MATH 415 Lecture 22

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Irreducibility

Example

$f(x)=x^4-2x^2+8x+1\in \mathbb{\mathbb{Q}}[x]$

If $f(x)=(x-a)g(x)$, where $\mathrm{deg}g=3$, we must have $a\mid 1$, so $a=\pm 1$. However, $f(\pm 1)\ne 0$

If $f(x)=(x^2+ax+b)(x^2+cx+d)$, then we must have $f(x)=x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$. In particular,

$\{\begin{array}{l}bd=1ad+bc=8b+d+ac=-2a+c=0\end{array}$

However, this system is inconsistent.

Theorem 23.5: Eisenstein Criterion

Let $p\in \mathbb{\mathbb{Z}}$ be a prime. Suppose that the polynomial $f(x)=a_n{x}^n+\dots +a_{1}x+a_0$ is in $\mathbb{\mathbb{Z}}[x]$, and $a_n\equiv ̸0(\mathrm{mod}p)$, but $a_i\equiv 0(\mathrm{mod}p)$ for all $i<n$, with $a_0\equiv ̸0(\mathrm{mod}{p}^2)$. Then $f(x)$ is irreducible over rationals.

Proof. Assume $f(x)=g(x)h(x)$ with $\mathrm{deg}g,\mathrm{deg}h\ge 1$ and $g(x),h(x)\in \mathbb{\mathbb{Q}}[x]$. Let $\mathrm{deg}g=r$ and $\mathrm{deg}h=s$ with coefficients $b_r\ne 0$ in $g(x)$ and $c_s\ne 0$ in $h(x)$:

$\begin{array}{rl}g(x)& =b_r{x}^r+\dots +b_0\\ h(x)& =c_r{x}^r+\dots +c_0\end{array}$

If $a_0\equiv ̸0(\mathrm{mod}{p}^2)$, then neither $b_0$ nor $c_0$ are congruent to 0 (mod $p$). We must have $a_0=b_0\cdot c_0$ and $a_n=b_r\cdot c_s$.

Let $m$ be the smallest $k$ such that $c_k\equiv ̸0(\mathrm{mod}p)$. We have

$a_m=b_0{c}_m+b_1{c}_{m-1}+\dots +\{\begin{array}{ll}{b}_m{c}_0& r\ge m\\ b_r\underset{\equiv 0(\mathrm{mod}p)}{\underbrace{c_{m-r}}}& r<m\end{array}$

This implies $a_m\equiv ̸0(\mathrm{mod}p)$, so $m=n$.

$s=n$

implies

$r=0$

, which means

$g(x)$

is a constant function. Contradiction!

quod erat demonstrandum

Example 1

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

1. $\sqrt{2}\in ̸\mathbb{\mathbb{Q}}$
2. $x^2-2=(x-a)(x-b)$ for $a,b\in \mathbb{\mathbb{Z}}$, then $a=\pm 1,\pm 2$
3. $p=2$, $p\nmid 1$, $p\mid 0$, $p\mid (-2)$
   • $1\equiv ̸0(\mathrm{mod}2)$, $0\equiv 0(\mathrm{mod}2)$, $-2\equiv 0(\mathrm{mod}2)$, and $-2\equiv ̸0(\mathrm{mod}{2}^2)$

Therefore this function is irreducible by Eisenstein Criterion.

Example 2

Show that $25x^5-9x^4-3x^2-12$ is irreducible over $\mathbb{\mathbb{Q}}$.

• $p=3$.
• $p^2\nmid 12$.

Irreducible

Let ${\varphi }_p(x)=\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+\dots +x+1$. This is called the $p$th cyclotonic polynomial.

${\varphi }_p(x)$ is irreducible over $\mathbb{\mathbb{Q}}$.

To prove this, we will need a well-defined homomorphism ${\phi }_{x+1}:\mathbb{\mathbb{Q}}[x]\to \mathbb{\mathbb{Q}}[x]$ (Take a look at the Remark after 22.5).

Since $\phi$ maps $\mathbb{\mathbb{Q}}[x]$ to itself, it is an automorphism.

Proof. Seeking a contradiction, assume ${\varphi }_p(x)=g(x)h(x)$, where $\mathrm{deg}g,\mathrm{deg}h\ge 1$. Because ${\phi }_{x+1}$ is a homomorphism, we have

${\phi }_{x+1}({\varphi }_p(x))={\phi }_{x+1}(g(x)h(x))={\phi }_{x+1}(g(x)){\phi }_{x+1}(h(x))$

Hence

$\begin{array}{rl}{\tilde{\varphi }}_p(x)={\varphi }_p(x+1)& =g(x+1)h(x+1)=\tilde{g}(x)\tilde{h}(x)\\ & =\frac{{(x+1)}^p-1}{(x+1)-1}=\frac{x^p+(\genfrac{}{}{0ex}{}{p}{1})+\dots +(\genfrac{}{}{0ex}{}{p}{i})x^{p-i}+\dots +(\genfrac{}{}{0ex}{}{p}{p})x+1-1}{x}\end{array}$

Since $p$ is prime, we have $(\genfrac{}{}{0ex}{}{p}{i})\equiv 0(\mathrm{mod}p)$ for $i\equiv ̸0(\mathrm{mod}p)$. Thus Eisenstein Criteroin applies, so the original function is irreducible.

Section 24: Noncommutative Rings

Rings of Endomorphisms

Let $(A,+)$ be an abelian group. A homomorphism $f:A\to A$ is called an endomorphism.

Note: An automorphism is an isomorphisom of a group onto itself, but an endomorphism is a homomorphism of a group onto itself.

Let $\mathrm{E}\mathrm{n}\mathrm{d}(A)$ be the set of endomorphisms of $A$.

We define the addition operation $(\phi +\psi )(a)=\phi (a)+\psi (a)$, and we get $⟨\mathrm{E}\mathrm{n}\mathrm{d}(A),+⟩$ is an abelian group with $\sigma :A\to A$ given by the trivial homomorphism $\sigma (a)=0$.

We construct the product $(\phi \circ \psi )(a)=\varphi (\psi (a))$. This satisfies the left and right distributive laws, so $⟨\mathrm{E}\mathrm{n}\mathrm{d}(A),+,\circ ⟩$ is a ring.