 mersenneforum.org > Math Basic Number Theory 21: ideals and homomorphisms
 Register FAQ Search Today's Posts Mark Forums Read 2017-06-03, 15:39 #1 Nick   Dec 2012 The Netherlands 5×353 Posts Basic Number Theory 21: ideals and homomorphisms This time, we look at two important concepts which initially appear unconnected but turn out to be strongly related. Let $$R$$ be a ring. An ideal of $$R$$ is a subset $$I\subset R$$ that has the following 2 properties: under addition, $$I$$ is a subgroup of $$R$$ (by proposition 78 this is equivalent to requiring that $$I$$ be non-empty and closed under subtraction); for all $$i\in I$$ and all $$r\in R$$, the products $$ri$$ and $$ir$$ are both elements of $$I$$. This is consistent with the definition of ideals which we gave for the Gaussian integers, extending it to cover ideals of any ring. Examples 1. The set $$2\mathbb{Z}$$ of all even numbers is an ideal of $$\mathbb{Z}$$: even numbers exist and, for any integers $$m,n$$, if $$m$$ and $$n$$ are even then $$m-n$$ is even, while if $$m$$ or $$n$$ is even then $$mn$$ is even. (Note that $$2\mathbb{Z}$$ is not an ideal of $$\mathbb{Q}$$, so in a situation with rings and subrings we must be careful which ring we are talking about.) 2. More generally, for any integer $$n\geq 0$$, the set $$n\mathbb{Z}$$ of all multiples of $$n$$ is an ideal of $$\mathbb{Z}$$. 3. In every ring $$R$$, the subsets $$\{0\}$$ and $$R$$ itself are both ideals of $$R$$. 4. For any real number $$a$$, the set $$\{f\in\mathbb{R}[X]:f(a)=0\}$$ of all polynomials over $$\mathbb{R}$$ having $$a$$ as a zero/root is an ideal of the ring $$\mathbb{R}[X]$$. Notation For any commutative ring $$R$$ and any element $$a\in R$$, we write $$aR$$ (or simply $$(a)$$ if $$R$$ is clear from the context) for the subset $$\{ar:r\in R\}$$. We now make rigorous the idea that $$aR$$ is the smallest ideal of $$R$$ containing $$a$$ (just as we did for Gaussian integers in proposition 53). Proposition 118 Let $$R$$ be a commutative ring and $$a\in R$$. Then $$aR$$ is an ideal of $$R$$ with $$a\in aR$$. If $$I$$ is also an ideal of $$R$$ with $$a\in I$$, then $$aR\subset I$$. proof We have $$0=a\cdot 0\in aR$$. For any $$x,y\in aR$$, $$x=ar$$ and $$y=as$$ for some $$r,s\in R$$, and $$r-s\in R$$ too so $$x-y=ar-as=a(r-s)\in aR$$. Thus $$aR$$ is non-empty and closed under subtraction, making it a subgroup of $$R$$ under addition by proposition 78. Take any $$r\in R$$ and any $$i\in aR$$. Then $$i=as$$ for some $$s\in R$$ and $$sr\in R$$ so $$ir=asr\in aR$$. As $$R$$ is commutative, we also have $$ri=ir\in aR$$. Hence $$aR$$ is an ideal of $$R$$, and $$1\in R$$ so $$a=a\cdot 1\in aR$$. Take any ideal $$I$$ of $$R$$ with $$a\in I$$. Then, for all $$r\in R$$, we have $$ar\in I$$ (as $$I$$ is an ideal) hence $$aR\subset I$$. ∎ For subsets $$A,B$$ of a ring $$R$$, we define their sum $$A+B$$ by $A+B=\{a+b:a\in A,b\in B\}$ (We have already used this notation with sets of numbers, but now extend it to any subsets of any ring.) Example Let $$I$$ be an ideal of a ring $$R$$. Then $$I$$ is closed under addition so $$I+I\subset I$$. And $$0\in I$$ so $$I=I+\{0\}\subset I+I$$ too, hence $$I+I=I$$. Let $$I$$ be an ideal of a ring $$R$$ and $$a$$ an element of $$R$$. We write $$a+I=\{a+i:i\in I\}$$ and call this subset a coset of $$I$$ in $$R$$. We may also do the same thing with $$a$$ on the right: $$I+a=\{i+a:i\in I\}$$ but, as addition is commutative, it is the same subset. We write $$R/I$$ for the set of all cosets of $$I$$ in $$R$$, i.e. $R/I=\{a+I:a\in R\}.