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Algebra-II: Companion notes

Abhay Soman

Section 4.8 Examples

Let \(F\) be a field and let \(a,b\in F\) with \(a\neq 0\text{.}\) We show that \(\varphi\colon F[x]\to F[x]\) given by \(\sum \alpha_ix^i\mapsto\sum \alpha_i(ax+b)^i\) is a ring automorphism.
We first show that \(\varphi\) is a ring homomorphism.
  1. \begin{equation*} \varphi\left(\sum\alpha_ix^i+\sum_i\beta_ix^i\right)=\varphi\left(\sum(\alpha_i+\beta_i)x^i\right)=\sum(\alpha_i+\beta_i)(ax+b)^i=\varphi\left(\sum\alpha_ix^i\right)+\varphi\left(\sum\beta_ix^i\right). \end{equation*}
  2. \begin{equation*} \varphi\left(\left(\sum\alpha_ix^i\right)\left(\sum\beta_jx^j\right)\right)=\varphi\left(\sum_k(\sum_{i=0}^k\alpha_i\beta_{k-i})x^k\right)=\sum_k\left(\sum_{i=0}^k\alpha_i\beta_{k-i}\right)(ax+b)^k\text{.} \end{equation*}
    While
    \begin{equation*} \varphi\left(\sum\alpha_ix^i\right)\varphi\left(\sum\beta_jx^j\right)=\left(\sum\alpha_i(ax+b)^i\right)\left(\sum\beta_j(ax+b)^j\right)=\sum_k\left(\sum_{i=0}^k\alpha_i\beta_{k-i}\right)(ax+b)^k. \end{equation*}
Consider the map \(\phi\colon F[x]\to F[x]\) given by
\begin{equation*} \sum\alpha_ix^i\mapsto\sum\alpha_i(x/a-b/a)^i. \end{equation*}
By the arguments in the last paragraph we see that \(\phi\) is a ring homomorphism. Furthermore, using the fact that \(\varphi\) and \(\phi\) are ring homomorphism we get the following.
\begin{equation*} \phi\circ\varphi\left(\sum\alpha_kx^k\right)=\phi\left(\sum\alpha_k(ax+b)^k\right)=\sum\alpha_k\phi\left(ax+b\right)^k=\sum\alpha_k\left(a(x/a-b/a)+b\right)^k=\sum\alpha_kx^k. \end{equation*}
\begin{equation*} \varphi\circ\phi\left(\sum\alpha_kx^k\right)=\varphi\left(\sum\alpha_k(x/a-b/a)^k\right)=\sum\alpha_k\varphi\left(x/a-b/a\right)^k=\sum\alpha_k\left((ax+b)/a-b/a\right)^k=\sum\alpha_kx^k. \end{equation*}
This shows that \(\varphi\) is a ring automorphism.
The polynomial \(x^4+1\in\Q[x]\) is irreducible in \(\Q[x]\text{.}\) Use the affine automorphism (refer Example 4.8.1) given by \(x\mapsto x+1\) and then apply Eisenstein’s criterion.
We show that \(x^4+4\) is reducible over \(\Q\text{.}\) We have
\begin{equation*} x^4+4=(x^4+4x^2+4)-(2x)^2=(x^2+2)^2-(2x)^2=(x^2+2x+2)(x^2-2x+2)\in\Q[x]. \end{equation*}
By Proposition 4.7.2 the polynomial \(x^4+4\) has no root in \(\Q\text{.}\) More directly, since for any \(r/s\in\Q\) we have \((r/s)^4\geq 0\) and hence \((r/s)^4+4\gt 0\text{.}\)
For a prime number \(p\) we consider \(\Phi_p(x)=x^{p-1}+x^{p-2}+\cdots+x+1\in\Q[x]\text{.}\) This polynomial is called cyclotomic polynomial. Note that
\begin{equation*} \Phi_p(x)=\frac{x^p-1}{x-1}. \end{equation*}
We show that \(\Phi_p(x)\) is irreducible over \(\Q\) using Eisenstein’s criterion. Consider
\begin{align*} \Phi_p(x+1)\amp=\frac{(x+1)^p-1}{x}\\ \amp=\frac{\left(\sum_{k=0}^p{p\choose k} x^k\right)-1}{x}\\ \amp= p+{p\choose 2}x+\cdots+x^{p-1} \end{align*}
For each \(k\in{1,2,\ldots,p-1}\) the prime \(p\) divides \({p\choose k}\text{.}\) Hence, Eisenstein’s criterion with prime \(p\) implies that \(\Phi_p(x+1)\) is irreducible over \(\Z[x]\) and hence \(\Phi_p(x)\) is irreducible over \(\Z[x]\) by Example 4.8.1. Now assume that \(\Phi_p(x)=gh\) with \(g,h\in\Q[x]\text{.}\) By clearing denominators we get \(r\in\Z\) such that \(r\cdot\Phi_p(x)=\tilde{g}\tilde{h}\in\Z[x]\text{.}\) Let \(p\) be a prime factor of \(r\text{.}\) Therefore, \(\overline{0}=\overline{\tilde{g}}\overline{\tilde{h}}\in\left(\Z/p\Z\right)[x]\text{.}\) Since \(\Z/p\Z\) is an integral domain we must have either all coefficients \(\tilde{g}\in p\) or all coefficients \(\tilde{h}\in p\text{.}\) Hence, we can cancel \(p\) from the equation \(r\cdot \Phi_p(x)=\tilde{g}\tilde{h}\text{.}\) Continuing in this way we can cancel all prime factors of \(r\text{,}\) and we get the equation \(\Phi_p(x)=G(x)H(x)\) with \(G,H\in\Z[x]\text{.}\) However, we already proved that \(\Phi_p(x)\) is irreducible over \(\Z\text{,}\) i.e., either \(G(x)=\pm 1\) or \(H(x)=\pm 1\text{.}\) Hence proved.
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