# Definition talk:Ring Extension

Should:

Then $R \subseteq S$ is a ring extension of $R$.

Then $S$ is a ring extension of $R$.

? --prime mover 17:01, 5 May 2011 (CDT)

Kind of... realy both rings are part of the extension; but good point, formally the ring extension should be a homomorphism:
$\phi: R \to S$ I'll change it --Linus44 17:07, 5 May 2011 (CDT)
okay, I'm seeing the notation but I can't help but think it's overcomplicated ... intuitively I see it that "if $R$ is a subring of $S$, then $S$ is an extension of $R$". Or is it that the "extension" is not the "thing" that is the super-ring, so much as the "process" which generates the super-ring? I've tried looking it up in various places and it appears to be something that is generally glossed over in the literature and we can add value to the site by including something nobody else does. --prime mover 00:31, 6 May 2011 (CDT)

I always thought of an extension as the latter: we cant just let $S$ be the extension, because $S$ has, in general, more than one subfield, and a homomorphism is a convenient way of identifying both with one object. But I guess it's just taste. Grillet discusses this a litte (he calls it surgery). The following is all for fields (it's the same for rings):$\newcommand{\im}{\operatorname{im}}$

(1): $K \subseteq L$

• Then there is a homomorphism $\iota: K\to L$ (standard inclusion)
• If $\phi:L \simeq M$ isomorphism, then $\phi\circ \iota : K \to M$ is a homomorphism.
• If $\psi : K \to N$ isomorphism, then $\iota \circ \psi^{-1} : N \to L$ is a homomorphism.

(2) $\phi : K \to L$

• Then $\im \phi \subseteq L$ satisfies (1)
• There exists a field $M \simeq L$ containing $K$. (*)

Proof of (*):

Let $M = K \cup (F \setminus \im \phi)$ (disjoint union).

Let $\theta : M \to L$, $K \owns x \mapsto \psi(x)$, and the identity on $(F \setminus \im \phi)$.

For $x,y\in M$ define

$x + y = \theta^{-1}(\theta(x) + \theta(y))$
$xy = \theta^{-1}(\theta(x)\theta(y))$

Then $K \subseteq M$ and $\theta :M \to L$ is an isomorphism.

I see where it's coming from. Along with the result Ring Homomorphism Preserves Subrings the two definitions are equivalent up to isomorphism, so ultimately they say the same thing, except that the definition as an isomorphism / injective homomorphism is more general.
Recommend we include both definitions with some weasel-words linking them. --prime mover 14:21, 6 May 2011 (CDT)
"Surgery for rings"? This site's for maths not proctology! Surely the existing results (extension theorems, injective homomorphisms, etc.) are sufficient? No matter, Grillet probably knows what he's doing ... I'll let you get on with it. --prime mover 15:18, 7 May 2011 (CDT)
It's no more silly than "contour surgery" in complex analysis. Admittedly I think it's just me thats uses that phrase too.
If you have a way to cover it, change it if you fancy; can't say i'm too excited about finishing the page --Linus44 15:46, 7 May 2011 (CDT)
Okay no worries ... I'll flag it up as WIP and when I get to that point I'll see what I can do about it. Just that I'm faiurly sure this has all been covered somewhere already, just a matter of working out where I put it. --prime mover 16:30, 7 May 2011 (CDT)