Abstract Nonsense

Crushing one theorem at a time

Third Ring Isomorphism Theorem


Point of Post: In this post we prove the third ring isomorphism theorem which takes the form \left(R/\mathfrak{b}\right)/\left(\mathfrak{a}/\mathfrak{b}\right)\cong R/\mathfrak{a}.

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Motivation

Similar to the second ring isomorphism theorem the third isomorphism theorem has an almost immediate proof from the first ring isomorphism theorem (which, of course, both really should harken back to the proofs of the first and second group isomorphism theorems since the ring isomorphism theorems are generalizations (for abelian groups at least) if we define the trivial product on G given by ab=0 for all a,b\in G). Thus, just like the second ring isomorphism theorem the real meat of the theorem, the real thing that need be considered is precisely how the question “is (R/\mathfrak{a})/(\mathfrak{a}/\mathfrak{b})\cong R/\mathfrak{b}?’ naturally comes up in our study of rings. So, suppose for a second that we have some ring R and two R-ideals \mathfrak{a} and \mathfrak{b} with \mathfrak{a}\supseteq\mathfrak{b}. We first note that \mathfrak{a}/\mathfrak{b}=\left\{a+\mathfrak{b}:a\in\mathfrak{a}\right\}\subseteq \left\{r+\mathfrak{b}:r\in R\right\}=R/\mathfrak{b} and moreover that \mathfrak{a}/\mathfrak{b} is an ideal of R/\mathfrak{b} since it’s the image of \mathfrak{a} under the canonical projection \pi:R\twoheadrightarrow R/\mathfrak{b} and we know that ideals are preserved under epimorphisms. So, it makes sense to consider the quotient ring \left(R/\mathfrak{b}\right)/(\mathfrak{a}/\mathfrak{b}). That said, the elements of this quotient ring are, for lack of a better word, horrific. I mean, the general element looks like \left(r+\mathfrak{b}\right)+\mathfrak{a}/\mathfrak{b}. A wish would be that there was some nice way to describe this ‘double quotient ring’ as something nicer, something less unwieldy. The third ring isomorphism theorem grants us this wish by telling us that the \mathfrak{b}‘s “cancel” in the sense that (R/\mathfrak{b})/(\mathfrak{a}/\mathfrak{b})\cong R/\mathfrak{a}. To see the intuition of this theorem let’s look at the case when R=\mathbb{Z}. We have by previous comment that the ideals of \mathbb{Z} are all of the form n\mathbb{Z} for some n\in\mathbb{N}\cup\{0\}. So, suppose we had two \mathbb{Z}-ideals, say n\mathbb{Z} and m\mathbb{Z}. We note that evidently n\mathbb{Z}\supseteq m\mathbb{Z} if and only if n\mid m.We then note that what n\mathbb{Z}/m\mathbb{Z} ‘looks like’ (by definition)  the set \left\{\cdots,-n+\mathbb{Z}/m\mathbb{Z},0+\mathbb{Z}/m\mathbb{Z},n+\mathbb{Z}/m\mathbb{Z},\cdots\right\}. So, now let’s list the elements of \left(\mathbb{Z}/m\mathbb{Z}\right)/\left(n\mathbb{Z}/m\mathbb{Z}\right). They look like:

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\left(0+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z},\;\left(1+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z},\;\cdots\;\left(m-1+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z}

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We note then that the first n of these are distinct since

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\left(k+\mathbb{Z}/m\mathbb{Z}\right)-\left(\ell+\mathbb{Z}/m\mathbb{Z}\right)=\left(\ell-k\right)+\mathbb{Z}/m\mathbb{Z}\notin n\mathbb{Z}/m\mathbb{Z}

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if \ell,k\in\{0,\cdots,n-1\}. That said, for elements beyond the first n listed we see the elements are of the form \left(jn+\ell+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z} for \ell\in\{0,\cdots,n-1\} in which case it is equal to \left(\ell+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z} since \left(jn+\ell\right)+\mathbb{Z} /m\mathbb{Z}-\left(\ell+\mathbb{Z}/m\mathbb{Z}\right)=jn+\mathbb{Z}\in n\mathbb{Z}/m\mathbb{Z}. In particular we see that \left(x+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z}=\left(\left(x\text{ mod }n\right)+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z}. Thus, from this we are able to conclude that as a set \left(\mathbb{Z}/m\mathbb{Z}\right)/\left(n\mathbb{Z}/m\mathbb{Z}\right)=\left\{\left(k+\mathbb{Z}/m\right)+n\mathbb{Z}/m\mathbb{Z}:k=0,\cdots,n-1\right\}.Moreover, it’s easy to extract from this that for k\in\{0,\cdots,n-1\} we have that

