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Algebra-II: Companion notes

Abhay Soman

Section 4.4 Division algorithm

If \(\deg(f)\lt\deg(g)\) then the resuLC follows by taking \(k=0\) and \(q(X)=0\text{.}\) So, assume that \(\deg(g)=m\leq\deg(f)\text{.}\) Suppose that
\begin{equation*} f(X)=a_0+a_1X+\cdots+a_nX^n\quad\text{and}\quad g(X)=b_0+b_1X+\cdots+b_mX^m. \end{equation*}
Put \(f_1(X)=b_mf(X)-a_nX^{n-m}g(X)\text{.}\) Thus
\begin{equation*} f_1(X)=b_ma_0+b_ma_1X+\cdots+b_ma_nX^n-a_nb_0X^{n-m}-a_nb_1X^{n-m+1}-\cdots-a_nb_mX^n. \end{equation*}
Therefore, \(\deg(f_1)\lt\deg(f)\text{.}\) We use induction on \(\deg(f)\text{.}\) Thus, by induction, there exists \(k_1\in\N\text{,}\) and \(q_1(X),r(X)\in A[X]\) with \(\deg(r)\lt\deg(g)\) such that
\begin{equation*} b_m^{k_1}f_1(X)=q_1(X)g(X)+r(X). \end{equation*}
Hence,
\begin{equation*} b_m^{k_1}f_1(X)=b_m^{k_1+1}f(X)-b_m^{k_1}a_nX^{n-m}g(X)=(q_1(X)+b_m^{k_1}a_nX^{n-m})g(X)+r(X). \end{equation*}

Remark 4.4.3.

The above Corollary 4.4.2 is particularly useful when we consider polynomial ring in several variables over a field.

Definition 4.4.4. (Divisor).

Let \(A\) be a commutative ring and let \(f,g\in A[X]\text{.}\) We say that \(g\) divides \(f\text{,}\) and write it as \(g\mid f\text{,}\) if there exists \(q(X)\in A[X]\) such that \(f(X)=q(X)g(X)\text{.}\)

Observation 4.4.5.

Let \(A\) be a commutative ring. Suppose that \(a\in A\) and \(f,g\in A[X]\text{.}\) Then
\begin{equation*} (fg)(a)=f(a)g(a)=g(a)f(a). \end{equation*}
In other words, the evaluation map \(\ev_a\colon F[X]\to F[X]\) is a ring homomorphism. We use this observation without explicitly mentioning it.
If ring is not commutative then \((fg)(a)\) need not be the same as \(f(a)g(a)\text{.}\) We first recall that \((fg)(a)\) is obtained as follows: First write \(fg\) in the form \(c_0+c_1X+\cdots+c_rX^r\) and then \((fg)(a)=c_0+c_1a+\cdots+c_ra^r\text{.}\)
Suppose that \(R=M_2(\R)\) and consider the following polynomials over \(M_2(\R)\text{:}\)
\begin{equation*} f(X)=X-\begin{pmatrix} 1\amp0\\0\amp-1 \end{pmatrix}\in M_2(\R)[X]\quad\text{and}\quad g(X)=X-\begin{pmatrix} 0\amp1\\-1\amp0 \end{pmatrix}\in M_2(\R)[X]. \end{equation*}
Then
\begin{equation*} (fg)(X)=X^2-\begin{pmatrix}1\amp1\\-1\amp-1\end{pmatrix}X+\begin{pmatrix} 0\amp1\\1\amp0 \end{pmatrix}\in M_2(\R)[X]. \end{equation*}
We can then see that
\begin{equation*} (fg)\left(\begin{pmatrix} 1\amp0\\0\amp-1 \end{pmatrix}\right)\neq 0\text{,} \end{equation*}
while
\begin{equation*} f\left(\begin{pmatrix} 1\amp0\\0\amp-1 \end{pmatrix}\right)g\left(\begin{pmatrix} 1\amp0\\0\amp-1 \end{pmatrix}\right)=0\text{.} \end{equation*}

Definition 4.4.6. (Root of a polynomial).

Let \(A\) be a commutative ring and \(f\in A[X]\text{.}\) An element \(a\in A\) is said to be a root of \(f(X)\) in \(A\) if \(f(a)=0\text{.}\)
Indeed, as \(LC(X-a)=1\in A\) is a unit, by Corollary 4.4.2, there exists \(q(X),r(X)\in A[X]\) with \(\deg(r)\lt 1\) (i.e., \(r(X)\) is a constant polynomial) such that \(f(X)=q(X)(X-a)+r(X)\text{.}\) Since \(f(a)=0\) we get \(0=f(a)=q(a)\cdot 0+r(a)\text{.}\) Thus \(r(X)\text{,}\) which is a constant polynomial, must be zero. Hence we get the resuLC.
The case when \(n=1\) is Lemma 4.4.7. We now assume that \(n\gt 1\text{.}\) By repeated application of Lemma 4.4.7, \(f(X)=(X-a_1)^rg(X)\) and \(g(a_1)\neq 0\text{.}\) Since \(A\) is an integral domain, \(\deg(g)\leq\deg(f)-r\lt\deg(f)\text{.}\) Moreover for \(2\leq i\leq n\text{,}\) \(0=f(a_i)=(a_i-a_1)^rg(a_i)\text{.}\) Since \(A\) is an integral domain and \(a_i\neq a_1\text{,}\) we have \(g(a_i)=0\text{.}\) By induction, \((X-a_2)\cdots(X-a_n)\) divides \(g(X)\text{.}\) Therefore, we get the resuLC.