$ (We did this earlier for groups in order to prove Lagrange's theorem (theorem 83). There, we wrote the group's binary operation as multiplication while here the ring $$R$$ is a group under its addition. The other difference is that addition in a ring is always commutative, so we don't have to distinguish between left and right cosets.) Proposition 119 Let $$I$$ be an ideal of a ring $$R$$ and define a relation $$\sim$$ on $$R$$ by $$a\sim b\Leftrightarrow a-b\in I$$. Then $$\sim$$ is an equivalence relation on $$R$$ and, for each $$a\in R$$, the equivalence class of $$a$$ is precisely the coset $$a+I$$. proof Exactly as for proposition 80, just using additive notation instead of multiplicative notation. ∎ Now let $$I$$ be an ideal of a ring $$R$$ again, and $$a,b$$ elements of $$R$$. We can use the subset sum defined above to add the cosets $$a+I$$ and $$b+I$$ together. As addition is commutative and $$I+I=I$$, it follows that their sum is again a coset of $$I$$ in $$R$$: $(a+I)+(b+I)=a+I+b+I=a+b+I+I=(a+b)+I.$ It would be nice to be able to multiply cosets together as well, and the obvious way to define this is by $(a+I)(b+I)=(ab)+I.$ We must check that this is well-defined, however. Take any elements $$a,a',b,b'\in R$$ and suppose that $$a+I=a'+I$$ and $$b+I=b'+I$$. Then $$a-a'\in I$$ and $$b-b'\in I$$. As $$I$$ is an ideal of $$R$$, it follows that $$(a-a')b$$ and $$a'(b-b')$$ are also elements of $$I$$. And $$I$$ is closed under addition so $$ab-a'b'=(a-a')b+a'(b-b')\in I$$, hence $$ab+I=a'b'+I$$ by proposition 119. Thus the product of two cosets does not depend on the elements we choose to represent them, making multiplication of cosets well-defined. Proposition 120 Let $$I$$ be an ideal of a ring $$R$$. Then the set $$R/I$$ of all cosets of $$I$$ in $$R$$ is itself a ring under addition and multiplication of cosets as defined above. proof For all $$a,b,c\in R$$, \begin{eqnarray*} ((a+I)+(b+I))+(c+I) & = & ((a+b)+I)+(c+I) = (a+b)+c+I \\ & = & a+(b+c)+I \mbox{ by associativity of addition in }R \\ & = & (a+I)+((b+c)+I) = (a+I)+((b+I)+(c+I)) \end{eqnarray*} so addition in $$R/I$$ is associative. All other required properties of $$R/I$$ follow from the corresponding properties of $$R$$ in the same way. The neutral elements are $$0+I$$ (which equals $$I$$ itself) under addition and $$1+I$$ under multiplication, and for each $$a\in R$$ the inverse of $$a+I$$ under addition is $$(-a)+I$$. ∎ We call $$R/I$$ the quotient ring of $$R$$ by $$I$$. Example For any positive integer $$n$$, the subset $$n\mathbb{Z}=\{nz:z\in\mathbb{Z}\}$$ is an ideal of $$\mathbb{Z}$$, and the quotient ring is $$\mathbb{Z}/n\mathbb{Z}$$, the integers modulo $$n$$. For each integer $$a$$, the coset $$a+n\mathbb{Z}$$ is what we write as $$\bar{a}$$. Thus what we have in fact done is take the way we constructed the integers modulo $$n$$ from the ordinary integers and generalized it to apply to any ideal of any ring. We now switch to our second important concept. Let $$R,S$$ be rings. A homomorphism from $$R$$ to $$S$$ is a function $$f:R\rightarrow S$$ satisfying the following 3 conditions: for all $$x,y\in R$$, $$f(x+y)=f(x)+f(y)$$; for