Math4302 Modern Algebra (Lecture 30)

Polynomials of degree $1$ are always irreducible.

Polynomials of degree $2$ is reducible if and only if $f$ haz zero in $F$.

$\impliedby$ If $a$ is a zero of $f$, then $(x-a)|f(x)\implies f(x)=(x-a)q(x)$, and $q(x)$ is of degree $\deg f(x)-1=2-1$. (also holds for all)

$\implies$ If $f(x)=g(x)h(x)$, and $\deg h+\deg g=2$, and $1\leq \deg h,\deg g<2$, so $\deg h=1$, $\deg g=1$, so $h(x)=a+bx$ where $b\neq 0$, so $-\frac{a}{b}$ is a zero fo $h$, so it is a zero of $f$.

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A polynomial of $\deg 3$ is reducible $\iff$ $f$ has a zero in $F$.

Same as the $\deg 2$ case.

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$x^2+2\in \mathbb{R}[x]$ is irreducible.

$x^2-2\in \mathbb{R}[x]$ is reducible.

$x^2-2\in \mathbb{Q}[x]$ is irreducible $\sqrt{2}\notin \mathbb{Q}$.

$f(x)=x^3+2x+1$ in $\mathbb{Z}_5[x]$ is irreducible.

• $f(0)=1$
• $f(1)=4$
• $f(2)=3$
• $f(3)=4$
• $f(4)=3$

no Zeros, therefore it is irreducible.

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Note that for degree greater than $4$ we cannot use the zero condition to determine if $f(x)$ is reducible or not.

Consider $(x^2+2)^2\in \mathbb{R}[x]$, with degree $4$, It is reducible but has no zeros in $\mathbb{R}$.

What are all irreducible polynomials of degree $2$ in $\mathbb{Z}_2[x]$?

$ax^2+bx+c\in \mathbb{Z}_2[x]$ with $a=1,b,c\in \mathbb{Z}_2$.

So there are only $4$ polynomials possible.

$x^2$, $x^2+1$, $x^2+x$, $x^2+x+1$.

$0$ is a root for $x^2$, $x^2+x$.

$1$ is a root for $x^2+1$.

So there is no root for $x^2+x+1$, so $x^2+x+1$ is irreducible.

Suppose $f(\frac{r}{s})=0$.

Then

$$\left(\frac{r}{s}\right)^n+a_{n-1}\left(\frac{r}{s}\right)^{n-1}+\cdots+a_1\left(\frac{r}{s}\right)+a_0=0$$

Multiply by $s^n$, we have

$$r^n+a_{n-1}r^{n-1}s+\cdots+a_1rs^{n-1}+a_0s^n=0$$

So $s|r^n$, $\operatorname{gcd}(r,s)=1$, therefore $s=1$.

So let $s=1$, we have

$$r^n+a_{n-1}r^{n-1}+\cdots+a_1r+a_0=0$$

So $r|a_0$, $r\in \mathbb{Z}$.

Our goal is to create contradiction between reducible and prime.

Suppose $f(x)=g(x)h(x)$ with $\deg g(x)=r$ and $\deg h(x)=s$. $g,h\in \mathbb{Q}[x]$. $1\leq r,s\leq n-1$.

Fact:

We can always assume $g(x),h(x)\in \mathbb{Z}[x]$.

Write $g(x)=b_r x^r+\cdots+b_0$, $h(x)=c_s x^s+\cdots+c_0$. Where $b_i,c_i\in \mathbb{Z}$.

Note that $a_0=b_0c_0$, therefore $p|b_0c_0=a_0$, but $p^2\nmid b_0c_0$. This implies that $p$ divides one of them, suppose $p|b_0$ but $p\nmid c_0$.

Let $i$ be the smallest index such that $p\nmid b_i$, $i\leq s<n$

$a_n=b_r c_s\implies p\nmid b_r$, $p\nmid a_n$.

Now look at $a_i$, we claim that $p\nmid a_i$.

$a_i=b_0c_i+b_1c_{i-1}+\cdots+b_{i-1}c_0+b_{i-1}c_1+\cdots+b_{i-1}c_{i-1}+b_ic_0$. But $p\nmid b_0$, $p\nmid c_0$. So $p\nmid a_i$.

This contradicts with out claim that $p|a_0,\cdots,a_{n-1}$.

$f(x)=x^4+8x+2\in \mathbb{Q}[x]$ ?

Apply the (weak) Eisenstein theorem, if $f(x)$ has a zero $\alpha\in \mathbb{Q}$, then $\alpha\in \mathbb Z$, then $\alpha|2$, so $\alpha=\pm1,\pm2$, in all $4$ cases, $f(x)\neq 0$. So there is no root of $f(x)$ in $\mathbb{Q}$.

Use the Eisenstein Criterion, $p=2$, $f(x)$ is irreducible.