Why I am getting
underfull hbox badness 10000 in paragraph at lines
Message even when I am not using \\
. After reading many questions on the same message, I eliminated all \\
that I used after line spacing and defined
\usepackage{parskip}
\setlength\parskip{\baselineskip}
\parindent=0pt
before
\begin{document}
and used a line gap whenever I wanted to skip base line.
Now I have two questions-
- I should use
\\
whenever I want to get to next line within a proof and do not want to skip a line, or is there something else for it? - After eliminating all
\\
except at line break, I still see some messages where I am using\[\]
inline equations. How can I eliminate them.
Below I am attaching a pic where it shows the badness message when I use inline. Another pic of preamble is attached too.
Error at Inline (line 90)-
Preamble-
Here is the code
\documentclass[12pt]{report}
\usepackage{epsfig,epic,eepic,units}
\usepackage{url}
\usepackage{longtable}
\usepackage{mathrsfs}
\usepackage{multirow}
\usepackage{bigstrut}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{centernot}
\usepackage{graphicx}
\usepackage{floatrow}
\usepackage{braket}
\usepackage{parskip}
\setlength\parskip{\baselineskip}
\parindent=0pt
\begin{document}
In this section we will describe the structure of group algebras over finite cyclic groups and will state the Main result which describe the structure of group algebras over any finite abelian group.
\vspace{1mm}\\
Let \(G\) be the cycle group of order \(n\), i.e. \(G=\langle a:a^n=1 \rangle\), and K be a field such that char\((K)\) doesn't divide \(|G|\). \\
Then considering the map \(\phi: K[x]\to KG\) given by:
\[f(x)\to f(a)\]
It is clear that \(\phi\) is an epimorphism and thus
\[\frac{K[x]}{\text{ker}(\phi)}\cong KG\] where Ker\((\phi)=\{f\in K[x]: f(a)=0\}\). Now, as \(K[x]\) is a P.I.D and \(x^n-1\in \text{Ker}(\phi)\), it easily follows that \(\text{Ker}(\phi)=\langle x^n-1 \rangle\) and thus \[KG \cong \frac{K[x]}{\langle x^n-1 \rangle}\]
Now decomposing \(x^n-1\) as product of irreducible factor polynomials in \(K[x]\), say, \(x^n-1=f_1f_2 \dots f_t\), and as \((x^n-1,nx^{n-1})=1 \) implies \(x^n-1\) is separable and thus for every \(i\neq j, f_i \neq f_j\). So now we can use Chinese Remainder Theorem, and express
$KG\cong \frac{K[x]}{\langle f_1 \rangle} \oplus \frac{K[x]}{\langle f_2 \rangle} \dots \oplus \frac{K[x]}{\langle f_t \rangle}$
Now \(\frac{K[x]}{\langle f_i \rangle} \cong K(\zeta_i)\) where \(\zeta_i\) is a root of \(f_i\) for \(1\le i \le t\). Therefore,\\
\[KG \cong K(\zeta_1) \oplus \dots \oplus K(\zeta_t)\]
as \(\zeta_i's\) are roots of \(x^n-1\) thus \(KG\) is isomorphic to a direct sum of \textit{cyclotomic extensions} of \(K\)\\
\textbf{Example-} Let \(G=C_5\) and \(K=\mathbb{Q}\). In this case
\[x^5-1=(x-1)(x^4+x^3+x^2+x+1)\]
and thus
\[\mathbb{Q}G\cong \mathbb{Q}\oplus \mathbb{Q}(\zeta)\]
where \(\zeta\) is a root of \(x^4+x^3+x^2+x+1\).\\
Notice that changing \(K=\mathbb{Q}\) to \(K=\mathbb{R}\) would similarly suggest that the group ring \(\mathbb{R}C_5 \cong \mathbb{R}\oplus \mathbb{R}(\zeta)\) as \(x^4+x^3+x^2+x+1\) is irreducible over \(\mathbb{R}\) too.\\
Similarly considering \(x^4-1=(x+1)(x-1)(x^2+1)\) , we see that \(\mathbb{Q}C_4 \cong \mathbb{Q}\oplus \mathbb{Q}\oplus \mathbb{Q}(i)\).\\
\end{document}
amsthm
package to type theorem like environments.\\
in running text. Mostly, a paragraph break (empty line in code) is what one wants. Judging by the PDF, you have\\
after abelian groups and after doesn't divide |G|`.\parskip
, you should have all paragraphs separated by this blank line. Your readers will be confused, otherwise. Just avoid it, like typography has done for five centuries. We need an example of code, anyway.