Remark 4.4.10.

Let \(A=\prod_{n\in\N}\Z/3\Z\text{,}\) a product ring as defined in Example 1.2.10. It is a commutative ring but not an integral domain. There are infinitely many roots of the polynomial \(X^2-1_A\in A[X]\) in \(A\text{.}\) For instance, let \(a_{n}=(1,\ldots,1,-1,1,\ldots)\in A\) be such that \(-1\in\Z/3\Z\) occurs at \(n\)-th place and all other entries are \(1\in\Z/3\Z\text{.}\) Then, \(a_{n}\) is a root of \(X^2-1_A\) for each \(1\leq n\lt\infty\text{.}\)
We show that \(\Z[i]\simeq\Z[X]/(X^2+1)\text{.}\) Consider the map \(\ev_i\colon\Z[X]\to\Z[i]\) defined by
\begin{equation*} \sum_ka_kX^k\mapsto\sum_ka_ki^k \end{equation*}
Check that the map \(\ev_i\) is a ring epimorphism. We compute its kernel, \(\ker(\ev_i)=\{f\in\Z[X]:f(i)=0\}\text{.}\) The ideal generated by \(X^2+1\text{,}\) \((X^2+1)\subseteq\ker(\ev_i)\text{.}\) Now suppose that \(f\in\ker(\ev_i)\text{.}\) Since the \(LC(X^2+1)=1\in\Z\) is invertible in \(\Z\text{,}\) we use Corollary 4.4.2, to obtain \(q(X),r(X)\in\Z[X]\) with \(\deg(r)\lt 2\) such that \(f(X)=q(X)(X^2+1)+r(X)\text{.}\) Hence,
\begin{equation*} 0=f(i)=q(i)(i^2+1)+r(i)\in\Z[i],\quad\text{i.e.,}\quad r(i)=0. \end{equation*}
If \(r(X)\neq 0\) then , by Lemma 4.4.7, \(r(X)=n(X-i)\in\Z[X]\) for some \(n\in\Z\text{.}\) This implies that \(i\in\Z\text{,}\) a contradiction. Hence \(r(X)=0\) and \(f\in(X^2+1)\text{.}\) Therefore, \(\ker(\ev_i)=(X^2+1)\) and by Theorem 3.2.3, we get the result.
The below example has connections to geometry. We will not make the connection explicit in this course. However, you may see the curve \(Y^2-X^3\) in \(\R^2\) here 1 .
Consider \(\Q[X,Y]\) and the ideal generated by \(Y^2-X^3\in\Q[X,Y]\text{,}\) \((Y^2-X^3)\text{.}\) We show that if \(f(X,Y)\in\Q[X,Y]\) is such that \(f(a^2,a^3)=0\) for any \(a\in\Q\) then \(f(X,Y)\in(Y^2-X^3)\text{.}\) We consider \(f(X,Y),Y^2-X^3\) as polynomials over \(\Q[X]\) in variable \(Y\text{.}\) Thus we can apply Corollary 4.4.2 to get the following.
\begin{equation*} f=q(X,Y)(Y^2-X^3)+r(X)Y+s(X),\quad\text{where}\quad r(X),s(X)\in\Q[X] \end{equation*}
By the assumption, \(f(a^2,a^3)=0\) for every \(a\in\Q\text{,}\) and hence we get that
\begin{equation*} r(a^2)a^3+s(a^2)=0\quad\text{for every}\quad a\in\Q. \end{equation*}
This shows that the polynomial \(r(X^2)X^3+s(X^2)\in\Q[X]\) has infinitely many roots in \(\Q\text{.}\) By Corollary 4.4.9, we must have \(r(X^2)X^3+s(X^2)=0\in\Q[X]\text{.}\) Comparing even degree and odd degree terms we get that \(r=s=0\in\Q[X]\text{.}\) Thus, \(f\in(Y^2-X^3)\) and the result is proved.
Let \(F\) be a field and let \(F[X_1,X_2,\ldots,X_n]\) be the polynomial ring over \(F\) in \(n\) variables. We give examples of maximal ideals in \(F[X_1,\ldots,X_n]\text{.}\) Fix \(a_1,a_2,\ldots,a_n\in F\text{.}\)
Consider the evaluation \(F\)-algebra homomorphism.
\begin{equation*} \ev_{(a_1,\ldots,a_n)}\colon F[X_1,\ldots,X_n]\to F\quad\text{given by}\quad \sum a_{i_1\cdots i_n}X_1^{i_1}\cdots X_n^{i_n}\mapsto\sum a_{i_1\cdots i_n}a_1^{i_1}\cdots a_n^{i_n} \end{equation*}
Thus the kernel of this map is