all $$x,y\in R$$, $$f(xy)=f(x)f(y)$$; f(1)=1. An isomorphism from $$R$$ to $$S$$ is a homomorphism from $$R$$ to $$S$$ that is also bijective. We say $$R$$ is isomorphic with $$S$$ and write $$R\cong S$$ if there exists an isomorphism from $$R$$ to $$S$$. (We already defined isomorphism for groups, and here we do the same thing for rings.) The first 2 conditions state that a homomorphism respects addition and multiplication: if $$x+y=z$$ in $$R$$ then $$f(x)+f(y)=f(z)$$ in $$S$$, and if $$xy=z$$ in $$R$$ then $$f(x)f(y)=f(z)$$ in $$S$$. The way in which elements of $$R$$ are related to each other by addition and multiplication in $$R$$ is reflected by the way in which the corresponding elements of $$S$$ are related to each other. The third condition says that a homomorphism maps the 1 of the first ring (i.e. its neutral element under multiplication) to the 1 of the second ring. You might expect to see the condition $$f(0)=0$$ here as well, but that follows automatically from condition (1), as we now show. Proposition 121 Let $$R,S$$ be rings and $$f:R\rightarrow S$$ a homomorphism. Then $$f(0)=0$$ and, for all $$x\in R$$, $$f(-x)=-f(x)$$. proof By condition (1), $$f(0)=f(0+0)=f(0)+f(0)$$ so $$f(0)=0$$. And, for all $$x\in R$$, $$f(x)+f(-x)=f(x+(-x))=f(0)=0=f(x)+(-f(x))$$ so $$f(-x)=-f(x)$$. ∎ We cannot deduce condition (3) from (2) in the same way. For example, the function $$f:\mathbb{Z}\rightarrow\mathbb{Z}$$ given by $$x\mapsto 0$$ (i.e. $$f(n)=0$$ for all integers $$n$$) satisfies conditions (1) and (2) but not (3). Examples 1. Let $$m,n$$ be integers with $$\gcd(m,n)=1$$ and $$f$$ the function from $$\mathbb{Z}/mn\mathbb{Z}$$ to $$\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}$$ given by $$f(a\bmod{mn})=(a\bmod{m},a\bmod{n})$$. Then $$f$$ is an isomorphism by the Chinese Remainder Theorem (theorem 29) so $$\mathbb{Z}/mn\mathbb{Z}\cong\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}$$. 2. Let $$f:\mathbb{C}\rightarrow\mathbb{C}$$ be the function $$z\mapsto\bar{z}$$, where $$\bar{z}$$ is the complex conjugate of $$z$$, i.e. for all $$x,y\in\mathbb{R}$$ we have $$f(x+yi)=x-yi$$. Then $$f$$ is an isomorphism (with $$f^{-1}=f$$) by proposition 43. 3. Let $$R$$ be an integral domain and $$Q(R)$$ its field of fractions, as in proposition 111. Then the function $$f:R\rightarrow Q(R)$$ given by $$a\mapsto\frac{a}{1}$$ is an injective homomorphism. It is an isomorphism if and only if $$R$$ is a field. 4. Let $$R$$ be a ring and $$f:R\rightarrow R[X]$$ the function mapping each element $$a\in R$$ to the constant polynomial $$a\in R[X]$$. Then $$f$$ is an injective homomorphism, but $$f$$ is not an isomorphism as it is not surjective. 5. Pick an integer $$n$$ and let $$\phi:\mathbb{Z}[X]\rightarrow\mathbb{Z}$$ be the function $$f\mapsto f(n)$$, i.e. we take each polynomial over the integers and evaluate it at $$n$$ (so $$n$$ is fixed but $$f$$ varies). Then $$\phi$$ is a homomorphism which is surjective but not injective. 