# Thread: Help wth rings and subrings

1. ## 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.

2. ## 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:
$\displaystyle \begin{pmatrix}a & b \\ 0 & a\end{pmatrix}+ \begin{pmatrix}x & y \\ 0 & x\end{pmatrix}= \begin{pmatrix}a+ x & b+ y \\ 0 & a+ x\end{pmatrix}$
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.

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

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

Now calculate $\displaystyle \begin{pmatrix}a & b \\ 0 & \end{pmatrix}\left(\begin{pmatrix} x & y \\ 0 & x\end{pmatrix}\begin{pmatrix}p & q \\ 0 & p\end{pmatrix}\right)$
and show that you get the same thing.

3. ## Re: Help wth rings and subrings

Originally Posted by HallsofIvy
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:
$\displaystyle \begin{pmatrix}a & b \\ 0 & a\end{pmatrix}+ \begin{pmatrix}x & y \\ 0 & x\end{pmatrix}= \begin{pmatrix}a+ x & b+ y \\ 0 & a+ x\end{pmatrix}$
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.

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

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

Now calculate $\displaystyle \begin{pmatrix}a & b \\ 0 & \end{pmatrix}\left(\begin{pmatrix} x & y \\ 0 & x\end{pmatrix}\begin{pmatrix}p & q \\ 0 & p\end{pmatrix}\right)$
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?

4. ## Re: Help wth rings and subrings

The function mapping $\displaystyle \begin{pmatrix}a & b \\ 0 & a\end{pmatrix}$ to a is certainly "onto". Now you need to show that it is a homomorphism.

5. ## 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 $M_2(\mathbb{R})$?

6. ## Re: Help wth rings and subrings

Originally Posted by gaussrelatz
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 $M_2(\mathbb{R})$?
$\displaystyle \phi: \begin{pmatrix}a & b \\ 0 & a\end{pmatrix} \mapsto a$
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 $\displaystyle \phi(A+B)=\phi(A)+\phi(B)$ and $\displaystyle \phi(AB)=\phi (A)\phi (B)$

7. ## 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?

8. ## Re: Help wth rings and subrings

Originally Posted by gaussrelatz
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.

9. ## Re: Help wth rings and subrings

Originally Posted by gaussrelatz
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 $\displaystyle \begin{pmatrix}a+ c & b+ d \\ 0 & a+ c\end{pmatrix}$ 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.