\begin{equation*} \ker(\ev_{(a_1,\ldots,a_n)})=\{f\in F[X_1,\ldots,X_n]:f(a_1,a_2,\ldots,a_n)=0\}. \end{equation*}
Hence the ideal generated by \(X_1-a_1,\ldots,X_n-a_n\text{,}\) \((X_1-a_1,\ldots,X_n-a_n)\subseteq\ker(\ev_{(a_1,\ldots,a_n)})\text{.}\) We claim that
\begin{equation*} (X_1-a_1,\ldots,X_n-a_n)=\ker(\ev_{(a_1,\ldots,a_n)}). \end{equation*}
Let \(f\in\ker(\ev_{(a_1,\ldots,a_n)})\text{.}\) We consider \(f\in F[X_1,\ldots,X_{n-1}][X_n]\text{.}\) By Corollary 4.4.2 there exists \(q_n,r_n\in F[X_1,\ldots,X_{n-1}][X_n]\) with \(\deg r_n\lt\deg (X_n-a_n)=1\text{,}\) i.e., \(r_n\) is a constant polynomial in \(F[X_1,\ldots,X_{n-1}][X_n]\text{,}\) i.e., \(r_n\in F[X_1,\ldots,X_{n-1}]\) such that
\begin{equation} f=q_n(X_n-a_n)+r_n.\tag{4.4.1} \end{equation}
We thus have \(0=f(a_1,\ldots,a_n)=0=r_n(a_1,\ldots,a_n)\text{.}\) We now consider \(r_n\in F[X_1,\ldots,X_{n-2}][X_{n-1}]\text{.}\) Again applying Corollary 4.4.2, there exists \(q_{n-1},r_{n-1}\in F[X_1,\ldots,X_{n-2}][X_{n-1}]\) with \(\deg r_{n-1}\lt\deg (X_{n-1}-a_{n-1})=1\text{,}\) i.e., \(r_{n-1}\) is a constant polynomial in \(F[X_1,\ldots,X_{n-2}][X_{n-1}]\) such that
\begin{equation*} r_n=q_{n-1}(X_{n-1}-a_{n-1})+r_{n-1}\text{.} \end{equation*}
Thus we also get \(0=r_{n}(a_1,\ldots,a_{n-1})=r_{n-1}(a_1,\ldots,a_{n-1})\text{.}\) Continuing in this way we get that \(r_2\in F[X_1]\) and, by Corollary 4.4.2, there exists \(q_1,r_1\in F[X_1]\) with \(\deg r_1\lt\deg(X_1-a_1)=1\text{,}\) i.e., \(r_1\in F\) such that
\begin{equation*} r_2=q_1(X_1-a_1)+r_1. \end{equation*}
Thus we have \(0=r_2(a_1)=r_1\text{,}\) i.e., \(r_2=q_1(X_1-a_1)\text{.}\) Putting successive expression of \(r_n,r_{n-1},\ldots,r_2\) in (4.4.1) we get that
\begin{equation*} f=q_n(X_n-a_n)+q_{n-1}(X_{n-1}-a_{n-1})+\cdots+q_2(X_2-a_1)+q_1(X_1-a_1). \end{equation*}
Hence \(\ker(\ev_{(a_1,\ldots,a_n)})\subseteq (X_1-a_1,\ldots,X_n-a_n)\) and we get the desired resuLC.
The \(\ev_{(a_1,\ldots,a_n)}\) is surjective. Hence, by Theorem 3.2.3, \(F[X_1,\ldots,X_n]\big/(X_1-a_1,\ldots,X_n-a_n)\simeq F\text{.}\) By Proposition 3.3.4 the ideal generated by \(X_1-a_1,\ldots,X_n-a_n\text{,}\) \((X_1-a_1,\ldots,X_n-a_n)\) is maximal.
In fact, by Hilbert’s Nullstellensatz, if \(F\) is algebraically closed then every maximal ideal in \(F[x_1,x_2,\cdots,x_n]\) is of the form \((x_1-a_1,\cdots,x_n-a_n)\) for some \(a_1,\ldots,a_n\in F\text{.}\)

Definition 4.4.14. (Reducible and irreducible polynomial in \(F[X]\)).

Let \(F\) be a field. A nonconstant polynomial \(f(X)\in F[X]\) is said to be an reducible polynomial if there exists nonconstant polynomials \(g,h\in F[X]\) with \(\deg(g)\lt\deg(f)\) and \(\deg(h)\lt\deg(f)\) such that \(f=gh\text{.}\)
A nonconstant polynomial in \(F[X]\) is called an irreducible polynomial in F[X] or an irreducible polynomial over \(F\) if it is not reducible.
If we consider \(X^2+1\) over \(\Q\) or \(\R\) then this polynomial is irreducible. However the same polynomial is reducible over \(\C\text{.}\) Indeed, \(X^2+1=(X-i)(X+i)\text{,}\) where \(i\in\C\) is such that \(i^2=-1\text{.}\)
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