MATH 415 Lecture 22

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Irreducibility

Example

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

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

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,

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

However, this system is inconsistent.

Theorem 23.5: Eisenstein Criterion

Let $p\in \mathbb {Z}$ be a prime. Suppose that the polynomial $f(x)=a_{n}\,x^{n}+\dots +a_{1}\,x+a_{0}$ is in $\mathbb {Z} [x]$, and $a_{n}\not \equiv 0{\pmod {p}}$, but $a_{i}\equiv 0{\pmod {p}}$ for all $i<n$, with $a_{0}\not \equiv 0{\pmod {p^{2}}}$. Then $f(x)$ is irreducible over rationals.

Proof. Assume $f(x)=g(x)\,h(x)$ with $\deg {g},\deg {h}\geq 1$ and $g(x),h(x)\in \mathbb {Q} [x]$. Let $\deg {g}=r$ and $\deg {h}=s$ with coefficients $b_{r}\neq 0$ in $g(x)$ and $c_{s}\neq 0$ in $h(x)$:

${\begin{aligned}g(x)&=b_{r}\,x^{r}+\dots +b_{0}\\h(x)&=c_{r}\,x^{r}+\dots +c_{0}\end{aligned}}$

If $a_{0}\not \equiv 0{\pmod {p^{2}}}$, 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 $c_{k}\not \equiv 0{\pmod {p}}$. We have

$a_{m}=b_{0}\,c_{m}+b_{1}\,c_{m-1}+\dots +{\begin{cases}b_{m}\,c_{0}&r\geq m\\b_{r}\,\underbrace {c_{m-r}} _{\equiv 0{\pmod {p}}}&r<m\end{cases}}$

This implies $a_{m}\not \equiv 0{\pmod {p}}$, 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 $\mathbb {Q} [x]$.

1. ${\sqrt {2}}\not \in \mathbb {Q}$
2. $x^{2}-2=(x-a)(x-b)$ for $a,b\in \mathbb {Z}$, then $a=\pm 1,\pm 2$
3. $p=2$, $p\nmid 1$, $p\mid 0$, $p\mid (-2)$
   • $1\not \equiv 0{\pmod {2}}$, $0\equiv 0{\pmod {2}}$, $-2\equiv 0{\pmod {2}}$, and $-2\not \equiv 0{\pmod {2^{2}}}$

Therefore this function is irreducible by Eisenstein Criterion.

Example 2

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

• $p=3$.
• $p^{2}\nmid 12$.

Irreducible

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

$\phi _{p}(x)$ is irreducible over $\mathbb {Q}$.

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

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

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

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

Hence

${\begin{aligned}{\tilde {\phi }}_{p}(x)=\phi _{p}(x+1)&=g(x+1)\,h(x+1)={\tilde {g}}(x)\,{\tilde {h}}(x)\\&={\frac {\left(x+1\right)^{p}-1}{(x+1)-1}}={\frac {x^{p}+{\binom {p}{1}}+\dots +{\binom {p}{i}}\,x^{p-i}+\dots +{\binom {p}{p}}\,x+1-1}{x}}\end{aligned}}$

Since $p$ is prime, we have ${\binom {p}{i}}\equiv 0{\pmod {p}}$ for $i\not \equiv 0{\pmod {p}}$. 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 $\mathrm {End} (A)$ be the set of endomorphisms of $A$.

We define the addition operation $(\varphi +\psi )(a)=\varphi (a)+\psi (a)$, and we get $\left\langle \mathrm {End} (A),+\right\rangle$ is an abelian group with $\sigma :A\to A$ given by the trivial homomorphism $\sigma (a)=0$.

We construct the product $(\varphi \circ \psi )(a)=\phi (\psi (a))$. This satisfies the left and right distributive laws, so $\left\langle \mathrm {End} (A),+,\circ \right\rangle$ is a ring.