9.5 Polynomials

Irreducible

$p(x)\in \mathbb{R}[x]$, we say that $p$ is irreducible if $p(x)=h(x)\cdot q(x),p(x)\in \R[x],q(x)\in \R[x]\Rightarrow h(x)$ is constant or $q(x)$ is constant

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Example

$p(x)=x^2-2\in Z[x]\subseteq Q[x]\subseteq R[x]\subseteq C[x]$

Note that $p(x)$ is not irreducible in $R[x]$

$p(x)=(x-\sqrt2)(x+\sqrt2)$

But $p(x)$ is irreducible in $Q[x]$

Suppose that $p(x)=(x-a)(x-b)$ with $a,b\in Q$

$\Rightarrow p(x)=x^2-(a+b)x+ab\Rightarrow a+b=0/ab=-2$, then we have $b=-a,a^2=2$

$\Rightarrow a\notin Q$ Contradiction

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Proposition=9=

$p(x)\in K[x]$,then

$deg(p)=1\Rightarrow p$ is irreducible

Suppose $p(x)=h(x)q(x)$

Then $1=deg(p)=deg(h)+deg(q)\Rightarrow deg(h)=0~or~deg(q)=0$
$deg(p)=2~or ~3$, then p is irreducible $\Leftrightarrow$ p has no roots in $K$

$\Rightarrow$) If $p$ has roots: $p(\alpha)=0\Rightarrow(x-\alpha)|p(x)\Rightarrow(x-\alpha)h(x)=p\left(x\right)\Rightarrow deg(h)>0$

then $p(x)$ has a nontrivial factorization

$\Leftarrow$) Hypothesis: $p(x)$ has no roots in $K$

Suppose $p(x)=h(x)q(x)$

Key observation: $deg(h)\neq 1,deg(q)\neq 1$

If $deg(h)=1\Rightarrow h(x)=ax+b$ with $a,b\in K$

Then $h(-\frac{b}{a})=0\Rightarrow p(-\frac{b}{a})=0\Rightarrow p$ has a root, contradiction

The same works to prove that $deg(q)\neq 1$

Then, if $p(x)=h(x)q(x)$

$deg(p)=2\Rightarrow$Since $2=2+0=0+2$, then either h or q is a constant

$deg(p)=3\Rightarrow$Since $3=3+0=0+3$, then either h or q is a constant

Therefore p is irreducible

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Proposition

Let $p(x)\in K[x]$,$deg(p(x))=n\in N_0$

Then $p(x)$ has at most n roots in K (No more than n roots)

Claim: If $\alpha_1\,...,\alpha_k$ are different roots of a polynomial $p(x)$

Then $p(x)=(x-\alpha_1)(x-\alpha_2)...(x-\alpha_k)c(x)$

Induction in $k$:

Base case: $k=1$

$p(\alpha) = 0 \implies (x - \alpha) | p(x) \implies p(x) = (x - \alpha) \cdot c(x)$

Inductive step:

Suppose that the claim is true for $k - 1$

Consider $\alpha_1, \alpha_2, \dots, \alpha_{k}$ roots of $p(x)$

Then $p(x) = (x - \alpha_k) \cdot q(x)$

Then $\alpha_1, \alpha_2, \dots, \alpha_{k-1}$ are roots of $q(x)$

$\implies q(x) = (x - \alpha_1) \cdot \dots \cdot (x - \alpha_{k-1}) \cdot c(x)$

Then, $p(x) = (x - \alpha_k) \cdot q(x) = (x - \alpha_k) \cdot (x - \alpha_1) \cdot \dots \cdot (x - \alpha_{k-1}) \cdot c(x)$.

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Eisenstein criterion for irreducibility=8=

Let $p(x)\in Z[x]$, $\begin{aligned}p(X)=\sum_{j=0}^na_jx^j\end{aligned}$

Suppose that there exists a prime number q

$q \mid a_j$ for $j = 0, 1, \dots, n-1$
$q^2 \nmid a_0$
$q \nmid a_n$

Then $p(x)$ is irreducible

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Before the proof, an example

$p(x) = 2x^5 - 6x^3 + 9x^2 - 15$.

