I am working the following example from Herstein's book "Topics in algebra" and some moments of the following excerpt seem very confusing and weird.
We are working with the field $\mathbb{Q}[x]/\langle x^3-2\rangle$. This is a field since the polynomial $p(x)=x^3-2$ is irreducible over $\mathbb{Q}$, then the ideal $A=\langle x^3-2\rangle$ is maximal which means that quotient ring $\mathbb{Q}[x]/\langle x^3-2\rangle$ is field.
Let me ask my questions:
1) How it follows that $f(x)+A=(a_0+A)+a_1(x+A)+a_2(x+A)^2 ?$
If we use the addition and multiplication in $\mathbb{Q}[x]/\langle x^3-2\rangle$ correctly we get the following expression. $$f(x)+A=a_0+a_1x+a_2x^2+A=(a_0+A)+(a_1x+A)+(a_2x^2+A)=$$$$=(a_0+A)+(a_1+A)(x+A)^2+(a_2+A)(x+A)^2.$$
2) Suppose we have $f(x)+A=(a_0+A)+a_1(x+A)+a_2(x+A)^2$ (although I do not understand why this is correct). If we put $t=x+A$ then we get $f(x)+A=(a_0+A)+a_1t+a_2t^2$ but NOT $f(x)+A=a_0+a_1t+a_2t^2$. Can anyone clarify this moment also?
3) Since we put $t=x+A$ we see that $t\in \mathbb{Q}[x]/\langle x^3-2\rangle$.
Consider $t^3-2=(x+A)^3-2=x^3-2+A=A$ since $A$ is zero element of the field $F=\mathbb{Q}[x]/\langle x^3-2\rangle $. In other words, $t^3-2=0_F$ but the expression $t^3=2$ seems to me meaningless. Am I right?
4) If $a_0+a_1t+a_2t^2=b_0+b_1t+b_2t^2$ then $(a_0-b_0)+(a_1-b_1)t+(a_2-b_2)t^2=0$ and we know that $t=x+A$. How it follows that $(a_0-b_0)+(a_1-b_1)x+(a_2-b_2)x^2$ is in $A=\langle x^3-2\rangle?$
Please help mo to answer my questions and clarify these moments! Would be very thankful for help.
As I said in the comments, what you want is the follwiwng result.
If $R$ is a ring and $I$ is an ideal in your ring then $R/I$ has a unique ring structure for which the natural map $\pi: R\to R/I$ mapping $r$ to $r+I$ is a ring morphism. In other words $\pi(x+y)=\pi(x)+\pi(y)$, $\pi(1)=1$ and $\pi(xy)=\pi(x)\pi(y)$.
This is straightforward and should be proven in your textbook whatever it is.
In any case let's see how this explains your computation.
You have as noted in your book, an expression of the form $f(x)=a_0+a_1x+a_2x^2+t(x)(x^3-2)$. You get appliying $\pi$, $$f(x)+A=\pi(f(x))=\pi(a_0+a_1x+a_2x^2)+\pi(t(x)(x^3-2))=\pi(a_0+a_1x+a_2x^2)=\pi(a_0)+\pi(a_1)\pi(x)+\pi(a_2)\pi(x)^2=(a_0+A)+(a_1+A)(x+A)+(a_2+A)(x+A)^2$$.
Notice that pretty commone to note $\pi(a_i)=a_i$ the reason for this is that the composition $\mathbb{Q}\to \mathbb Q[X]\to Q[X]/(x^3-2)$ is injective as $\mathbb{Q}$ is a field. But even when it's not, we may use the same abuse of notation.
Now to prove the result I have mentionned well since the ring structure is putatively unique you have no choice. You need to define $(x+A)(y+A)$ to be $x.y+A$ and you need to check that this is well defined. In other words, take $x+A=x'+A$ of course this means that $x-x'\in A$ and thus $x'.y+A=x.y+A+(x'-x)A=x.y+A$. And ditto for $y$.
One remark, I don't think using the notation $x+A$ is in any case a good idea. What I prefer personnaly is to use representativess to define things. Which make thigs clearer I think.
In other words two elements $x,y$ in $R$ have the same image in $R/I$ iff they differ by an element of $I$. What we're proving here is that if $x,x'$ differ by an element of $I$ and $y,y'$ differ by an element of $I$ then $xy,x'y'$ differ by an element of $I$ and therefore you can define the multiplication of two classes by the class of the multiplication of any two representatives of the classes. The result won't depend on the particular choice of the representatives.
It should solve all your troubles.
I suggest you to use the map $\pi$ rather that the cumbersone notation $+A$, and in a little while you'll be able to drop the $\pi$ altogether and to pretend it is there even though not written explicitely.