6. Let $$I$$ be an ideal of a ring $$R$$ and define $$f:R\rightarrow R/I$$ by $$a\mapsto a+I$$. Then $$f$$ is a surjective homomorphism. Suppose a ring $$R$$ is isomorphic with a ring $$S$$. If we restrict our attention to the ring properties of $$R$$, i.e. the number of elements it has and the way they combine under addition and multiplication, ignoring any other properties its elements might have, then anything we deduce about $$R$$ will also be true about $$S$$. In that sense, $$R$$ and $$S$$ are two instances of the same underlying abstract ring, and we may think of isomorphic rings as being essentially the same ring (just as we did with groups). Suppose instead we have a homomorphism $$f:R\rightarrow S$$ that is not an isomorphism. Then $$f$$ is not surjective or it is not injective. Recall that we write im($$f$$) for the subset $$\{f(x):x\in R\}$$ of $$S$$, known as the image of the function $$f$$. If $$f$$ is not surjective then its image is not the whole of $$S$$, but it is a subring, as we now prove. Proposition 122 Let $$R,S$$ be rings and $$f:R\rightarrow S$$ a homomorphism. Then im($$f$$) is a subring of $$S$$. proof By condition (3), 1 is an element of im($$f$$). Take any elements $$x,y$$ of im($$f$$). Then $$x=f(a)$$ and $$y=f(b)$$ for some $$a,b\in R$$. By proposition 121 and condition (1), $$x-y=f(a)-f(b)=f(a)+f(-b)=f(a-b)$$ so $$x-y$$ is also an element of im($$f$$). And by condition (2), $$xy=f(a)f(b)=f(ab)$$ so $$xy$$ is an element of im($$f$$) as well. It follows by proposition 109 that im($$f$$) is a subring of $$S$$. ∎ What happens if $$f$$ is not injective? Let $$a,b$$ be elements of $$R$$ and suppose we also have an element $$k\in R$$ with $$k\neq 0$$ but $$f(k)=0$$. Then $$0$$ and $$k$$ are two distinct elements of $$R$$ which get mapped to 0. Moreover, $$f(a+k)=f(a)+f(k)=f(a)+0=f(a)$$ (as $$f$$ is a homomorphism) so $$a$$ and $$a+k$$ are two distinct elements of $$R$$ that get mapped to $$f(a)$$. Similarly, $$b$$ and $$b+k$$ get mapped to $$f(b)$$. Conversely, for any $$x,y\in R$$, if $$f(x)=f(y)$$ then, putting $$k=y-x$$, we get $$y=x+k$$ with $$f(k)=f(y-x)=f(y)-f(x)=0$$ (by proposition 121). So, for each $$x\in R$$, the elements of $$R$$ which get mapped to $$f(x)$$ are precisely those of the form $$x+k$$ where $$k$$ gets mapped to 0. We now express this idea more formally. Let $$R,S$$ be rings and $$f:R\rightarrow S$$ a homomorphism. The kernel of $$f$$, denoted $$\ker(f)$$, is the subset of $$R$$ given by $\ker(f)=\{x\in R:f(x)=0\}.$ Proposition 123 Let $$R,S$$ be rings and $$f:R\rightarrow S$$ a homomorphism. Then $$\ker(f)$$ is an ideal of $$R$$. Let $$I=\ker(f)$$. Then, for all $$a\in R$$, the set $$\{x\in R:f(x)=f(a)\}$$ is precisely the coset $$=a+I$$. proof By proposition 121, $$0\in\ker(f)$$ and, for all $$x,y\in\ker(f)$$, $$f(x-y)=f(x)-f(y)=0-0=0$$. So $$\ker(f)$$ is non-empty and closed under subtraction, making it a subgroup of $$R$$ under addition by proposition 78. For all $$i\in I$$ and all $$r\in R$$, we have $$f(ri)=f(r)f(i)=f(r)\cdot 0=0$$ so $$ri\in\ker(f)$$, and similarly $$ir\in\ker(f)$$ hence $$\ker(f)$$ is an ideal of $$R$$. Take any $$a\in R$$, and let $$S=\{x\in R:f(x)=f(a)\}$$. For all $$x\in a+I$$, we have $$x=a+i$$ for some $$i\in I=\ker(f)$$ so $$f(x)=f(a+i)=f(a)+f(i)=f(a)$$ as $$f$$ is a homomorphism and $$f(i)=0$$, so $$x\in S$$. Thus $$a+I\subset S$$. Conversely, take any $$x\in S$$ and let $$i=x-a$$. Then $$x=a+i$$ and $$f(i)=f(x-a)=f(x)-f(a)=0$$ as $$f(x)=f(a)$$ so $$i\in\ker(f)=I$$, giving $$x\in a+I$$. Hence $$S\subset a+I$$ as well, and therefore $$S=a+I$$. ∎ Corollary 124 Let $$R,S$$ be rings and $$f:R\rightarrow S$$ a homomorphism. Then $$f$$ is injective if and only if $$\ker(f)=\{0\}$$. proof By proposition 121, we have $$0\in\ker(f)$$. If $$f$$ is injective then, for all $$x\in\ker(f)$$, $$f(x)=0=f(0)$$ and therefore $$x=0$$. So $$\ker(f)=\{0\}$$. Conversely, suppose $$\ker(f)=\{0\}$$ and take any $$a\in R$$. By proposition 123, $$\{x\in R:f(x)=f(a)\}=a+\ker(f)=a+\{0\}=\{a\}$$ hence $$f$$ is injective. ∎ Example Let $$I$$ be an ideal of a ring $$R$$ and $$f:R\rightarrow R/I$$ the homomorphism $$a\mapsto a+I$$. Then, for all $$a\in R$$, we have $a\in\ker(f)\Leftrightarrow a+I=0+I\Leftrightarrow a\in I$ by proposition 119, so $$\ker(f)=I$$. This reveals the connection between our 2 new concepts: the kernel of any homomorphism is an ideal (by proposition 123) and, conversely, for any ideal there exists a homomorphism for which it is the kernel (as in the above example). So the ideals of a ring $$R$$ are precisely the subsets which occur as kernels of homomorphisms from $$R$$ to other rings. Theorem 125 First Isomorphism Theorem Let $$R,S$$ be rings and $$f:R\rightarrow S$$ a homomorphism. Then $$R/\ker(f)\cong im(f)$$. proof Let $$I=\ker(f)$$ and $$S=im(f)$$. Define $$\phi:R/I\rightarrow S$$ by $$a+I\mapsto f(a)$$. We must check that this is well-defined. Take any $$a,b\in R$$ and suppose that $$a+I=b+I$$. Then $$a-b\in I$$ by proposition 119 and $$I=\ker(f)$$ so, by proposition 121, $$f(a)-f(b)=f(a-b)=0$$ hence $$f(a)=f(b)$$. Thus $$\phi$$ is well-defined. For all $$a,b\in R$$, $\phi((a+I)+(b+I))=\phi((a+b)+I)=f(a+b)=f(a)+f(b)=\phi(a+I)+\phi(b+I)$ and $\phi((a+I)(b+I))=\phi(ab+I)=f(ab)=f(a)f(b)=\phi(a+I)\phi(b+I).$ And $$\phi(1+I)=f(1)=1$$ so $$\phi$$ is a homomorphism. For all $$a,b\in R$$, if $$\phi(a+I)=\phi(b+I)$$ then $$f(a)=f(b)$$ so $$f(a-b)=f(a)-f(b)=0$$ and therefore $$a-b\in\ker(f)=I$$ hence $$a+I=b+I$$ by proposition 119. Thus $$\phi$$ is injective. And every element of $$S$$ has the form $$f(a)$$ for some $$a\in R$$ so $$\phi(a+I)=f(a)$$, making $$\phi$$ surjective as well. Hence $$\phi$$ is an isomorphism from $$R/\ker(f)$$ to im($$f$$). ∎ Exercises 96. Let $$R$$ be the set of all 2x2 matrices of the form $\left(\begin{array}{rr} x & -y \\ y & x \end{array}\right)$ where $$x,y\in\mathbb{R}$$. Show that $$R$$ is a subring of the ring $$M_2(\mathbb{R})$$ of all 2x2 matrices over $$\mathbb{R}$$, and that the function $$f:R\rightarrow\mathbb{C}$$ given by $\left(\begin{array}{rr} x & -y \\ y & x \end{array}\right)\mapsto x+yi$ is an isomorphism. 96. Let $$I,J$$ be ideals of a ring $$R$$. Show that $$I+J$$ is also an ideal of $$R$$. 97. Prove that $$\cong$$ is an equivalence relation on any set of rings. 98. Show that, for every ring $$R$$, there exists a unique homomorphism from $$\mathbb{Z}$$ to $$R$$. 99. Let $$I$$ be an ideal of a ring $$R$$ containing a unit of $$R$$. Prove that $$I=R$$. 100. Let $$R$$ be a commutative ring. Prove that $$R$$ is a field if and only if $$R$$ has precisely 2 ideals. (You may want to use exercise 99.) 101. Let $$K,L$$ be fields and $$f:K\rightarrow L$$ a homomorphism. Prove that $$f$$ is injective. (You may want to use exercise 100.) 102. Let $$f:\mathbb{Q}\rightarrow\mathbb{Q}$$ be a homomorphism. Show that $$f$$ is just the identity map, i.e. $$f(x)=x$$ for all $$x\in\mathbb{Q}$$. 103. (Only if you want a challenge!) Prove that the identity map is the only homomorphism from $$\mathbb{R}$$ to $$\mathbb{R}$$.   