Note that $q = 3$ works:
a) $3 \mid -15$, $3 \mid 9$, $3 \mid -6$ ✓
b) $9 \nmid -15$ ✓
c) $3 \nmid 2$ ✓

Then: $p(x)$ cannot be factored as $p(x) = h(x) \cdot q(x)$ with a non-trivial decomposition.

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The proof:

Suppose we have $p(x) = h(x) \cdot q(x)$, where $\deg(h) > 0$ and $\deg(q) > 0$.

Write: $\begin{aligned}p(x) = \sum_{j=0}^{n} a_j x^j = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0\end{aligned}$ $\begin{aligned}h(x) = \sum_{j=0}^{d} b_j x^j\end{aligned}$ $\begin{aligned}q(x) = \sum_{j=0}^{e} c_j x^j\end{aligned}$

We know that $n = d + e \quad (d < n)$.

Now, since $q \mid a_0$$\implies q \mid b_0 c_0$

$p(x) = \left(b_d x^d + b_{d-1} x^{d-1} + \dots + b_1 x + b_0\right)\left(c_e x^e + c_{e-1} x^{e-1} + \dots + c_1 x + c_0\right)$

$\implies q \mid b_0$ or $q \mid c_0$ due to the hypothesis ( $q^2 \nmid a_0$)

Also, $q^2 \nmid a_0 \implies (q \mid b_0 \land q \nmid c_0)$ or $(q \nmid b_0 \land q \mid c_0)$

Suppose that the case is $q \mid b_0 \land q \nmid c_0$.

Consider $k = \min \{ j \in \mathbb{N} : q \nmid b_j \}~~~0<k\leq deg(h)=d<n$

i.e., $q \nmid b_k$, $q \mid b_{0}, q \mid b_{1}, \dots, q \mid b_{k-1}$

Consider the coefficient $a_k$ from $p(x)$:

$\begin{aligned}a_k = \sum_{i+j=k} b_i c_j = b_k c_0 + b_{k-1} c_1 + \dots + b_1 c_{k-1} + b_0 c_k\end{aligned}$

By hypothesis, $q \mid a_k$ and $q \mid b_0, b_1, \dots, b_{k-1}$.

$\implies q \mid b_k c_0$. Then $q \mid b_k$: Contradiction!

Conclusion: $h(x)$ is constant or $q(x)$ is constant

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Polynomials in $\mathbb{Z}[x]$ and rational roots=8=

Let $m \in \mathbb{N}$, $\begin{aligned}p(x) = \sum_{j=0}^{n} a_j x^j\end{aligned}$. Consider the fraction $\frac{r}{s} \in \mathbb{Q}$ with $(r:s) = 1,r,s\in \Z$.

If $p\left( \frac{r}{s} \right) = 0 \implies r \mid a_0 \ \land \ s \mid a_n$

Proof: We have that $p\left( \frac{r}{s} \right) = 0$.

$\begin{aligned}0 = \sum_{j=0}^{n} a_j \left( \frac{r}{s} \right)^j\end{aligned}$

Multiplying by $s^m$:

$\begin{aligned}0 = \sum_{j=0}^{n} a_j r^j s^{n-j}\end{aligned}$

$0 = a_0 s^n + a_1 r s^{n-1} + a_2 r^2 s^{n-2} + \dots + a_{n-1} r^{n-1} s + a_n r^n$

$-a_0 s^n = r \left( a_1 s^{n-1} + a_2 r s^{n-2} + \dots + a_n r^{n-1} \right)$

$\implies r \mid a_0$ since the right hand can be divisible by r, so the left hand is same, due to $(r:s)=1$

Similarly

$-a_n r^n = s \left( a_0 s^{n-1} + a_1 r s^{n-2} + \dots + a_{n-1} r^{n-1} \right)$

