$A=K[x]$,$\mathfrak{m}$ is a maximal ideal containing a principle ideal of $A$.
every element of $K[x]/\mathfrak{m}$ can be described by form $f+\mathfrak{m}$.
every element of $K$ is the coefficient of $f$.
What's the mean of "$A/\mathfrak{m}$ is an extension field of K in which each $f\in\Sigma$ has a root"?
$A/\mathfrak{m}$ is an extension field of $K$, $\Rightarrow\quad$ $K$ is a subfield of $A/\mathfrak{m}$, $\Rightarrow$ every element in K is in $A/\mathfrak{m}$.
but it's mean the coefficient of f can be described by form f+m.
What 's the mean of it?It doesn't make sense.
what exactly is my problem is :
the elements in K are a1,a2,....ai
the elemt in K[x]/m are (a1,a2,....ai)+a set of (a1,a2,....ai)).
they are totally different.How can K be a subset of K[x]/m?
There is an embedding $K\hookrightarrow A/{\frak m}$ given by $a\mapsto a+{\frak m}$. If you want to be technical, you'd say that it's an isomorphic copy of $K$ sitting inside $A/{\frak m}$ rather than $K$ itself, where the elements look like $a+{\frak m}$ for $a\in K$ instead of the $a$s themselves. So in particular if $f(T)=a_nT^n+\cdots+a_1T+a_0$, when we say $x_f$ is a root of $f$ we mean $x_f+\frak m$ is a root of $(a_n+{\frak m})T^n+\cdots+(a_0+\frak m)$.
To see why this is true, observe that
$$\begin{array}{ll} \bar{f}(\bar{x_f})& = (a_n+{\frak m})(x_f+{\frak m})^n+\cdots+(a_1+{\frak m})(x_f+{\frak m})+(a_0+{\frak m}) \\ & = (a_nx_f^n+\cdots+a_1x_f+a_0)+{\frak m} \\ & = f(x_f)+{\frak m} \\ & = 0+{\frak m}\end{array}$$
because $f(x_f)\in{\frak a}\subseteq{\frak m}$ as elements of $A$. Above by $\bar{f}$ we mean the image of $f\in K[T]$ in the polynomial ring $(A/{\frak m})[T]$ (by applying $K\to A/{\frak m}$ coefficientwise) and by $\bar{x_f}$ the image of the element $x_f\in K$ in $A/{\frak m}$ (again under the embedding $K\to A/{\frak m}$).
The whole point of this construction was to find an extension field of $K$ in which the original irreducible polynomials would now have roots. Constructing (or tacitly invoking constructions of) extension fields is a ubiquitous happening in field theory. Therefore it is not useful to think of $A/\frak m$ being an extension "of a copy of $K$ sitting inside it," but rather as a bona fide extension field of $K$, a field containing $K$ as an actual subset. Thus, it is useful to identify the elements $a\in K$ with the elements $a+{\frak m}$; we think of them as one and the same, and when we write $a$ (where $a\in K$) in expressions involving elements of $A/\frak m$ we really mean $a+{\frak m}$, its corresponding image in $A/\frak m$.
You are probably already familiar with this kind of choice to identify an object with a copy of it sitting inside a construction, and tacitly approve of it where you know it happens, only you don't realize it. Consider how the integers $\Bbb Z$ are constructed from the natural numbers $\Bbb N$. We define tuples $(n,m)$ subject to $(a,b)+(c,d):=(a+c,b+d)$. Then we quotient by the congruence relation $(a+b,c+b)\sim(a,c)$ to obtain $\Bbb Z$. There is an embedding $\Bbb N\to\Bbb Z$ given by $n\mapsto [(n,0)]$, where we use brackets to denote "the equivalence class thereof." (Or, if your $\Bbb N$ doesn't have a $0$, you can use $n\mapsto[(n+1,1)]$.) Technically $n$ and $[(n,0)]$ are different things and look different; but we don't want to go around our entire life writing down $[(1,0)]$ when we simply mean $1$ do we?
Of course not. So we identify the natural number $n$ with the corresponding integer $[(n,0)]$. The same thing applies with constructing the rationals. Write $\Bbb Z=\{\cdots,-2,-1,0,1,2,\cdots\}$ for our convenience now. The tuple $(n,m)$ now stands for what we could call $n/m$ (where $m\ne0$), and we subject these tuples to $(a,b)+(c,d)=(ad+cb,bd)$ and quotient by $(ac,bc)\sim(a,b)$. We don't want to write $[(n,1)]$ our whole lives for the integer $n$. And the same thing goes for constructing the real numbers, using either equivalence classes of Cauchy sequences or Dedekind cuts.