont have to find a new mapping? But also havent we already shown that it is a homomorphism when showing that it is a subring of $M_2(\mathbb{R})$?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2010
    Posts
    3,014
    Thanks
    1152

    Re: Help wth rings and subrings

    Quote Originally Posted by gaussrelatz View Post
    So essentially the given mapping in the problem is the actual mapping itself? So we dont have to find a new mapping? But also havent we already shown that it is a homomorphism when showing that it is a subring of $M_2(\mathbb{R})$?
    \phi: \begin{pmatrix}a & b \\ 0 & a\end{pmatrix} \mapsto a
    Is a function that takes a matrix to a real number. It is not a set of matrices. To prove it is a homeomorphism, you need to show that if A,B are matrices then \phi(A+B)=\phi(A)+\phi(B) and \phi(AB)=\phi (A)\phi (B)
    Thanks from gaussrelatz
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2013
    From
    Hong Kong
    Posts
    38

    Re: Help wth rings and subrings

    Could we also show that the above is a subring by the subring test? For instance:

    i) to show $S \ne \emptyset$,$ \begin{pmatrix}
    a & b \\
    0 & a \end{pmatrix}$ is indeed not the null set.
    ii) for all $ \begin{pmatrix}
    a & b \\
    0 & a \end{pmatrix}$ and ,$ \begin{pmatrix}
    c & d \\
    0 & c \end{pmatrix}$, we have ,$ \begin{pmatrix}
    ac & ad \\
    0 & ac \end{pmatrix}$ which is indeed in $R$.
    iii) for all for all $ \begin{pmatrix}
    a & b \\
    0 & a \end{pmatrix}$ and ,$ \begin{pmatrix}
    c & d \\
    0 & c \end{pmatrix}$ we have $ \begin{pmatrix}
    a-c & b-d \\
    0 & a-c \end{pmatrix}$ which is indeed in $R$.

    Does this work?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,462
    Thanks
    2893

    Re: Help wth rings and subrings

    Quote Originally Posted by gaussrelatz View Post
    So essentially the given mapping in the problem is the actual mapping itself?
    What "given mapping"? There was no mapping given, only a set of matrices.

    So we dont have to find a new mapping? But also havent we already shown that it is a homomorphism when showing that it is a subring of $M_2(\mathbb{R})$?
    And this makes no sense. Your two uses of "it" refer to two different things. the first "it" refers to a homomorphism while the second "it" refers to a subring of matrices.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,462
    Thanks
    2893

    Re: Help wth rings and subrings

    Quote Originally Posted by gaussrelatz View Post
    Could we also show that the above is a subring by the subring test? For instance:

    i) to show $S \ne \emptyset$,$ \begin{pmatrix}
    a & b \\
    0 & a \end{pmatrix}$ is indeed not the null set.
    ii) for all $ \begin{pmatrix}
    a & b \\
    0 & a \end{pmatrix}$ and ,$ \begin{pmatrix}
    c & d \\
    0 & c \end{pmatrix}$, we have ,$ \begin{pmatrix}
    ac & ad \\
    0 & ac \end{pmatrix}$ which is indeed in $R$.
    We have that for what? The sum of those two matrices is \begin{pmatrix}a+ c & b+ d \\ 0 & a+ c\end{pmatrix} and the product is $\begin{pmatrix}ac & ad+ bc \\ 0 & ac\end{pmatrix}$. Neither of those is what you give.

    iii) for all for all $ \begin{pmatrix}
    a & b \\
    0 & a \end{pmatrix}$ and ,$ \begin{pmatrix}
    c & d \\
    0 & c \end{pmatrix}$ we have $ \begin{pmatrix}
    a-c & b-d \\
    0 & a-c \end{pmatrix}$ which is indeed in $R$.

    Does this work?
    That last is the difference between two matrices but that is not really relevant. The sum of two matrices is more fundmental.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rings, subrings problem
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Apr 21st 2013, 05:39 PM
  2. Rings and subrings
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Mar 7th 2013, 03:49 PM
  3. Abstract Algebra - Rings and Subrings
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Feb 19th 2013, 08:28 PM
  4. Rings and subrings
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Feb 7th 2010, 01:30 PM
  5. Rings and Subrings
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Nov 24th 2007, 04:19 PM

/mathhelpforum @mathhelpforum