$\implies s \mid a_n$

Example:

$p(x) = 3x^3 + 2x^2 - 2x - 8$

If $\frac{r}{s} \in \mathbb{Q}$ is a root, $p\left( \frac{r}{s} \right) = 0$, then

$\frac{r}{s} \in \left\{ \pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{4}{3}, \pm \frac{8}{3} \right\}$

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Polynomials in $\mathbb{R}[x]$ and complex roots:

Proposition

Let $p(x) \in \mathbb{R}[x]$, $\begin{aligned}p(x) = \sum_{j=0}^{m} a_j x^j\end{aligned}$.

If $z \in \mathbb{C}$ is a root $(p(z) = 0)$, then $\bar{z}$ is also a root: $p(\bar{z}) = 0$.

Example

$p(x) = x^2 + 4$

$z = 2i$ is a root, $p(2i) = 4i^2 + 4 = 0$

Then $\bar{z} = -2i$ is also a root: $p(-2i) = 4i^2 + 4 = 0$

Proof

$\begin{aligned} p(\bar{z}) &= \sum_{j=0}^{n} a_j \bar{z}^j = \sum_{j=0}^{n} {a_j} \overline{(z)^j} = \sum_{j=0}^{n} \bar{a_j} \overline{z^j} \quad \text{(since $a_j \in \mathbb{R} \implies \bar{a_j} = a_j$)} \end{aligned} $

$\begin{aligned} = \sum_{j=0}^{n}\overline{ a_j z^j} = \overline{\sum_{j=0}^{n} a_j z^j} = \overline{p(z)} = \overline{0} = 0\end{aligned} $

Therefore, $p \in \mathbb{R}[x] \implies$ complex roots come by pairs.

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Consequence

$p(x) \in \mathbb{R}[x]$, $\deg(p(x)) = 2k+1$, $k \in \mathbb{N}_0$ (odd degree).

Then, $\exists \alpha \in \mathbb{R}$ such that $p(\alpha) = 0$.

complex roots are pair so they are even

Fundamental Theorem of Algebra (without proof)

$p(x) \in \mathbb{C}[x]$, $\deg(p(x)) = n \in \mathbb{N}_0$.

Then $p(x)$ has exactly $n$ roots in $\mathbb{C}$.

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Corollary

$p(x) \in \mathbb{C}[x]$. Then $p(x)$ is irreducible in $\mathbb{C}[x]$ $\iff \deg(p(x)) = 1$.
$p(x) \in \mathbb{C}[x]$, then it can be expressed as $p(x) = (x - c_1)(x - c_2) \dots (x - c_m)$ with $C_j \in \mathbb{C}$ and $n = \deg(p(x))$.

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Note 9.9

For a polynomial $p(x) \in R[x]$ (remember $R \in \{ \mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C} \}$),

We say that $p(x)$ is irreducible if:

\[ p(x) = h(x) \cdot q(x) \implies h(x) \text{ or } q(x) \text{ is an invertible constant } c \in R. \]

Note: $c \in \mathbb{Q}, \mathbb{R}, \mathbb{C}, (c \neq 0)$

Then, $\exists c^{-1}:c^{-1}\cdot c=1$

The difference appears in $\mathbb{Z}[x]$:

\[ p(x) = 2x + 4 = 2(x + 2) \]

Not irreducible in $\mathbb{Z}[x]$.

