1. Rings


The idea is that we take as an example the set of integer numbers, and we write out some of their properties. Next, we say that any set P along with some defined operations + and \cdot and chosen elements denoted as 0 and 1, is a ring (commutative with an identity), if these properties are met. These properties are, for any a,b,c\in P:

  • a+b=b+a, a\cdot b=b\cdot a,
  • (a+b)+c=a+(b+c), (a\cdot b)\cdot c=a\cdot (b\cdot c),
  • a\cdot (b+c)=a\cdot b+a\cdot c,
  • 0+a=a, 1\cdot a =a,
  • for any a, there exists b (unique, denoted by -a), such that a+b=0.

The integer numbers \mathbb{Z} is obviously an example of a ring, and so is \mathbb{Z}_n=\{0,\ldots, n-1\} with operations defined modulo n. Notice that every field is a ring as well. Indeed, if we add a condition about inverses to the definition of a ring we get the definition of a field!

If P is a ring that obviously P[x], i.e. the sets of all polynomials with coefficients in P is a ring with operations defined in a natural way.


A subset S\subseteq P of a ring (commutative with identity) is called a subring if it includes 0 and 1 and is closed under addition, multiplication and taking negatives. In other words, it is a subring if it is a ring with the same operation as in the bigger ring. E.g. \mathbb{Z} is a subring in \mathbb{R}. On the other hand, notice that even numbers are not a subring in \mathbb{\Z}, because it is closed under addition, multiplication and taking negatives, but does not contain one.

If S is a subring of P and a\in S\setminus P, we can consider the ring S[a]. Then S[a] is also a subring of P. E.g., \mathbb{Z}[i]=\{a+bi\colon a,b\in \mathbb{Z}\} is a subring of the complex numbers.

Properties if elements of a ring, domains

We sat that a\in P:

  • is a divisor of zero, if there exists b\in P, b\neq 0, such that ab=0,
  • is a unit, if there exists b\in P, że ab=1,
  • is a nilpotent, if multiplied a number of times by itself it gives 0 (i.e. a^n=0 for some natural number n).

E.g 3 is a divisor of zero in \mathbb{Z}_{12}, because in this ring, 3\cdot 4 =0. On the other hand, 5 is a unit there, since 5\cdot 5=1. Finally, 6 is a nilpotent, because 6\cdot 6=0.

Notice, that every nilpotent is a divisor of zero, but no divisor of zero can be a unit (in a non-zero ring).

One can also show that in a finite ring every element is either a unit or a divisor of zero.

A ring is called a integral domain, if it has no divisors of zero. Notice that this means that in a finite integral domain all the non-zero elements are units, and thus every finite integral domain is a field!

Moreover, the degree of a product of two polynomials over a integral domain is not less than the bigger degree of the multiplied polynomials.

Homomorphisms of rings

Given two rings (commutative with identity) P and S, a function h\colon P\to S is a homomorphism if for any a,b\in P: h(a+b)=h(a)+h(b), h(a\cdot b)=h(a)\cdot h(b) and h(1_P)=1_S. It is easy to check that a homomorphism maps a zero onto zero and a negative onto negative.

E.g. for any integer a, h\colon \mathbb{Z}[x]\to \mathbb{Z} given as h(w)=w(a) is a homomorphism which can be easily verified.

An isomorphism of rings is a homomorphism which is a bijection. The kernel of a homomorphism is the set \ker h=\{a\in P\colon h(a)=0_S\}.


A subset I of a ring P is and ideal, if

  • for any a,b\in I, a+b\in I,
  • for any a\in R and b\in I, a\cdot b\in I.

We are going to denote this fact by I\trianglelefteq P.

E.g. even number is an ideal in the ring of integer number, because a sum of even number is even and an even number multiplied by any number is an even number.

The whole ring and the zero ideal \{0\} are the most trivial examples of ideals. An ideal is proper, if I\neq P. Notice that an ideal is proper if and only if it does not contain 1. An ideal is prime if for any a,b\in P, if a\cdot b\in I, then a\in I or b\in I. It is maximal if there is no ideal J, such that I\subsetneq J. Soon we are going to get to corollary that in the case of finite rings, prime and maximal ideals are the same thing.

Notice that the image of a ring under a homomorphism in a subring, and the image of an ideal is its ideal. Morover, a counterimage of an ideal is an ideal as well. Finally, a kernel of a homeomorhpism is an ideal.

Finally, notice that the only ideals in a field and the zero ideal and the whole field.

Factorization, domains with unique factorization

Assume now that P is an integral domain (so there are no non-zero divisors of zero). A non-zero element a\in P is reducible, if it can be written as a=b\cdot c, where b and c are not units. If a non-zero element a is not a unit and is not reducible it is called irreducible.

E.g. it is clear that (x^2+1) is irreducible in \mathbb{Z}[x]. But (x^2-1)=(x+1)(x-1) is reducible. Similarly 5 is irreducible in \mathbb{Z}, but 6 is not.

Sometimes an element can be factorized into a product of irreducible elements in more then one way. E.g. 10=(3-i)(3+i)=2\cdot 5 in \mathbb{Z}[i]. We shall say that an element a has a unique factorization, if whenever a=b_1\cdot\ldots \cdot b_n=c_1\cdot\ldots \cdot c_m, where b_1,\dlots, b_n,c_1,\ldots, c_m are irreducible, we have n=m and elements b_i and c_j can be set into pairs such that b_i=d_ic_j for a unit d_i.

Integral domains in which all non-zero element have unique factorization are called domains with unique factorization. An example of such a domain are the integrals themselves, or e.g. \mathbb{Z}[x].

