Rings and Fields

Study Sheet

Rings and Fields

Two operations at once: rings, fields, integral domains and the characteristic

Rings

Concept
The definition of a ring

A ring (R,+,⋅)(R,+,\cdot) is a set with two binary operations such that: (R,+)(R,+) is an abelian group (identity 00, inverses −a-a); multiplication is associative; and the distributive laws a(b+c)=ab+aca(b+c)=ab+ac and (a+b)c=ac+bc(a+b)c=ac+bc hold. The ring is commutative if ab=baab=ba always, and has a unity if some 1≠01\ne0 satisfies 1⋅a=a⋅1=a1\cdot a=a\cdot1=a.

Everything learned about groups applies to the addition of a ring: it is an abelian group. What is new is the second operation, multiplication, which is only required to be associative and to interact with addition through the distributive laws. Multiplication does NOT have to have inverses, so a ring is usually not a group under multiplication. An element uu that does have a multiplicative inverse is called a unit, and a nonzero aa with ab=0ab=0 for some nonzero bb is called a zero divisor. These two kinds of element are what decide, later in this topic, whether a ring is a field.

Example
Clock arithmetic is a ring

Z6={0,1,2,3,4,5}\mathbb Z_6=\{0,1,2,3,4,5\} with addition and multiplication mod 66 is a commutative ring with unity 11. For instance 4⋅5=20≡2(mod6)4\cdot5=20\equiv2\pmod6 and 4+5=9≡3(mod6)4+5=9\equiv3\pmod6. It has zero divisors: 2⋅3=6≡02\cdot3=6\equiv0 although neither 22 nor 33 is 00.

Example
Units and zero divisors in Z12\mathbb Z_{12}

In Z12\mathbb Z_{12} the units are the residues coprime to 1212: 1,5,7,111,5,7,11 (each is its own inverse, e.g. 5⋅5=25≡15\cdot5=25\equiv1). Every other nonzero residue is a zero divisor: 2,3,4,6,8,9,102,3,4,6,8,9,10, since each shares a factor with 1212 (for example 8⋅3=24≡08\cdot3=24\equiv0). So the 1111 nonzero elements split into 44 units and 77 zero divisors.

Example
A ring that is not commutative

The 2×22\times2 real matrices form a ring under matrix addition and multiplication. With A=(0100)A=\begin{pmatrix}0&1\\0&0\end{pmatrix} and B=(0010)B=\begin{pmatrix}0&0\\1&0\end{pmatrix}: AB=(1000)AB=\begin{pmatrix}1&0\\0&0\end{pmatrix} but BA=(0001)BA=\begin{pmatrix}0&0\\0&1\end{pmatrix}, so multiplication is not commutative. Also A2=(0000)A^2=\begin{pmatrix}0&0\\0&0\end{pmatrix} with A≠0A\ne0: a nonzero element whose square is zero.

Side note
Where it goes wrong

A common slip is to expect a ring to be a group under multiplication. It is not: 00 never has a multiplicative inverse (0⋅b=0≠10\cdot b=0\ne1), and in Z\mathbb Z even 22 has none. Another slip is cancelling in a ring with zero divisors: in Z6\mathbb Z_6, 2⋅1=2⋅4=22\cdot1=2\cdot4=2, yet 1≠41\ne4. Cancelling a nonzero factor is only safe when there are no zero divisors.

Try it
Try it: units of Z10\mathbb Z_{10}

List the units of Z10\mathbb Z_{10} and the inverse of each.

Show the answer

The units are the residues coprime to 1010: 1,3,7,91,3,7,9. Inverses: 1⋅1=11\cdot1=1, 3⋅7=21≡13\cdot7=21\equiv1, 9⋅9=81≡19\cdot9=81\equiv1. So 1−1=11^{-1}=1, 3−1=73^{-1}=7, 7−1=37^{-1}=3, 9−1=99^{-1}=9.

