Homomorphic images of nth power of a division ring

Question

Let $R$ be a division ring and $n \geq 2$. I need to identify up to isomorphism all homomorphic images of the $n$-th power of $R$:

$$ R^{n} = R \oplus R \oplus \cdots \oplus R\quad (\text{for}\,n\, \text{summands}\, R)$$

(Note that $R$ is not given to be a ring with identity. I don't use the ring/rng terminology. BUT, see next parenthetical statement.)

Now, recall that a division ring $R$ is a ring such that the set $R\setminus\{ 0\}$ of nonzero elements of $R$ is a group under multiplication, meaning every nonzero element has a multiplicative inverse (so does this mean then that $R$ IS a ring with identity??), but $R$ is not necessarily commutative.

Another thing I know is that every division ring is an integral domain, but I also know that the property of not having zero divisors is not preserved under the direct sum.

Finally, by the First Isomorphism Theorem, I know that if $f:R\to S$ is a ring homomorphism, then $R/\ker f \simeq \text{Im} f$.

From this point, I do not know how to proceed. How do I put all of these things together to prove what I need to prove? Is induction required here?

Please help - and I thank you for your time and patience

Answer 1

A division ring $R$ necessarily has an identity: as you say, $R\setminus\{0\}$ is a group, so it has an identity.

Hint: an ideal of a direct sum of rings with identity $R\oplus S$ has the form $I\oplus J$, where $I$ and $J$ are ideals of $R$ and $S$ respectively.

Use induction for the general case of $n$ rings. What are the (two-sided) ideals of a division ring?

An ideal of $R^n$ is of the form $I_1\oplus I_2\oplus\dots\oplus I_n$, where $I_k=\{0\}$ or $I_k=R$. If $t$ of them are equal to $\{0\}$, the quotient is isomorphic to $R^t$.

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Proposition. Let $R$ and $S$ be rings with identity and let $K\subseteq R\oplus S$. Then $K$ is an ideal if and only if $K=I\oplus J$, where $I$ and $J$ are ideals in $R$ and $S$ respectively.

Proof. Suppose $I$ and $J$ are ideals in $R$ and $S$ respectively. Then $I\oplus J$ is an ideal of $R\oplus S$: for $a,c\in I$ and $b,d\in J$ we have $(a,b)+(c,d)=(a+c,b+d)\in I\oplus J$; for $a\in I$, $b\in J$, $x\in R$ and $y\in S$ we have $(a,b)(x,y)=(ax,by)\in I\oplus J$ and $(x,y)(a,b)=(xa,yb)\in I\oplus J$.

Conversely, let $K$ be an ideal in $R\oplus S$. Set $I=\{a\in R:(a,0)\in K\}$ and $J=\{b\in S:(0,b)\in K\}$. It's easy to prove that $I$ and $J$ are ideals in $R$ and $S$ respectively. If $(a,b)\in K$, then $$ (a,0)=(1,0)(a,b)\in K,\qquad (0,b)=(0,1)(a,b)\in K, $$ so $a\in I$ and $b\in J$. Therefore $(a,b)\in I\oplus J$. QED

By induction, this extends to a direct sums of $n$ rings.

Answer 2

I presume everything is associative; you are not dealing with the octonions or worse.

If you have a group, then $R-\{0\}$ had better have an identity, so $R$ is a ring with $1$.

If you could classify all (two-sided) ideals in $R^n$ you'd be in business. For each subset $A$ of $ [1,\ldots,n]$ there's an ideal $I_A$ of $R^n$, which consists of all vectors in $R^n$ whose only nonzero entries are contained within the positions indexed by $A$. Then $R^n/I_A\cong R^{n-|A|}$ (why?). If you could show that the $I_A$ are all the ideals, you'd be in business.