Skip to main content

Algebra-II: Companion notes

Abhay Soman

Section 5.1 Unique Factorization Domains

Let \(A\) be an integral domain. An element \(b\in A\) is said to divide an element \(a\in A\) if there exists \(x\in A\) such that \(b=ax\text{.}\) In this case \(b\) is called a divisor or a factor of \(a\) and we write \(b\mid a\text{.}\) We write \(c\nmid a\) to denote that \(c\) is not a factor of \(a\text{.}\) An element \(u\in A\) is called an unit if \(u\mid 1\text{.}\) Units are factors of every element. Two elements \(a,b\in A\) are called associates if \(a\mid b\) and \(b\mid a\text{.}\) We denote associates by \(a\sim b\text{.}\) It is easy to see that \(a\sim b\) if and only if \(b=au\) and \(a=bv\) for some units \(u,v\in A\text{.}\) If \(b\mid a\) and \(a\nmid b\) then \(b\) is said to be a proper factor of \(a\text{.}\) An element \(a\in A\) is said to be irreducible if \(a\) is not a unit and \(a\) has no proper factors other than units. It is easy to verify that if \(a\) is irreducible and \(a\sim b\) then \(b\) is also irreducible. Suppose that \(a\in A\) is such that \(a=p_1p_2\cdots p_n\) where \(p_i\) are irreducibles. This is called a factorization of \(a\) into irreducibles. A factorization of \(a=p_1p_2\cdots p_n\) is said to be essentially unique if for any other factorization \(a=q_1q_2\cdots q_m\) (\(q_i\) are irreducibles) we have \(m=n\) and \(p_i\sim q_j\) for a suitable permutation \(\sigma\in S_n\) with \(\sigma(i)=j\text{.}\)

Definition 5.1.1. (Factorial domain or Unique factorization domain).

An integral domain \(A\) is called factorial or unique factorization domain if every nonzero non-unit in \(A\) has an essentially unique factorization into irreducibles.

Definition 5.1.2. (Divisor Chain Condition).

Let \(A\) be an integral domain. We say that \(A\) satisfy the divisor chain condition if \(a_1,a_2,\ldots\) is a sequence of elements of \(A\) such that \(a_{i+1}\mid a_i\) then there exists an integer \(N\) such that \(a_N\sim a_{N+1}\sim\cdots\text{.}\)

Definition 5.1.3. (Primeness condition).

An integral domain is said to satisfy primeness condition if every irreducible is prime.

Definition 5.1.5. (Greatest Common Divisor).

Let \(A\) be an integral domain. An element \(d\in A\) is called a greatest common divisor or g.c.d. of \(a,b\in A\) if \(d\mid a\) and \(d\mid b\text{;}\) and if \(e\mid a\) and \(e\mid b\) then \(e\mid d\text{.}\)

Remark 5.1.6.

The g.c.d., if exists, is determined up to a unit multiplier.

Definition 5.1.7.

Let \(A\) be an integral domain. We say that \(A\) satisfy the GCD condition if any two elements of \(A\) have a g.c.d.
We will show in the later part that the ring of Gaussian integers \(\Z[i]=\{a+bi:a,b\in\Z\}\) is a UFD. Here we only show that its subring
\begin{equation*} \Z[2i]=\{a+2bi:a,b\in\Z\} \end{equation*}
is not a UFD by showing that primeness condition does not hold.
We show that \(2i\) is an irreducible element \(\Z[2i]\) which is not a prime element. Suppose that \(2i=(a+2bi)(c+2di)\) for some \(a,b,cd\in\Z\text{.}\) Considering the square of the norm in \(\C\) we have
\begin{equation*} 4=(a^2+4b^2)(c^2+4d^2)=(a^2c^2+16b^2d^2)+4(b^2c^2+a^2d^2). \end{equation*}
Note that all terms are non-negative integers. Hence if both \(b,d\) are nonzero then RHS will be strictly greater than \(4\text{.}\) Therefore either \(b=0\) or \(d=0\text{.}\) Suppose that \(b=0\text{.}\) Then the above equation becomes
\begin{equation*} 4=a^2c^2+4a^2d^2=a^2(c^2+4d^2) \end{equation*}
Thus \(a\neq 0\) and either \(d=0\) or \(d=\pm 1\text{.}\) If \(d=0\) then \(4=a^2c^2\text{,}\) i.e, \(2=\pm ac\in\Z\text{,}\) i.e., either \(a=\pm 2\) or \(c=\pm 2\text{.}\) This will contradiction to the assumption, because in this case \(b=d=0\) \(2i=(a+2bi)(c+2di)=2\text{.}\) Other case is left to the reader. Thus \(2i\) is irreducible. Note that \(2i\mid 2\cdot 2\) however \(2i\nmid 2\text{.}\) Hence \(\Z[2i]\) does not satisfy the primeness condition and therefore it is not a UFD.
In this example we use the fact that if \(A\) is UFD then the polynomial ring over \(A\text{,}\) \(A[x]\) is also a UFD. This fact will be proved in the next chapter. Thus if \(F\) is a field then \(F[x,y]\) is a UFD. Consider the following subring of \(F[x]\text{:}\)
\begin{equation*} R=F+x^2F[x]=\left\{\sum_{i=0}^{n}a_ix^i:n\in\N\text{ and }a_1=0\right\}. \end{equation*}
Verify that the units of \(R\) are same as the units of \(F[x]\text{,}\) i.e., units of \(R\) are nonzero elements of \(F\text{,}\) \(F^\times\text{.}\) We claim that \(x^2\) is irreducible in \(R\) but it is not a prime element. Indeed, suppose that \(r,s\in R\) are such that \(x^2=rs\text{.}\) Since \(R\) is a subring of \(F[x]\) the same factorization can be considered in \(F[x]\text{.}\) However as \(F[x]\) is a UFD we must have \(r\sim x\in F[x]\) and \(s\sim x\in F[x]\text{.}\) Since \(x\not\in R\) and units of \(R\) are \(F^\times\) we get \(x^2\) is irreducible in \(R\text{.}\) Furthermore, as \(x\not\in R\) we have \(x^2\nmid x^3\text{.}\) However, \(x^2\mid(x^3)(x^3)\text{.}\) Hence \(x^2\) is irreducible but not a prime in \(R\text{.}\)
Consider a ring homomorphism \(\ev\colon F[x,y]\to F[x]\) given by
\begin{equation*} f(x,y)\mapsto f(x^2,x^3) \end{equation*}
The image of this map is the ring \(R\text{.}\) We now compute its kernel. Note that \((y^2-x^3)\subseteq\ker(\ev)\text{.}\) Now suppose that \(f\in\ker(\ev)\text{.}\) We use division algorithm in \(F[x][y]\text{:}\)
\begin{equation*} f(x,y)=q(x,y)(y^2-x^3)+(a(x)y+b(x))\quad\text{with}\;a(x),b(x)\in F[x]. \end{equation*}
Hence
\begin{equation*} 0=f(x^2,x^3)=a(x^2)x^3+b(x^2),\quad{i.e.,}\quad b(x^2)=-a(x^2)x^3. \end{equation*}
Comparing degree on the both side we must have \(a(x)=0=b(x)\text{.}\) Therefore, \((y^2-x^3)=\ker(\ev)\text{.}\) By Theorem 3.2.3
\begin{equation*} F[x,y]\big/(y^2-x^3)\simeq R\text{.} \end{equation*}
Thus quotient of a UFD need not be a UFD.
<Prev^TopNext>