Math4302 Modern Algebra (Lecture 31)

a) $f(x)$ does not factor as a product of smaller degree polynomials in $\mathbb{Z}[x]$.

From last lecture.

b) if $f(x)=g(x)h(x)$, $g(x),h(x)\in \mathbb{Q}[x]$, $\deg g=r$ and $\deg h=s$. Then there are $\tilde{g}(x),\tilde{h}(x)\in \mathbb{Z}[x]$ with $f(x)=\tilde{g}(x)\tilde{h}(x)$. $\deg \tilde{g}=r$, $\deg \tilde{h}=s$.

$q(x)=b_mx^m+b_{m-1}x^{m-1}+\cdots+b_1x+b_0\in \mathbb{Z}[x]$, $c(q)=\operatorname{gcd}(b_m,\cdots,b_0)$

Fact, $c(pq)=c(p)c(q)$, $p(x),q(x)\in \mathbb{Z}[x]$. (Proof omitted)

Now assume: $f=gh$. We clear the denominators of $g,h$. Let $m$ be a common denominators. multiply by $m$, $mf=g_1h_1$ such that $g_1h_1\in \mathbb{Z}[x]$.

$\deg g_1=\deg_g$, and $\deg h_1=\deg_h$.

Let $p$ be a prime factor of $m$, then $p|c(mf)$. By our Fact. $p|c(mf)\implies p|c(g_1)$ or $p|c(h_1)$.

Suppose $p|c(g_1)$, so $g_1(x)=pg_2(x)$ for some $g_2(x)\in \mathbb{Z}[x]$ with $\deg g_2=\deg g_1=\deg g$.

$mf=pg_2h_1\implies \frac{m}{p}g=g_2h$. Now we can continue with a prime factor of $\frac{m}{p}$.

$f(x)=x^4+x^3+x^2+x+1$ is irreducible in $\mathbb{Q}[x]$.

Apply the Eisenstein criterion.

If $f(x)=g(x)h(x)$ is reducible, then $f(x+1)=g(x+1)h(x+1)$ is also reducible.

So we show $f(x+1)=(x+1)^4+(x+1)^3+(x+1)^2+(x+1)+1$ is irreducible in $\mathbb{Q}[x]$.

$f(x+1)=(x^4+4x^3+6x^2+4x+1)+(x^3+3x^2+3x+1)+(x^2+2x+1)+x+1=x^4+5x^3+10x^2+10x+5$

By the Eisenstein criterion, $f(x+1)$ is irreducible by taking $p=5$.

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More generally, $f(x)=x^{p-1}+x^{p-2}+\cdots+x+1$ is irreducible in $\mathbb{Q}[x]$.

In general if $f(x)=x^{p-1}+x^{p-2}+\cdots+x+1$, $p$ is a prime, then $f(x+1)$ is irreducible by Eisenstein criterion. So $f(x)$ is also irreducible.

$$\begin{aligned} f(x)&=\frac{x^p-1}{x-1}\\ f(x+1)&=\frac{(x+1)^p-1}{x}\\ &=\left (-1+\sum_{k=0}^{p}\binom{p}{k}x^{p-k}\right)/x\\ &=\sum_{k=0}^{p-1}\binom{p}{k}x^{p-k-1}\\ \end{aligned}$$

Note that $\binom{p}{k}$ is $\frac{p!}{k!(p-k)!}$, where $1\leq k\leq p-1$. Where $p$ in $p!$ but $p$ not in $k!(p-k)$, so it is a multiple of $p$.

So we can apply the criterion again.

Let $|F^*|=m$, then (F^*,\cdot)\simeq \mathbb_Z_{m_1}\times\cdots\times\mathbb_Z_{m_r} where $m=m_1\cdots m_r$. $m_i=p_i^{d_i}$.

If $m=18$, then (F^*,\cdot)\simeq \mathbb_Z_{2}\times\mathbb_Z_{3}\times\mathbb_Z_{3} or \mathbb_Z_{9}\times\mathbb_Z_{2} (cyclic).

So it is enough to show that $p_1,\ldots p_r$ are distinct.

So (F^*,\cdot)\simeq \mathbb_Z_{m_1,\ldots,m_r} is cyclic.

Suppose $t=\operatorname{lcm}(m_1,\ldots,m_r)$, then everything in $(F^*,\cdot)$ has order $\leq t$.

But $x^t-1\in F[x]$ has at most $t$ zeros, so $m\leq t$, where $m$ is the size of $(F^*,\cdot)$. Therefore $m_1,\ldots,m_r\leq \operatorname{lcm}(m_1,\ldots,m_r)$. So the $p_i$ are distinct.

So (F^*,\cdot)\simeq \mathbb_Z_{m_1,\ldots,m_r} is cyclic.