We offer exercises on polynomials with detailed proofs. In fact, our goal is to show the student how to calculate the greatest common divisor of two polynomials, and how to use the Bezout relation between polynomials.

## Definition and properties of a polynomial

In the sequel,  $\mathbb{K}$ is a commutative ring.

A polynomial $P$ is an infinite sequance $(a_n)_{n\ge 0}$ such that there exist an integer $p\ge 0$ with $a_k=0$ for any $k\ge p+1$. The smallest element $p$ such that $a_k\neq 0$ for $k\ge p+1$ is called the degree of $P$ and will be denoted by ${\rm deg}(P)$.

Let $p$ be the degree of $P,$ we can write \begin{align*} P=a_0(1,0,\cdots)+a_1(0,1,0,\cdots)+\cdots+a_p (0,\cdots,0,1,0,\cdots).\end{align*} If we select \begin{align*} 1=X^0=(1,0,\cdots),\;X=(0,1,0,\cdots),\; \cdots,X^p=(0,\cdots,0,1,0,\cdots),\end{align*} then the polynomial $P$ takes the form \begin{align*}P=a_0+a_1 X+\cdots+a_p X^p.\end{align*} The set of all polynomials with coefficients in $\mathbb{K}$ is denoted by $\mathbb{K}[X]$ and the set of all polynomials of degrees less or equals $p$ is denoted by $\mathbb{K}_p[X]$.

Addition of polynomials: Let $P=(a_n)_n$ and $Q=(b_n)_n$ be two polynomials. Then $P+Q$ is a polynomial with coefficients $(a_n+b_n)_n$.

Scalar multiplication of a polynomial: Let $P=(a_n)_n$ be a polynomial with coefficients in $\mathbb{K},$ and $\lambda\in\mathbb{K}$ be a scalar. Then $\lambda P$ is a polynomial with coefficients $(\lambda a_n)_n$.

Product of polynomials: Take two polynomials $P=(a_n)_n$ and $Q=(b_n)_n$ then $PQ=(c_n)_n$ is a polynomial with \begin{align*} c_n=\sum_{k=0}^n a_k b_{n-k}.\end{align*}

Proposition: $(\mathbb{K},+,\cdot)$ is a commutative ring.

## A selection of exercises on polynomials

Exercise: Determine the greatest common divisor (gcd) and a Bezout relation between the polynomials $P=X^3+X^2+X-3$ and $Q=X^2-3X+2$.

Solution: Here we shall apply Euclid’s algorithm: we select $P_0=P$ and $P_1=Q$. Moreover, we put $U_0=1,\,U_1=0$ and $V_0=0,\,V_1=1$ in order to have $PU_0+QV_0=P_0$ and $PU_1+QV_1=P_1$. The Euclidean division of $P_0$ by $P_1$ is \begin{align*} P_0=P_1(X+4)+(11 X-11). \end{align*} We set $P_2=11 X-11$. Then \begin{align*} P_2&=P_0-P_1(X+4)\cr &= PU_0+Q V_0-(PU_1+QV_1)(X+4)\cr &= P(U_0-(X+4)U_1)+Q(V_0-(X+4)V_1)\cr &= P U_2+Q V_2, \end{align*} where \begin{align*} U_2=U_0-(X+4)U_1=1,\quad V_2=V_0-(X+4)V_1=-X-4. \end{align*} The Euclidean division of $P_1$ by $P_2$ is \begin{align*} P_1=P_2 \left(\frac{1}{11}X-\frac{2}{11}\right)+0 \end{align*} The GCD of $P$ and $Q$ is \begin{align*} \frac{1}{11}P_2=X-1. \end{align*} The Bezout relation is then \begin{align*} \frac{1}{11} P-\frac{1}{11}(X+4) Q=X-1. \end{align*}

Exercise: Does the polynomial $P=X^4+1$ is irreducible in $\mathbb{C}[X]$ ? in $\mathbb{R}[X]$ ? in $\mathbb{Q}[X]$ ?

Solution: It is not irreducible neither in $\mathbb{C}[X]$ nor in $\mathbb{R}[X]$ as it is not either of degree 1 or of degree 2.

Assume that $P$ is reducible in $\mathbb{Q}[X]$. Remark that $P$ does not have rational roots “and also real roots”, $P$ can not admit a divisor of degree $1$. Then $P$ is the product of two polynomials of degree 2. We then can assume that \begin{align*} P=(X^2+aX+b)(X^2+\alpha+\beta)\end{align*} with $(a,b,\alpha,\beta)\in\mathbb{Q}^4$. By identification of coefficients, we obtain \begin{align*} a+\alpha=0,\quad a\alpha+b+\beta=0,\quad a\beta+\alpha b=0,\quad b\beta=1. \end{align*}

From this, we deduce that $b$ and $\beta$ are not null and have the same sign so that $b+\beta\neq 0$. Consequently, $\alpha=-a\neq 0$, and the third equation gives $b=\beta$. Finally, $b=\beta=\pm 1,$ and hence $a^2=\pm 2$ this is not possible if $a\in \mathbb{Q}$. Thus $P$ is irreducible in $\mathbb{Q}[X]$.

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