# The Unapologetic Mathematician

## The First Isomorphism Theorem (for rings)

Just like we had for groups, there is an isomorphism theorem for rings. In fact, the demonstration goes much the same as it did there.

Any subring $S$ of a ring $R$ comes equipped with an inclusion homomorphism $\iota_{(R,S)}:S\rightarrow R$. Any quotient of a ring $R$ by and ideal $I$ comes with a projection homomorphism $\pi_{(R,I)}:G\rightarrow G/I$. For both of these, just construct the inclusion or projection homomorphism for the underlying abelian groups and check that it preserves the multiplication. Just like before, the inclusion is a monomorphism, the projection is an epimorphism, and the kernel of the projection — the set of elements of $R$ that get sent to zero — is the ideal $I$.

Now given any ring homomorphism $f:R\rightarrow R'$, the kernel is an ideal. Indeed, if $f(x)=0$ and $f(x')=0$ then
$f(x+x')=f(x)+f(x')=0+0=0$
$f(rx)=f(r)f(x)=f(r)0=0$
$f(xr)=f(x)f(r)=0f(r)=0$
So the set of elements that get sent to zero is closed under addition and under left and right multiplication by any element of the ring.

Also, the image of $f$ is a subring of $R'$. Given elements $f(r)$ and $f(r')$ in the image of $f$ we have $f(r)f(r')=f(rr')$ and $f(r)+f(r')=f(r+r')$, so sums and products of elements of the image are again in the image.

Now we want to take $f$ and make an isomorphism $\bar{f}:R/{\rm Ker}(f)\rightarrow{\rm Im}(f)$. For any coset in $R/{\rm Ker}(f)$, pick a representative element $r$ of $R$ and define $\bar{f}(r+{\rm Ker}(f))=f(r)$. Clearly this lands in ${\rm Im}(f)$, but does it really define a homomorphism from $R/{\rm Ker}(f)$? Indeed it does, because any other representative of the same coset looks like $r+x$ for some element $x$ of the kernel. Then
$\bar(f)(r+x+{\rm Ker}(f))=f(r+x)=f(r)+f(x)=f(r)+0=f(r)$
so we get the same answer — the value of $\bar{f}$ doesn’t depend on the choice of representative we make.

Is $\bar{f}$ a monomorphism? Yes, because if $\bar{f}(r+{\rm Ker}(f)=0$ then $f(r)=0$, so $r$ is a representative of ${\rm Ker}(f)$, which takes the place of ${}0$ in $R/{\rm Ker}(f)$. Is it an epimorphism? Yes, because every element of ${\rm Im}(f)$ comes from some element $r$ of $R$, so we can hit it by taking $\bar{f}(r+{\rm Ker}(f))$.

Putting it all together we can factor any homomorphism $f$ into the composition of an epimorphism $\pi_{(R,{\rm Ker}(f))}$, an isomorphism $\bar{f}$, and a monomorphism $\iota_{(R',{\rm Im}(f))}$. All homomorphisms of rings work this way: factor out some kernel, then send the quotient isomorphically to some subring of the target ring. Again, all the interesting stuff really happens in the first step. Studying homomorphisms from a given ring really comes down to studying the possible ideals of a given ring. In particular, if a ring has no ideals but the whole ring itself and the ideal consisting only of ${}0$ we call it “simple”. Every homomorphic image of a simple ring is either zero or the whole ring itself.

As I said above, this really looks a lot like what we did for groups, and there’s actually a very good reason why that I want to put off a while longer. Essentially, the fact that we have an isomorphism theorem like this doesn’t depend on the “groupiness” or the “ringiness” of the objects we’re studying, but on deeper structure shared by both groups and rings — or rather shared by group and ring homomorphisms.

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Posted by John Armstrong | Ring theory

## 2 Comments	»

1. [...] the first isomorphism theorem for rings tells us that we can impose our relation by taking the quotient ring . But what we just discussed [...]

Pingback by | August 7, 2008 | Reply

2. [...] object, kernels, and cokernels, which is enough to get the first isomorphism theorem, just like for rings. Specifically, if is any homomorphism of Lie algebras then we can factor it as [...]

Pingback by | August 15, 2012 | Reply

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