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Algebra-II: Companion notes

Abhay Soman

Section 4.5 Arithmetic in \(F[X]\)

Observation 4.5.2.

Let \(F\) be a field and let \(f,g\in F[X]\) be nonzero polynomials. Suppose that \(f\mid g\) and \(g\mid f\text{.}\) Thus there exists \(h_1,h_2\in F[X]\) such that \(g=fh_1\) and \(f=gh_2\text{.}\) Therefore, \(f=fh_1h_2\text{,}\) i.e., \(f(1-h_1h_2)=0\in F[X]\text{.}\) As \(F[X]\) is an integral domain and \(f\neq 0\text{,}\) we have \(h_1h_2=1\text{,}\) i.e., \(h_1,h_2\) are units in \(F[X]\text{.}\) Hence, \(h_1,h_2\) are constant polynomials.

Definition 4.5.3. (Associates).

Two polynomials \(f,g\in F[X]\) are said to be associates if there is a \(u\in F^\times\) such that \(f=ug\text{.}\)
We assume that \(I\) is a nonzero ideal. Let \(f\in I\) be a nonzero element of least degree among all elements of \(I\text{.}\) We claim that \(I=(f)\text{.}\) Let \(h\in I\text{.}\) We apply Corollary 4.4.2 to get \(q,r\in F[X]\) with \(\deg(r)\lt\deg(f)\) such that \(h=qf+r\text{.}\) Now if \(r\neq 0\) then we get that \(r=h-qf\in I\text{.}\) This is a contradiction to the minimality of \(\deg(f)\text{.}\) Therefore, \(r=0\) and \(f\mid h\text{.}\) Hence the result is proved.
Let \(f\in F[X]\) be an irreducible polynomial. Suppose that \(I\) is an ideal containing \((f)\text{.}\) By Theorem 4.5.5, there exists \(A\in F[X]\) such that \(I=(A)\text{.}\) Since \(f\in I\) we have \(A\mid f\text{.}\) However as \(f\) is irreducible we must have \(A\) is a unit or \(A\sim f\text{.}\) In other words, \(I=(f)\) or \(I=F[X]\text{.}\)
Suppose that \(f\nmid g\text{,}\) i.e., \(g\not\in (f)\text{.}\) Consider the ideal \((f,g)\text{.}\) By Lemma 4.5.6, \((f,g)=F[X]\text{,}\) i.e., there exist \(A,B\in F[X]\) such that \(Af+Bg=1\text{.}\) Therefore \(Afh+Bgh=h\text{.}\) By the assumption \(gh\in (f)\) and hence \(h\in (f)\text{,}\) i.e., \(f\mid h\text{.}\)
If \(f\) is irreducible then we are done. So, assume that \(f\) is reducible. Let \(f=f_1f_2\) be a factorization of \(f\) and neither \(f_i\) is a constant polynomial. If both \(f_i\) are irreducible polynomials we are done. Now assume that one of the polynomials, say \(f_1\) is reducible. So, there are \(f_{1i}\in F[X]\text{,}\) not both constant, such that \(f_1=f_{11}f_{12}\text{.}\) Say \(f_{11}\) is reducible. Then we continue the process. We claim that this process will eventually terminate. Suppose not. Then we get the following infinite chain of ideals:
\begin{equation*} I_0=(f)\subsetneq I_1=(f_1)\subsetneq I_2=(f_{11})\subsetneq\cdots\subsetneq F[X] \end{equation*}
Since \(f_2\) is not a constant polynomial the first inclusion is strict. Similar reasoning implies that all inclusion are strict. Consider the ideal \(I=\cup_{i=0}^{\infty} I_i\text{.}\) Since every ideal of \(F[X]\) is generated by a single element, there is \(g\in F[X]\) with \((g)=I\text{.}\) Therefore, there is \(n\) such that \(g\in I_n\text{.}\) Hence, \((g)=I\subseteq I_n\subseteq I\text{,}\) i.e., \(I_n=I_{k}\) for any \(k\geq n\text{.}\) Thus this shows that every non-constant polynomial is a product of irreducible factors. Now assume that
\begin{equation*} f=p_1p_2\cdots p_n=q_1q_2\cdots q_m\quad\text{with}\;n\leq m \end{equation*}
Since \(p_1\) is irreducible it divides \(q_i\) for some \(i\text{.}\) By renumbering we assume that \(p_1\) divides \(q_1\text{.}\) As \(q_1\) is irreducible there is \(u_1\in F^\times\) such that \(u_1p_1=q_1\text{.}\) Therefore, we have the following.
\begin{equation*} f=p_1p_2\cdots p_n=u_1p_1q_2\cdots q_m \end{equation*}
Thus \(p_1(p_2\cdots p_n-u_1q_2\cdots q_m)=0\text{.}\) Therefore, as \(F[X]\) is an integral domain, \(p_2p_3\cdots p_n=q_2q_3\cdots q_m\text{.}\) We repeat the above argument to get \(1=u_1u_2\cdots u_nq_{n+1}\cdots q_m\text{,}\) i.e., \(q_{n+1},\ldots,q_m\) are units in \(F[X]\text{.}\) Hence, \(q_{n+1},\ldots,q_m\) are constant polynomials. Therefore, \(m=n\) and the last statement of the theorem is proved.
By Theorem 4.5.5, there exists \(f\in F[X]\) be such that \(I=(f)\text{.}\)
Suppose that \(I\) is a nonzero prime ideal. Let \(g,h\in F[X]\) be such that \(f=gh\text{.}\) Hence \(gh\in I\text{.}\) Since \(I\) is a prime ideal, either \(g\in I\) or \(h\in I\text{.}\) If \(g\in I\) then, using Observation 4.5.2, we get that \(g=uf\) for some unit \(u\text{.}\) Similar arguments shows that if \(h\in I\) then \(h=u^\prime f\) for some unit \(u^\prime\text{.}\) Hence, \(f\) is irreducible.
Conversely, assume that \(f\) is irreducible. If \(gh\in (f)\) then \(gh=fk\) for some \(k\in F[X]\text{.}\) By the Theorem 4.5.8, either \(g\) or \(h\) has \(f\) as one of the irreducible factors. Hence, either \(g\in(f)\) or\(h\in(f)\text{.}\)