Fields

Concept
The definition of a field

A field is a commutative ring with unity 1≠01\ne0 in which every nonzero element is a unit. Equivalently: (F,+)(F,+) is an abelian group, (F∖{0},⋅)(F\setminus\{0\},\cdot) is an abelian group, and multiplication distributes over addition.

A field is the setting where all four arithmetic operations work: you can add, subtract, multiply, and divide by anything except 00. The rationals Q\mathbb Q, the reals R\mathbb R and the complex numbers C\mathbb C are fields; the integers Z\mathbb Z are not, because 22 has no integer inverse. The second half of this course studies fields the way the first half studied groups — and it uses groups constantly, since a field contains two of them: its additive group and its group of nonzero elements.

Example
Building a field inside R\mathbb R

Let Q(2)={a+b2:a,b∈Q}\mathbb Q(\sqrt2)=\{a+b\sqrt2:a,b\in\mathbb Q\}. Sums and products stay in the set: (a+b2)(c+d2)=(ac+2bd)+(ad+bc)2(a+b\sqrt2)(c+d\sqrt2)=(ac+2bd)+(ad+bc)\sqrt2. Inverses do too, by "rationalizing": (1+2)(2−1)=2−1=1(1+\sqrt2)(\sqrt2-1)=2-1=1, so (1+2)−1=−1+2(1+\sqrt2)^{-1}=-1+\sqrt2. In general (a+b2)−1=a−b2a2−2b2(a+b\sqrt2)^{-1}=\dfrac{a-b\sqrt2}{a^2-2b^2}, and a2−2b2≠0a^2-2b^2\ne0 unless a=b=0a=b=0 because 2\sqrt2 is irrational. So Q(2)\mathbb Q(\sqrt2) is a field.

Example
The field Z5\mathbb Z_5

In Z5\mathbb Z_5 every nonzero element has an inverse: 1⋅1=11\cdot1=1, 2⋅3=6≡12\cdot3=6\equiv1, 4⋅4=16≡14\cdot4=16\equiv1. So 1−1=11^{-1}=1, 2−1=32^{-1}=3, 3−1=23^{-1}=2, 4−1=44^{-1}=4, and Z5\mathbb Z_5 is a field with 55 elements. Equations can now be solved by division: 3x=43x=4 gives x=3−1⋅4=2⋅4=8≡3x=3^{-1}\cdot4=2\cdot4=8\equiv3, and indeed 3⋅3=9≡43\cdot3=9\equiv4.

Side note
Where it goes wrong

It is tempting to call Z6\mathbb Z_6 a field because addition and multiplication mod 66 "work". But 22 has no inverse mod 66 (the products 2⋅x2\cdot x are only 0,2,40,2,4), and 2⋅3=02\cdot3=0. A field can never have zero divisors: if ab=0ab=0 with a≠0a\ne0, multiply by a−1a^{-1} to get b=0b=0.

Try it
Try it: an inverse in Q(2)\mathbb Q(\sqrt2)

Find (3+2)−1(3+\sqrt2)^{-1} in the form p+q2p+q\sqrt2 with p,qp,q rational.

Show the answer

Multiply top and bottom by 3−23-\sqrt2: (3+2)(3−2)=9−2=7(3+\sqrt2)(3-\sqrt2)=9-2=7. So (3+2)−1=3−27=37−172(3+\sqrt2)^{-1}=\dfrac{3-\sqrt2}{7}=\mathbf{\tfrac37-\tfrac17\sqrt2}.

Integral Domains and Zp\mathbb Z_p

Concept
Integral domains, and when Zn\mathbb Z_n is a field

An integral domain is a commutative ring with unity and no zero divisors. Every field is an integral domain. Two key facts: Zn\mathbb Z_n is a field exactly when nn is prime, and every FINITE integral domain is a field.