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Primitive

Definition: Let $p(x) \in \mathbb{Z}[x]$,

\[ p(x) = \sum_{j=0}^{n} a_j x^j, \ a_j \in \mathbb{Z} \]

We say that $p(x)$ is primitive if:

\[ \gcd(a_0, a_1, \dots, a_n) = 1 \]

Equivalently: $q \mid a_j \ \forall j = 0, \dots, m \implies q = \pm 1$

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Proposition

Then, if $p(x) \in \mathbb{Z}[x]$, primitive, then:

$p(x)$ irreducible $\Leftrightarrow p(x)$ has no proper factorizations

Gauss' Lemma=8=

Consider $f(x) \in \mathbb{Z}(x)$ such that $f(x) = h(x) \cdot g(x) \text{ with } g(x), h(x) \in \mathbb{Z}(x)$

Let

\[ h(x) = \sum_{j=0}^{n} a_j x^j, \quad g(x) = \sum_{j=0}^{m} b_j x^j,\quad f(x) = \sum_{k=0}^{n + m} c_k x^k \]

Then: If $p \in \mathbb{Z}$ is a prime such that $p \mid c_k \ \forall k = 0, \dots, m+n$,

\[ \implies p \mid a_j \text{ for } j = 0, \dots, n \text{ or } p \mid b_j \text{ for } j = 0, \dots, m. \]

Proof: Let $p \in \mathbb{Z}$ be a prime such that $p \mid c_k, \ k = 0, \dots, m+n.$

Suppose the lemma is not true:

\[ \exists \ r = \min\{j = 0, \dots, m : p \nmid a_j\} \text{ and } s = \min\{j = 0, \dots, m : p \nmid b_j\}. \]

This means that $p \mid a_0, p \mid a_1, \dots, p \mid a_{r-1}, p \nmid a_r$

$p \mid b_0, p \mid b_1, \dots, p \mid b_{s-1}, p \nmid b_s$

Consider the coefficient $c_{r+s}$ in $f(x)$:

\[ c_{r+s} = \sum_{i+j = r+s} a_i b_j = \sum_{k=0}^{r+s} a_k b_{r+s-k} \]

Thus,

\[ c_{r+s} = a_0 b_{r+s} + a_1 b_{r+s-1} + \dots + a_{r+s} b_0. \]

\[ c_{r+s} = \sum_{k=0}^{r+s} a_k b_{r+s-k} = a_r b_s + \sum_{k=0}^{r-1} a_k b_{r+s-k} + \sum_{k=r+1}^{r+s} a_k b_{r+s-k} \]

By hypothesis, the entire sum $c_{r+s}$ is a multiple of $p$. Breaking this down:

$a_r b_s$: This term is the only one where neither $a_r$ nor $b_s$ is divisible by $p$.
$\sum_{k=0}^{r-1} a_k b_{r+s-k}$: Each $a_k$ for $k = 0, \dots, r-1$ is divisible by $p$ (multiple of $p$).
$\sum_{k=r+1}^{r+s} a_k b_{r+s-k}$: Each $b_j$ for $j = 0, \dots, s-1$ is divisible by $p$ (multiple of $p$).

This implies that $p \mid a_r \cdot b_s \implies p \mid a_r \text{ or } p \mid b_s$

Contradiction!

Equivalently: $f(x), g(x), h(x) \in \mathbb{Z}(x)$. If $g(x), h(x)$ are primitive and $f(x) = g(x) \cdot h(x)$, then $f(x)$ is primitive.

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Some remarks on primitive polynomials:

Let $p(x) \in \mathbb{Z}[x]$. Then there exists a constant $d \in \mathbb{Z}$ such that $p(x) = d \cdot p_0(x)$ with $p_0(x)$ primitive.
Let $f(x) \in \mathbb{Q}[x]$. Then $\exists \ a \in \mathbb{Z}$ such that $f(x) = \frac{1}{a} f_0(x)$ with $f_0(x) \in \mathbb{Z}[x]$.

For example: $f(x) = \frac{1}{2} x^2 + \frac{2}{3} x + \frac{1}{4} = \frac{1}{2} (6x^2 + 8x + 3)$

Let $f(x) \in \mathbb{Q}[x]$. There exist $a \in \mathbb{Q}$ such that $f(x) = a \cdot f_0(x)$ with $f_0(x) \in \mathbb{Z}[x]$ primitive.