2017-06-03, 17:06 #2 LaurV Romulan Interpreter   "name field" Jun 2011 Thailand 5·2,003 Posts For sure you have a lot of patience to type so much, and we are grateful for that! I stopped somewhere in the middle, 12:05 AM here, I am quite tired and the letters were jumping in front of me... I had to work today (my working Saturday, here we are a bunch of people that have to work on alternating Saturdays by rotation, to ensure production support)... I am going to bed, but be sure that I will come back in the morning with a cup of coffee to continue the reading....   2017-06-03, 18:20   #3
Nick

Dec 2012
The Netherlands

110111001012 Posts Quote:
 Originally Posted by LaurV I am going to bed, but be sure that I will come back in the morning with a cup of coffee to continue the reading....
Thank you - your detailed checking is much appreciated!   2017-06-03, 19:57 #4 ET_ Banned   "Luigi" Aug 2002 Team Italia 29·167 Posts Sorry for looking possibly rude, it's not my intention at all... I fell into this thread when RS and others were discussing about teaching some number theory, then had some real life(C)TM issues, and found the course had started already. Now I found some spare time, and would like to (re-)start from chapters 1 & 2. My question: is this course based on some specific book? I remember we were choosing Edgar, "A first course in number theory", but then I had to quit and don't know if the title was finally accepted or not... Thank you for your lessons.   2017-06-03, 21:49   #5
Nick

Dec 2012
The Netherlands

6E516 Posts Quote:
 Originally Posted by ET_ My question: is this course based on some specific book? I remember we were choosing Edgar, "A first course in number theory", but then I had to quit and don't know if the title was finally accepted or not...
That book is no longer in print and, after several forum members bought second-hand copies of it, the price rose very quickly and it was no longer worth buying.
So we considered various alternatives before finally deciding to write it ourselves on the forum:

At LaurV's suggestion, we start a new thread for each new topic so that people can post their questions at any time in the appropriate thread.
That way, new people can join in whenever they want.   2017-06-04, 12:36   #6
ET_
Banned

"Luigi"
Aug 2002
Team Italia

29×167 Posts Quote:
 Originally Posted by Nick That book is no longer in print and, after several forum members bought second-hand copies of it, the price rose very quickly and it was no longer worth buying. So we considered various alternatives before finally deciding to write it ourselves on the forum: http://www.mersenneforum.org/showthread.php?t=21514 At LaurV's suggestion, we start a new thread for each new topic so that people can post their questions at any time in the appropriate thread. That way, new people can join in whenever they want.
That's quite a nice decision Thank you to all the people who are collaborating, I will join the study!  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Nick Number Theory Discussion Group 17 2017-12-23 20:10 Nick Number Theory Discussion Group 5 2017-04-25 14:32 Nick Number Theory Discussion Group 0 2016-12-29 13:47 Nick Number Theory Discussion Group 6 2016-10-14 19:38 Nick Number Theory Discussion Group 5 2016-10-08 09:05

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