1

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-

  1. 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?
  2. 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)-

enter image description here

Preamble-

enter image description here

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}
  • I recommend you to use amsthm package to type theorem like environments. – Sigur Jul 20 '15 at 10:53
  • 4
    1) Generally, please don't post code as screenshots, makes it hopeless it we want to test your code. 2) Generally, you should never use \\ 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|`. – Torbjørn T. Jul 20 '15 at 10:53
  • 1
    If you choose to have a large \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. – egreg Jul 20 '15 at 10:54
  • @TorbjørnT. plz see the code in edit – Bhaskar Vashishth Jul 20 '15 at 10:59
  • @egreg plz see edit – Bhaskar Vashishth Jul 20 '15 at 10:59
3

You still had some double back-slashed where they should not be. In front of display math, for example. And some in-line math had to be broken over the line which could cause such errors.

You should not use double back-slashes at all. If you want to have a new paragraph, do a blank line (no matter, if you use parskip or parindent then). If you want to generate more breaks as you do here, you may put such stuff in align environments.

The nicest would be to stick to one paragraph style. And if there is not enough reason for a blank line, then there is maybe no reason for a new paragraph as well. Right now, you break after almost each sentence... why?

For stuff as the math in your example, you should have a look on theorems and examples as defined by amsthm.

Here is a quick (not complete) code review of yours:

% arara: pdflatex

\documentclass[12pt]{report}
\usepackage{amssymb}
\usepackage{mathtools}
\usepackage[parfill]{parskip}

\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.

    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 
        \begin{align*}
            \frac{K[x]}{\text{ker}(\phi)}&\cong KG\\
            \shortintertext{where}
            (\phi)&=\{f\in K[x]: f(a)=0\}.\\
            \intertext{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\\
            \shortintertext{and thus}
            KG &\cong \frac{K[x]}{\langle x^n-1 \rangle}.           
        \end{align*}        
    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
    \begin{align*}
        x^5-1&=(x-1)(x^4+x^3+x^2+x+1\\
        \shortintertext{and thus}
        \mathbb{Q}G&\cong \mathbb{Q}\oplus \mathbb{Q}(\zeta)
    \end{align*}
    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}
2

In addition to replacing all remaining \\ (double backslash) instances with a simple blank line, I also recommend you

  • replace all instances of : (colon) in math mode with \colon
  • make use of pre-defined "math operators" such as \ker
  • replace one or the other inline-math expression with a display-math construct. Most importantly, you should convert

    \(KG\cong \frac{K[x]}{\langle f_1 \rangle} \oplus 
    \frac{K[x]}{\langle f_2 \rangle} \dots \oplus 
    \frac{K[x]}{\langle f_t \rangle}\) 
    

    to a display-math statement

  • don't leave spaces between \) (closing inline math) and subsequent punctuation marks (such as ,, and .).

enter image description here enter image description here

\documentclass[12pt]{report}
%% I've omitted all unneeded packages from this preamble
\usepackage{amssymb,amsmath}
\DeclareMathOperator{\chr}{char}
\usepackage{parskip}
\setlength\parskip{\baselineskip}

\begin{document}

In this section we will describe the structure of group algebras over finite cyclic 
groups and state the main result, which describes the structure of group algebras 
over any finite abelian group.

Let \(G\) be the cycle group of order \(n\), i.e.\ \(G=\langle a\colon a^n=1 \rangle\), 
and \(K\) be a field such that \(\chr(K)\) doesn't divide \(|G|\). 

Then consider the map \(\phi\colon K[x]\to KG\) given by
\[
f(x)\to f(a).
\]
It is clear that \(\phi\) is an epimorphism and thus 
\[
\frac{K[x]}{\ker(\phi)}\cong KG,
\] 
where \(\ker(\phi)=\{f\in K[x]\colon f(a)=0\}\). Now, as \(K[x]\) is a P.I.D 
and \(x^n-1\in \ker(\phi)\), it easily follows that 
\(\ker(\phi)=\langle x^n-1 \rangle\) 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, \(x^n-1=f_1f_2 \dotsm f_t\), 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 the \(\zeta_i\)'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}

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