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\begin{aligned}\left(\left(x+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z}\right)\left(\left(y+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z}\right) &= \left(\left(x+\mathbb{Z}/m\mathbb{Z}\right)\left(y+\mathbb{Z}/m\mathbb{Z}\right)\right)+n\mathbb{Z}/m\mathbb{Z}\\ &= \left(xy+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z}\\ &= \left(\left(xy\text{ mod }n\right)+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z}\end{aligned}

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and similarly

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\begin{aligned}\left(\left(x+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z}\right)+\left(\left(y+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z}\right) &= \left(\left(x+\mathbb{Z}/m\mathbb{Z}\right)+\left(y+\mathbb{Z}/m\mathbb{Z}\right)\right)+n\mathbb{Z}/m\mathbb{Z}\\ &= \left((x+y)+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z}\\ &= \left(\left(x+y\text{ mod }n\right)+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z}\end{aligned}

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Thus, f:\left(\mathbb{Z}/m\mathbb{Z}\right)/\left(n\mathbb{Z}/m\mathbb{Z}\right)\to\mathbb{Z}_n:\left(x+\mathbb{Z}/m\mathbb{Z}\right)+n\mathbb{Z}/m\mathbb{Z}\mapsto x\text{ mod }n is an isomorphism, and since \mathbb{Z}_n\cong\mathbb{Z}/n\mathbb{Z} the conclusion follows.

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This example should give you an intuition then why the full theorem is true, with the intuitive idea that what makes this theorem true is precisely the reason why \left(a\text{ mod }n\right)\text{ mod }2n=a\text{ mod }n–once the first hits, the second isn’t affected.

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Third Ring Isomorphism Theorem

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We now actually prove the third ring isomorphism theorem. The proof is much, much simpler than the above fully-worked example for why \left(\mathbb{Z}/m\mathbb{Z}\right)/\left(n\mathbb{Z}/m\mathbb{Z}\right)\cong \mathbb{Z}/n\mathbb{Z} if n\mid m. Most of this is masked because we’ll use the first isomorphism theorem

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Theorem: Let R be a ring with ideals \mathfrak{a} and \mathfrak{b} with \mathfrak{a}\supseteq\mathfrak{b}. Then, 

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\displaystyle \frac{\displaystyle \left(\frac{R}{\mathfrak{b}}\right)}{\displaystyle \left(\frac{\mathfrak{a}}{\mathfrak{b}}\right)}\cong \frac{R}{\mathfrak{b}}

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Proof: Define f:R/\mathfrak{b}\to R/\mathfrak{a} by r+\mathfrak{b}\mapsto r+\mathfrak{a}. It’s not evident that this map is well-defined–to see this suppose that r+\mathfrak{b}=s+\mathfrak{b} then r-s\in\mathfrak{b}\subseteq\mathfrak{a} and so by definition f(r)=r+\mathfrak{a}=s+\mathfrak{a}=f(s), from where well-definedness follows. We now claim that f is a morphism. To see this let r+\mathfrak{b},s+\mathfrak{b}\in R/\mathfrak{b} be arbitrary, we see then that

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\begin{aligned}f\left(\left(r+\mathfrak{b}\right)\left(s+\mathfrak{b}\right)\right) &=f\left(rs+\mathfrak{b}\right)\\ &=rs+\mathfrak{a}\\ &=\left(r+\mathfrak{a}\right)\left(s+\mathfrak{a}\right)\\ &=f(r)f(s)\end{aligned}

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and

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\begin{aligned}f\left(\left(r+\mathfrak{b}\right)+\left(s+\mathfrak{b}\right)\right) &=f\left((r+s)+\mathfrak{b}\right)\\ &=(r+s)+\mathfrak{a}\\ &=\left(r+\mathfrak{a}\right)+\left(s+\mathfrak{a}\right)\\ &=f(r)+f(s)\end{aligned}

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Evidently f is epic (an epimorphism) and by definition we see that f(r+\mathfrak{b})=r+\mathfrak{a}=0+\mathfrak{a} if and only if r\in\mathfrak{a}, or in oher words r\in\ker f if and only if r\in \mathfrak{a}/\mathfrak{b}. The conclusion then follows from the first ring isomorphism theorem. \blacksquare

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

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July 1, 2011 - Posted by | Algebra, Ring Theory | , , , , , , , ,

2 Comments »

  1. […] from the fourth ring isomorphism theorem that the ideals of are precisely with . But, by the third ring isomorphism theorem we know that and so, in particular, we have that is an integral domain if and only if is prime. […]

    Pingback by Prime Ideals (Pt. II) « Abstract Nonsense | August 17, 2011 | Reply

  2. […] theorems we are going to discuss have the same feel and motivation as the first, second, third, and fourth ring isomorphism theorems with “ring” replaced by “left […]

    Pingback by The Module Isomorphism Theorems « Abstract Nonsense | November 11, 2011 | Reply


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