Integral domains are the rings where cancellation works: if ab=acab=ac and a≠0a\ne0, then a(b−c)=0a(b-c)=0 forces b=cb=c. The integers are the model example — an integral domain that is not a field. For Zn\mathbb Z_n the picture is sharp: if n=abn=ab with 1<a,b<n1<a,b<n then a⋅b=0a\cdot b=0 is a zero divisor pair, and if n=pn=p is prime then every nonzero residue is coprime to pp and so has an inverse. We write Zp\mathbb Z_p (sometimes Fp\mathbb F_p) for this field with pp elements.

Example
Solving a linear equation in Z7\mathbb Z_7

Solve 4x+3=54x+3=5 in Z7\mathbb Z_7. Subtract 33: 4x=24x=2. Since 4⋅2=8≡14\cdot2=8\equiv1, 4−1=24^{-1}=2, so x=2⋅2=4x=2\cdot2=4. Check: 4⋅4+3=19≡5(mod7)4\cdot4+3=19\equiv5\pmod7.

Example
Zero divisors in Z8\mathbb Z_8

In Z8\mathbb Z_8 the zero divisors are 2,4,62,4,6 (for example 2⋅4=8≡02\cdot4=8\equiv0 and 6⋅4=24≡06\cdot4=24\equiv0), and the units are 1,3,5,71,3,5,7. Since zero divisors exist, Z8\mathbb Z_8 is not an integral domain, let alone a field — as expected, because 88 is not prime.

Example
The Gaussian integers

Z[i]={a+bi:a,b∈Z}\mathbb Z[i]=\{a+bi:a,b\in\mathbb Z\} is an integral domain: it sits inside the field C\mathbb C, so a product of nonzero elements is never 00. It is not a field: 22 has no inverse, since 12∉Z[i]\tfrac12\notin\mathbb Z[i]. Its units are exactly 1,−1,i,−i1,-1,i,-i.

Side note
Where it goes wrong

Students sometimes think "no zero divisors" is the same as "field". It is not: Z\mathbb Z and Z[i]\mathbb Z[i] have no zero divisors but are not fields. The two notions agree only for FINITE rings, which is why Zp\mathbb Z_p is a field while Z\mathbb Z is not.

Try it
Try it: field or not?

Which of Z11\mathbb Z_{11}, Z15\mathbb Z_{15}, Z17\mathbb Z_{17}, Z21\mathbb Z_{21} are fields?

Show the answer

Zn\mathbb Z_n is a field exactly when nn is prime: Z11\mathbb Z_{11} and Z17\mathbb Z_{17} are fields; Z15\mathbb Z_{15} (3⋅5=03\cdot5=0) and Z21\mathbb Z_{21} (3⋅7=03\cdot7=0) are not.

Characteristic

Concept
The characteristic of a ring

The characteristic of a ring with unity is the least n≥1n\ge1 with 1+1+⋯+1⏟n=0\underbrace{1+1+\cdots+1}_{n}=0, or 00 if no such nn exists. The characteristic of a field (indeed of any integral domain) is either 00 or a prime pp.

Q\mathbb Q, R\mathbb R, C\mathbb C have characteristic 00; Zp\mathbb Z_p has characteristic pp. A field of characteristic pp contains a copy of Zp\mathbb Z_p (the multiples of 11), and one of characteristic 00 contains a copy of Q\mathbb Q. Arithmetic in characteristic pp has a famous surprise, the freshman's dream: (a+b)p=ap+bp(a+b)^p=a^p+b^p, because every middle binomial coefficient (pk)\binom pk with 0<k<p0<k<p is a multiple of pp and so vanishes.