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Remark=8=

If there is also another way of writing this, $f(x) = b \cdot g_0(x)$ with $g_0(x) \in \mathbb{Z}[x]$

Then $a = \pm b, \ f_0(x) = \pm g_0(x)$ and $f_0(x)$ is primitive.

To see this, consider: $a \cdot f_0(x) = b \cdot g_0(x)$ with $f_0(x), g_0(x)$ primitive.

Equivalently: $f_0(x) = \frac{b}{a} \cdot g_0(x)$

Write $\frac{b}{a} = \frac{r}{s}$ with $r, s \in \mathbb{Z}$ and $\gcd(r, s) = 1$.

Suppose $r > s > 0$ (the other cases are similar).

Now consider: $\frac{b}{a} = \frac{r}{s}$ with $f_0(x), g_0(x)$ primitive.

If $p \mid r$ and $p \mid s \implies p$ divides all the coefficients in $s \cdot f_0(x)$.

Then $p$ is a divisor of all coefficients in $f_0(x)$. Contradiction.

Thus, $a = \pm b$.

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Theorem=8=

Let $p(x) \in \mathbb{Z}[x]$ be primitive.

$p(x)$ irreducible in $\mathbb{Z}[x] \Leftrightarrow p(x)$ irreducible in $\mathbb{Q}[x]$.

$\Leftarrow$ Suppose that $p(x)$ is not irreducible in $\mathbb{Z}[x]$.

Then $p(x) = h(x) \cdot q(x)$ with $\deg(h(x)) \geq 1$, $\deg(q(x)) \geq 1$.

Since $h(x), q(x) \in \mathbb{Z}[x] \subseteq \mathbb{Q}[x] \implies p(x)$ can be reduced in $\mathbb{Q}[x]$.

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\\\\h(x) g(x) 不一定 primitive

$\Rightarrow$ Suppose that $p(x) = h(x) \cdot q(x)$ with $h(x), q(x) \in \mathbb{Q}[x]$, $\deg(h(x)) \geq 1$, $\deg(q(x)) \geq 1$.

Let $a, b \in \mathbb{Z}$ such that:

$a \cdot h(x) \in \mathbb{Z}[x]$, so $a \cdot h(x) = \bar{h}(x)$,
$b \cdot q(x) \in \mathbb{Z}[x]$, so $b \cdot q(x) = \bar{q}(x)$.

Thus $a \cdot b \cdot p(x) = a \cdot \bar{h}(x) \cdot b \cdot \bar{q}(x) = \bar{h}(x) \cdot \bar{q}(x)$

Let $\alpha, \beta \in \mathbb{Z}$ such that:

$\bar{h}(x) = \alpha \cdot h_0(x)$, with $h_0(x)$ primitive,
$\bar{q}(x) = \beta \cdot q_0(x)$, with $q_0(x)$ primitive.

Thus $a \cdot b \cdot p(x) = \alpha \cdot \beta \cdot h_0(x) \cdot q_0(x)$

Therefore $p(x) = \frac{\alpha \cdot \beta}{a \cdot b} \cdot h_0(x) \cdot q_0(x)$

Since $h_0(x) \cdot q_0(x)$ is primitive (by Gauss' Lemma), $p(x)$ is primitive.

Then (we saw it before): $p(x) = \pm h_0(x) \cdot q_0(x)$

$\Rightarrow p(x)$ is not irreducible in $\mathbb{Z}[x]$.

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9.5 Polynomials

Irreducible

Example

Proposition=9=

Proposition

Eisenstein criterion for irreducibility=8=

Before the proof, an example

Polynomials in \(\mathbb{Z}[x]\) and rational roots=8=

Polynomials in \(\mathbb{R}[x]\) and complex roots:

Proposition

Example

Proof

Consequence

Fundamental Theorem of Algebra (without proof)

Corollary

Note 9.9

Primitive

Proposition

Gauss' Lemma=8=

Some remarks on primitive polynomials:

Remark=8=

Theorem=8=