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LECTURE 14

The Isomorphism Theorems

The idea of quotient spaces developed in the last lecture is fundamental to modern mathematics. Indeed,

the basic idea of quotient spaces, from a suitably abstract perspective, is just as natural and important as

the notion of a subspace. Below we give the three theorems, variations of which are foundational to group

theory and ring theory. (A vector space can be viewed as an abelian group under vector addition, and a

vector space is also special case of a ring module.) Theorem14.1 (First Isomorphism Theorem).Let:V!Wbe a homomorphism between two vector spaces over a eldF. (i)The kernel ofis a subspace ofV: (ii)The image ofis a subspace ofW. (iii)The image ofis isomorphic to the quotient spaceV=ker().

Proof.We have proved (i) and (ii) early on in our initial discussion of linear transformations between vector

spaces. IfVis nitely generated (iii) is pretty simple. LetB=fb1;:::;bng. Then, by Theorem 11.6 in

Lecture 11, we have

Im() =span((b1);:::;(bn))

and by Corollary 11.7 in the same lecture dimV= dimIm() + dimker()

On the other hand, at the end of the preceding lecture we had the result that ifSis a subspace ofVthen

dimV= dim(S) + dim(V=S)

Using ker() forS, we conclude that

dim(V=ker(T)) = dim(Im) Theorem 12.6 says that if two nite-dimensional vector spaces have the same dimension then they are isomorphic. Therefore,

V=ker(T)=Im()

However, the statement of Theorem 14.1 is true even whenVandWare ininite dimensional. Again,

since the proofs of (i) and (ii) used only the dening properties of vector space homomorphisms, we have

eectively already demonstrated the validity of (i) and (ii) in the innite-dimensional setting. To prove (iii) in the situation where both the domainVand the codomainWmight be innite-dimensional, we'll display the isomorphism betweenV=ker() and Im() explicitly.

Before proving (iii), let me rst establish an importantuniversality propertyof a canonical projection

p S:V7!V=S. This property deals with the situation where one has both a linear transformation :V!Wand a particular subspace ofV. We thus have two linear transformations V !W p S# V=S 1

14. THE ISOMORPHISM THEOREMS 2

Theorem14.2.:V!Wis a linear transformation and thatSis a subspace of a vector spaceV contained in the kernel of. Then there is a unique linear transformation:V=S!Wwith the property that (i)pS= :

Moreover,

(ii) ker() = ker()=S and (iii)im() =im() Proof.The stipulation (i) amounts the condition that the following diagram V !W p S# % V=S commutes; in other words, (v) =(pS(v)) =(v+S) Now the functionwould be well dened onV=Sif and only if (a)v+S=u+S)(v+S) =(u+S) (because to compute the value of(v+S), we rst need to choose a representativevofv+SinV). But now, on the one hand, (a) is equivalent to each of the following statements v+S=u+S)(v+S) =(u+S) ()(v) =(u) ()(vu) =0W

So suppose

v+S=u+S

Then, since by hypothesisSker(),

vu2S)(vu) =0W )(v) =(u) )(v+S) =(u+S)

Note also that

im() =f(v+S)jv2Vg=f(v)jv2Vg=im() and ker() =fv+Sj(v+S) =0Wg =fv+Sj(v) = 0g =fv+Sjv2kerg = ker()=S

The preceding theorem now enables a simple proof of statement (iii) of the First Isomorphism Theorem.

Let:V!Wbe a vector space homomorphism and chooseS= ker(). Then by the preceding theorem we have an induced homomorphism :V=ker!W

14. THE ISOMORPHISM THEOREMS 3

such that im() =im() ker() = ker()=ker() =f0g In other words,is a surjective homomorphism fromV=kerontoim() that has has kernelf0g.is thus an isomorphism and Theorem 14.1 (iii) is established.

There are two other isomorphism theorems

Theorem14.3 (Second Isomorphism Theorem).LetVbe a vector space and letSandTbe subspaces of

V. Then

(i)S+T=fv2Vjv=s+t ; s2Sandt2Tgis a subspace ofV. (ii)The intersectionS\Tis a submodule ofS: (iii)The quotient modules(S+T)=TandS=(S\T)are isomorphic.

Proof.

(i) Letw=s+t,w0=s0+t0be generic elements ofS+Tand let;2F. Then w+w0=(s+t) +(s0+t0) = (s+s0) + (t+t0)2S+T SoS+Tis closed under linear combinations, and henceS+Tis a subspace ofV. (ii) Letw;w02S\T. Thenw;w02Sandw;w02Tand w;w

02Sandw;w02T)w+w02Sandw+w02T

)w+w02S\T and soS\Tis a submodule ofS(and is also a submodule ofTand a submodule ofV) (iii) Leti:S!S+T, be the natural embedding i(s) =s+ 0V2S+T

It is easy to check this is a linear transformation. When we compose it with the canonical projection

p:S+T!(S+T)=T, we obtain a linear transformationf=pi:S!(S+T)=T. S i !S+T p!(S+T)=T)f=pi:S!(S+T)=T The kernel offis going to consist of those elements ofSthat get sent byito the kernel ofp; in other words, ker(f) =fs2Sji(s)2ker(p)g =fs2Sjs+ 0V2Tg =fs2Sjs2Tg =S\T So ker(f) =S\T

I claim that

Range(f) = (S+T)=T

14. THE ISOMORPHISM THEOREMS 4

Indeed, supposew2(S+T)=T, then sincepTis surjective,w=p(s+t) for somes+t2S+T. But then w=p(s+t) =p(s) +p(t) =p(s) =p(s+ 0V) =p(i(s)) =f(s) Thus, every element of (S+T)=Tis in the range off. Having displayedf:S!(S+T)=Tas a surjective linear transformation, the First Isomorphism Theorem now implies (S+T)=T=Range(f)S=ker(f) =S=(S\T) Theorem14.4 (Third Isomorphism Theorem).LetVbe a vector space and letSandTbe subspaces ofV withTSV. Then (i)The quotient spaceS=Tis a submodules of the quotientV=T. (ii)The quotient(V=T)=(S=T)is isomorphic toV=S.

Proof.

(i) This is more-or-less self evident from the dentions. Each elements+T2S=Tcan be identied with the elements+T2V=Tsinces2V(asSis a subspace ofV). Thus,S=Tis naturally a subset ofV=T.

It isS=Tis also closed under linear combinations (with the rules for addition and scalar multiplication it

inheritsfromV), and so it is, in fact, a subspace ofV. (ii) Let's just construct the space (V=T)=(S=T). An element ofV=Tis a hyperplane of form v+T fv+tjt2Tg An element of (V=T)=(S=T) is a hyperplane of hyperplanes (v+T) +S=T=f(v+t1) + (s+t2)js2S ; t1;t22Tg =fv+s+t1+t2js2S ; t1;t22Tg However, sincet1;t22TS, we haves+t1+t22Sand so we have (v+T) +S=T=fv+s0js02Sg 2V=S Thus, the two quotient spaces (V=T)=(S=T) andV=Sidentify with exactly the same set of hyperplanes in

V. That shows that (V=T)=(S=T) andV=Scoincide as sets; however, a vector space is a set with additional

structures; namely, scalar multiplication and vector addition. So to identify (V=T)=(S=T) withV=Sas

vector spaces, we need to check that their respective rules for scalar multiplication and vector addition

coincide as well. But this, in fact, will be the case, because in both quotient spaces the rules for scalar

multiplication and vector are inherited from those ofV.quotesdbs_dbs12.pdfusesText_18

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