Definition 4.5.10.

Let \(F\) be a field and let \(f,g\in F[X]\text{.}\)
The greatest common divisor of \(f,g\) is a polynomial \(d\in F[X]\) such that \(d\mid f\) and \(d\mid g\) and if \(e\in F[X]\) divides both \(f,g\) then \(e\mid d\text{.}\) It is denoted by \(\gcd(f,g)\text{.}\)
The least common multiple of \(f,g\) is a polynomial \(\ell\in F[X]\) such that \(m\mid\ell\) and \(n\mid\ell\) and if \(e\in F[X]\) is divisible by both \(f,g\) then \(\ell\mid e\text{.}\)
We have following properties of ideals corresponding to GCD and LCM.
By Theorem 4.5.5, every ideal of \(F[X]\) is principal. Therefore, we can assume that \(I+J\) and \(I\cap J\) are principal.
We first show that \(I+J\) is generated by \(d=\gcd(f,g)\text{.}\) Since \((d)=I+J\text{,}\) we have \(d\mid f\) and \(d\mid g\text{.}\) If \(\ell\in F[X]\) is such that \(e\mid f\) and \(e\mid g\) then \(e\mid fx+gy\) for every \(x,y\in F[X]\text{,}\) i.e., \(\ell\mid d\text{.}\) Hence \(d=\gcd(f,g)\text{.}\)
Suppose that \((e)=I\cap J\text{.}\) We show that \(e=lcm(f,g)\text{.}\) Since \(e\in I\) and \(e\in J\) we have \(f\mid e\) and \(g\mid e\text{.}\) Suppose that \(\ell\in F[X]\) is such that \(f\mid\ell\) and \(g\mid\ell\text{.}\) Thus \(\ell\in I\cap J=(e)\text{,}\) i.e., \(e\mid\ell\text{.}\) Hence \(e=lcm(f,g)\text{.}\)
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