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November 24th, 2017, 06:03 AM  #1 
Newbie Joined: Oct 2017 From: sweden Posts: 14 Thanks: 0  Understanding ring homomorphisms
So trying to study chapters on ring homomophisms and integral domains, and there is this one problem in the book where I just don't get how to show the kind of problems yet (as I'm not yet used to it); the problem is: Show that the matrices $R=\begin{pmatrix} a & b \\ 0 & a \\ \end{pmatrix}$ 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 onto. I will appreciate the help as I am very new to ring homomoprhisms. Last edited by skipjack; November 24th, 2017 at 12:41 PM. 
November 24th, 2017, 10:54 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,919 Thanks: 785 
You know what a ring is don't you? To show that the set of all matrices of the form $\displaystyle \begin{pmatrix}a & b \\ 0 & a\end{pmatrix}$ is a ring, you need to show 1) It is closed under matrix addition $\displaystyle \begin{pmatrix}p & q\\ 0 & p \end{pmatrix}+ \begin{pmatrix}r & s \\ 0 & t\end{pmatrix}= \begin{pmatrix}p+ r & q+ s \\ 0 & p+ r\end{pmatrix}$ is of that same form. 2) It contains the additive identity: $\displaystyle \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}$ is of this form with $\displaystyle a= b= 0$. 3) Addition is associative. $\displaystyle \left(\begin{pmatrix}a & b \\ 0 & a \end{pmatrix}+ \begin{pmatrix}c & d \\ 0 & c\end{pmatrix}\right)+ \begin{pmatrix}e & f \\ 0 & e \end{pmatrix}= \begin{pmatrix}a+ c & b+ d \\ 0 & a+ c\end{pmatrix}+ \begin{pmatrix}e & f \\ 0 & e\end{pmatrix}= \begin{pmatrix} (a+ c)+ e & (b+ d)+ f \\ 0 & (a+ c)+ e\end{pmatrix}$ $\displaystyle \begin{pmatrix}a & b \\ 0 & a \end{pmatrix}+ \left(\begin{pmatrix}c & d \\ 0 & c\end{pmatrix}+ \begin{pmatrix}e & f \\ 0 & e \end{pmatrix}\right)= \begin{pmatrix}a & b \\ 0 & a\end{pmatrix}+ \begin{pmatrix}c+e & d+f \\ 0 & c+e\end{pmatrix}= \begin{pmatrix} a+ (c+ e) & b+ (d+ f) \\ 0 & a+ (c+ e)\end{pmatrix}$ Essentially, that follows from the fact that matrix addition is "element wise" and addition of numbers is "associative". For multiplication, you only have to prove that 4) It is closed under matrix multiplication. $\displaystyle \begin{pmatrix}a & b \\ 0 & a \end{pmatrix}\begin{pmatrix}c & d \\ 0 & c \end{pmatrix}= \begin{pmatrix}ac & ad+ bc \\ 0 & ac\end{pmatrix}$ In a ring, there is not necessarily a multiplicative identity nor multiplicative inverses, so we don't need to show that. (Actually, this is a "ring with identity: $\displaystyle \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$ with $\displaystyle a= 1,\ b= 0$ is the multiplicative identity. The matrix $\displaystyle \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}$ is of this form, but does not have an inverse.) 5) Multiplication is associative: $\displaystyle \left(\begin{pmatrix}a & b \\ 0 & a \end{pmatrix}\begin{pmatrix}c & d \\ 0 & c\end{pmatrix}\right)\begin{pmatrix}e & f \\ 0 & e\end{pmatrix}= \begin{pmatrix}ac & ad+ bc \\ 0 & ac\end{pmatrix}\begin{pmatrix}e & f \\ 0 & e\end{pmatrix}= \begin{pmatrix} ace & acf+ ade+ bce \\ 0 & ace\end{pmatrix}$ A ring homomorphism that is "onto" (but not onetoone) maps $\displaystyle \begin{pmatrix} a & b \\ 0 & a\end{pmatrix}$ is simply "a". Last edited by skipjack; November 24th, 2017 at 11:26 PM. 
November 24th, 2017, 04:05 PM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,919 Thanks: 785 
Above I neglected the second part of "multiplication is associative". $\displaystyle \begin{pmatrix}a & b \\ 0 & a\end{pmatrix}\left(\begin{pmatrix}c & d \\ 0 & c\end{pmatrix}\begin{pmatrix}e & f \\ 0 & e\end{pmatrix}\right)$$\displaystyle = \begin{pmatrix}a & b \\ 0 & a\end{pmatrix}\begin{pmatrix}ce & cf+ de \\ 0 & ce\end{pmatrix}$$\displaystyle = \begin{pmatrix}ace & acf+ ade+ bce \\ 0 & ace\end{pmatrix}$. 
November 24th, 2017, 04:23 PM  #4 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 112 Thanks: 35 Math Focus: Algebraic Number Theory, Arithmetic Geometry  You still need to show 2') $R$ contains additive inverses 6) addition is commutative 7) multiplication and addition distribute over each other Note that 2') and 1) immediately imply 2). Also, the fact that $R$ is a subset of the ring $M_2(\mathbb{R})$ and has the same addition/multiplication (except restricted to $R$) gives 3), 5), 6) and 7) for free. 
December 31st, 2017, 08:51 AM  #5 
Senior Member Joined: Oct 2009 Posts: 180 Thanks: 70  
December 31st, 2017, 10:03 AM  #6  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 112 Thanks: 35 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
I've noticed a mistake in my post, though  conditions 2)' and 1) only give 2) (existence of additive identity) if we know the set is nonempty. And to show the set is nonempty, often the best way to go is to show 2) directly!  
January 2nd, 2018, 08:25 AM  #7  
Senior Member Joined: Oct 2009 Posts: 180 Thanks: 70  Quote:
 
January 2nd, 2018, 05:49 PM  #8  
Member Joined: Aug 2011 From: Nouakchott, Mauritania Posts: 84 Thanks: 14 Math Focus: Algebra, Cryptography 
Salam ! Quote:
$\displaystyle \begin{array}{cccc} \Phi:&R &\rightarrow&\mathbb{R}\\ &M=\begin{pmatrix}a&b\\0&a\end{pmatrix}&\mapsto & \Phi(M)=a\end{array}$. Now you can try to check if :  $\displaystyle \Phi(M+N)=\Phi(M)+\Phi(N)$  $\displaystyle \Phi(MN)=\Phi(M)\Phi(N)$  $\displaystyle \Phi(1_R)=1$ ( Some references don't require this condition ) for $\displaystyle M,N\in R$. In this case, $\displaystyle \Phi$ will be a ring homomorphism. Finally, $\displaystyle \Phi$ is a surjective (onto) application because : $\displaystyle \forall a\in\mathbb{R}, \exists M=aI_2\in R,\quad\phi(M)=a.$  

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