Generated ideals and quotient rings

The ideal generated by A\subseteq P is the least ideal I\trianglelefteq P, such that A\subseteq I. We denote it by (A). Notice that (A)=\{a_1x_1+\ldots +a_nx_n\colon n\in \mathbb{N}, a_i\in A, x_i\in P\}. If A=\{a\} is a one-element set, then (A) is called a principal ideal and we write simply (a)=(\{a\}). Obviously, then (a)=\{ax\colon x\in P\}.

It is easy to prove that P is a field if and only if it is non-zero and the only ideals in it are \{0\} and P.

If I is an ideal in P and a\in P, we define a coset a+I=\{a+x\colon x\in I\}\subseteq P. Notice that a,b\in P are in the same coset of I is and only if a-b\in I. The set of all cosets is denoted by P/I and we can introduce a structure of a ring in it stipulating that zero is 0+I, identity is 1+I, and (a+I)+(b+I)=(a+b)+I, (a+I)\cdot (b+I)= a\cdot b+I and -(a+I)=-a+I. Such a ring is called the quotient ring.

E.g. take 2\mathbb{Z}=(2) the ideal of even numbers in \mathbb{Z}. Then \mathbb{Z}/2\mathbb{Z} is isomorphic to \mathbb{Z}_2. Indeed, two numbers are in the same coset of the ideal 2\mathbb{Z} if and only if both are even or both are odd, so there are only two cosets, and the operations on these cosets are consistent with operations modulo 2. Thus h\colon \mathbb{Z}_2\to \mathbb{Z}/2\mathbb{Z} defined as h(0)=0+2\mathbb{Z}, h(1)=1+2\mathbb{Z} is an isomorphism of rings.

One can also prove the following important theorem, called the theorem about a homomorphism. If h\colon P\to R is a homomorphism, then there exists exactly one homomorphism \phi\colon P/\ker h\to R such that h=\phi\circ \pi, where \pi\colon P\to P/\ker h is given by \pi(a)=a+\ker h. Moreover, \phi is an isomorphism of P/\ker h and \text{im} (h) and thus we have a one-to-one correspondence between ideals in the ring \text{im} (h) and ideals of P containing \ker h.

This leads to the following characterization:

  • P/I is a field if and only if the ideal I is maximal,
  • P/I is an integral domain if and only if the ideal I is prime.

In particular, every maximal ideal is prime. By Zorn Lemma, every proper ideal is contained in a maximal ideal. Thus, every ring can be mapped onto a field.

Divisibility, domains of proper ideals

In a domain P, a|b, if there exists c\in P such that ac=b. If a|b and b|a, then we say that a is associated to b, and write a\sim b. It is easy to prove that a|b if and only if (b)\subseteq (a). Moreover, a\sim b if and only if there exists a unit c such that a=bc.

An element p\in P is a prime, if and only if for any ab such that p|ab, p|a or p|b. It is clear that p is a prime if and only if (p) is a prime ideal, and that every prime ideal is irreducible. This implication can be reversed in domains with unique factorization. In such domains every irreducible element is a prime. Actually, if in a domain every element which is nonzero and is not a unit can be factorized into irreducible elements, and every irreducible element is a prime, then it is a domain with unique factorization.

In a domain P, an element d is a greatest common divisor of a,b\in P (we write d\sim \gcd(a,b)), if d|a, d|b, and for every c such that c|a and c|b, also c|d. In general, the greatest common divisor may not exist. It is clear that it always exists in a domain with unique factorization.

A ring is called a ring of principal ideals, if every ideal in it is principal. A domain which is a ring of principal ideals is called a domain of principal ideals. Notice that \mathbb{Z} and K[x] for any field K are domains of principal ideals, but \mathbb{Z}[x] is not.

Every ideal generated by an irreducible element in a domain of principal ideals is maximal, thus every prime ideal in such a domain is either zero ideal or maximal. From this it follows that every domain of principal ideals is a domain with unique factorization.

Moreover, in a domain of principal ideal it is easy to see that every two elements have their gcd. Indeed, d\sim \gcd(a,b) if and only if (a,b)=(d).

Euclidean domains

There is this known algorithm of finding the gcd of two integers. If we want to find \nwd(a,b) (assume, that |a|>|b|, b\neq 0) we find c,r such that a=bc+r, where |r|<|b| and then, if r\neq 0, then \nwd(a,b)=\nwd(b,r), so we can repeat the step for smaller numbers. If, on the other hand r=0, the divisor we are looking for is b. E.g. we can find \nwd(64,24) in the following way 64=24\cdot 2+16, 24=16\cdot 1 + 8, 16=2\cdot 8, thus \nwd(64,24)=8.

This algorithm ends because in every step the rest in in some sense smaller — in this case in the sense of absolute value. We may use this algorithm in other rings as well provided that in the given ring we find a property (called a norm) according to which the rest descreases in every step, because only this guarantees that the algorithm eventually stops.

An norm in a domain P is a function N\colon P\to \mathbb{N} such that N(a)=0 is and only if a=0 and N(ab)=N(a)N(b). A norm is eauclidean (thus, it allows to use the Euclid algorithm to find gcd), if morover for any non-zero a,b there exist c and r, such that a=b\cdot c +r and N(r)<N(b). The domains for which there exists an euclidean norm is called euclidean domains. One can prove that every euclidean domani is a domain of principal ideals. Both \mathbb{Z} with the norm being absolute value, and K[x] for any field K with the norm being the degree of a polynomial +1 are euclidean domains. There exists principal ideal domains which are not euclidean, but they are not very easy to find, e.g. \mathbb{R}[X,Y]/(X^2+Y^2+1).

It is worth noticing that every field is an euclidean domain in a not very interesting and quite trivial way — with the norm equal to 0 for 0 and 1 for any other element.