2
$\begingroup$

For which positive integer $n$ does there exist a $\Bbb R$-linear ring homomorphism $f:\Bbb C \to M_n(\Bbb R)?$

If I send everything to zero matrix that will give trivial ring homomorphism for every $n$ .

Now nontrivial ring homomorphism: for $n=1$ there is no such homomorphism from $\Bbb C$ to $M_n(\Bbb R)$ as there is no ring homomorphism from $\Bbb C$ to $\Bbb R$. Now $M_n(\Bbb R)$ is not an integral domain for $n\ge2$, so $f(1)$ will be an idempotent element. I am not getting any idea to proceed further.

Thanks in advance.

  • 0
    no .f(1) may not be 1 @JonasMeyer – 2017-01-31
  • 1
    If you don't require $f(1)=1$ then you can get all $n\geq 2$, just by embedding in the top left corner after you figure out how to get $n=2$. If you do require $f(1) = 1$ then not all dimensions are possible because of the equation $f(i)^2=-1$. – 2017-01-31
  • 0
    is it true that, if f(1)=1 then it may be possible to find such homomorphism for n= even case only@JonasMeyer – 2017-01-31
  • 0
    If you are looking for a ring homomorphism $f(1)=id$ is a necessarily conditions: because every ring homomorphism $f:A\to B$ between unitary ring send $1_A$ to $1_B$ – 2017-01-31
  • 0
    no i think this is not true always.there are ring homomorphism between ring with unit but does not send id to id @Legoman – 2017-01-31
  • 0
    Could you give an example of such homomorphism? – 2017-01-31
  • 0
    http://math.stackexchange.com/questions/179842/ring-homomorphism-with-phi1-r-neq1-s look at this math stack question@Legoman – 2017-01-31
  • 0
    Mmmm I am so confused now, because if you want to include a ring $R$ in $R\times R$ the best homomorphism is $r\rightarrow (r,r)$ and here again $1\longrightarrow (1,1)$. If $r$ is such that $r^2=1_R$ it is not true that $(r,r)\dot(r,r)=(1,1)$ because the second component is zero. So it don't preserve the structure. I know that Atyiah and MacDonald use the condition that $f(1)=1$, and I have never seen people that doesn't use it. I am confused now. – 2017-02-03
  • 0
    Most people require ring homomorphisms to send the identity element of one to the identity element of the other, but this requirement can be dropped as desired. @Legoman If $r^2=1$ then we have $(r,r)(r,r)=(r^2,r^2)=(1,1)$, so I'm not sure what you're talking about with your example. – 2017-02-14

2 Answers 2

2

When $n=2$, you have $f(a+ib) = \begin{bmatrix} a & -b \\ b&a\end{bmatrix}$ as an example of a unital $\mathbb R$-algebra homomorphism. This 2-by-2 matrix representation of complex numbers is mentioned in Wikipedia.

For $n>2$, if you do not require $f(1)=1$ (as is a common requirement in some contexts), then the easiest way to proceed is to put the $n=2$ example in the top left corner with zeros elsewhere.

If you do have $f(1)=1$, one restriction comes from $f(i)^2 = -1$, which requires $n$ to be even. When $n$ is even, the easiest way to proceed is to put the $n=2$ example $n/2$ times into block diagonal matrices. When $n=4$ this gives a restriction to $\mathbb C$ of a real matrix representation of quaternions.

2

If there is an $\mathbb R$-linear homomorphism $\mathbb C\to M_n(\mathbb R)$, then $\mathbb R^n$ can be turned into a complex vector space compatibly with its usual real vector space structure. Of course, this implies that $n$ is even.

Now, if $n$ is even, then we may identify $\mathbb R^n$ with $\mathbb C^{n/2}$, and there is a obvious may to let $\mathbb C$ act on the latter. Picking a real basis, this gives a map $\mathbb C\to M_n(\mathbb R)$.

  • 0
    I just want to clear my doubt. If f:C→Mn(R), is an R-linear homomorphism, then Rn can be turned into a complex vector space by defining multiplication i.(a1,a2....an) as f(i).(a1,a2..an). I am really sorry but it is not clear to me what is the meaning of this complex v.sp compatible with real v.sp structure.would you please explain it @ Mariano Suárez-Álvarez – 2017-02-14
  • 0
    Indeed, the R-linear morphism of rings makes R^n into a compex vector space in the way you describe. What I mean by compatibility is: every complex vector space is a rea vector space by restriction of scalar. Now R^n es on one hand a real vector space in the usual way, and, on the other hand, a complex vector space so that it is by restriction of scalars a real vector space in a possibly different way: compatibility means that in fact these two real vector space structures are the same. – 2017-02-14