Example
Characteristics of familiar rings

Z7\mathbb Z_7 has characteristic 77, since 7⋅1=7≡07\cdot1=7\equiv0 and no smaller multiple of 11 is 00. Z12\mathbb Z_{12} has characteristic 1212. Q\mathbb Q has characteristic 00. The ring Z2×Z3\mathbb Z_2\times\mathbb Z_3 (componentwise operations) has characteristic 66: (1,1)(1,1) added to itself returns to (0,0)(0,0) first after lcm⁡(2,3)=6\operatorname{lcm}(2,3)=6 steps — and indeed it has zero divisors such as (1,0)⋅(0,1)=(0,0)(1,0)\cdot(0,1)=(0,0).

Example
The freshman's dream in Z5\mathbb Z_5

In Z5\mathbb Z_5: (2+4)5=65≡15=1(2+4)^5=6^5\equiv1^5=1, while 25+45=32+1024=1056≡1(mod5)2^5+4^5=32+1024=1056\equiv1\pmod5. They agree, as the dream predicts, because (51),(52),(53),(54)=5,10,10,5\binom51,\binom52,\binom53,\binom54=5,10,10,5 are all multiples of 55.

Side note
Where it goes wrong

The dream is only true in characteristic pp and only for the exponent pp (or a power of pp). Over R\mathbb R, (1+1)2=4≠2=12+12(1+1)^2=4\ne2=1^2+1^2. And in characteristic 55, (1+1)2=4≠2(1+1)^2=4\ne2 as well — the identity is for the fifth power, not the square.

Try it
Try it: the characteristic of a product

What is the characteristic of Z4×Z6\mathbb Z_4\times\mathbb Z_6? Is it a field?

Show the answer

The unity is (1,1)(1,1); n(1,1)=(0,0)n(1,1)=(0,0) needs 4∣n4\mid n and 6∣n6\mid n, so the characteristic is lcm⁡(4,6)=12\operatorname{lcm}(4,6)=\mathbf{12}. It is not a field: 1212 is not prime, and (1,0)⋅(0,1)=(0,0)(1,0)\cdot(0,1)=(0,0) is a zero divisor pair.

Why It Works

If n=abn=ab with 1<a,b<n1<a,b<n, then a⋅b≡0a\cdot b\equiv0 with a,b≠0a,b\ne0, so aa is a zero divisor and has no inverse (if ac=1ac=1 then b=(ac)b=c(ab)=0b=(ac)b=c(ab)=0, a contradiction). If n=pn=p is prime and 1≤a<p1\le a<p, then gcd⁡(a,p)=1\gcd(a,p)=1, so Bezout gives integers x,yx,y with ax+py=1ax+py=1, i.e. ax≡1(modp)ax\equiv1\pmod p: every nonzero element is a unit.

Let DD be a finite integral domain and a≠0a\ne0. The map x↦axx\mapsto ax is one-to-one, since ax=ayax=ay gives a(x−y)=0a(x-y)=0 and hence x=yx=y. A one-to-one map from a finite set to itself is onto, so some xx has ax=1ax=1. Thus every nonzero aa is a unit.

Suppose the characteristic is n=abn=ab with 1<a,b<n1<a,b<n. Then 0=n⋅1=(a⋅1)(b⋅1)0=n\cdot1=(a\cdot1)(b\cdot1), and since there are no zero divisors, a⋅1=0a\cdot1=0 or b⋅1=0b\cdot1=0 — contradicting the minimality of nn. So nn is prime (or 00).

In a commutative ring of characteristic pp, the binomial theorem gives (a+b)p=∑k=0p(pk)akbp−k(a+b)^p=\sum_{k=0}^p\binom pk a^kb^{p-k}. For 0<k<p0<k<p, (pk)=p!k!(p−k)!\binom pk=\frac{p!}{k!(p-k)!}: the prime pp divides the numerator but not k!(p−k)!k!(p-k)!, so p∣(pk)p\mid\binom pk and those terms are 00. What remains is ap+bpa^p+b^p.

Reminder — The binomial theorem:  (x+y)n=∑k=0n(nk)x n−kyk(x+y)^{n}=\sum_{k=0}^{n}\binom{n}{k}x^{\,n-k}y^{k}