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I have started teaching myself Galois Theory. I have a problem understanding a part of the proof of the following proposition : Let $K\subseteq L$ be a field extension and $l\in L$ an element which is algebraic over $K$. Let $p(X)$ be the minimal polynomial of $l$. Then the smallest subfield $K(l)$ of $L$ containing $K$ and $l$ is isomorphic to the quotient $K[X]/\langle p(X) \rangle$, where $\langle p(X) \rangle\subseteq K[X]$ is the principal ideal generated by $p(X)$. The proof begins with (and this is what I do not understand): Writing $p(X)=X^n +a_{n-1}X^{n-1}+\cdots +a_1 X+a_0$, it follows at once that we have an isomorphism $K[X]/\langle p(X) \rangle\;\; \cong \lbrace k_{n-1}X^{n-1}+\cdots +k_1 X+k_0 : k_i\in K\rbrace$, where the operations on the right hand side are defined modulo the relation $X^n= -a_{n-1}X^{n-1}-\cdots -a_1 X-a_0$.

How does this isomorphism follow at once? And what does the author mean about the operations on the right hand side? I have spent a lot of time going over the definition of quotient rings and trying to understand what goes on but I cannot. I think in general I have a problem with quotient constructions. Can someone please give me a way to look correctly at these objects? Thank you very much for any help...

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    $\begingroup$ By the division algorithm every element in $K[x]$ can be written as $f(x) = p(x)q(x) + r(x)$ where $0 \leq \deg r < \deg p$. Notice that $p(x)q(x) \in \langle p(x)\rangle$. So what happens when you quotient out by this ideal? $\endgroup$
    – user38268
    Feb 19 '13 at 10:29
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When we take a quotient structure, then we basically keep the same elements, but modify the equality between them. In case of $R/I$ for a ring $R$ and its ideal $I$, we force all elements of $I$ to be equal to $0$ in the new ring. Note that the properties of an ideal are exactly the properties of a collection of all elements who 'want to be zero somewhere'. (E.g. if $x$ is zero 'somewhere' then at the same place all $rx$ and $xr$ also has to be zero for all $r$...)

Now, in case of $K[x]/(p)$ exactly the same happens as in the modulo $m$ arithmetic, $\Bbb Z/(m)$: two polynomials, $u$ and $v$ are equal in the quotient ring iff $u-v\in (p)$, that is, iff $p|\, u-v$. Luckily, as BenjaLim noted, among polynomials the division algorithm works fine, just as in $\Bbb Z$, giving a remainder mod $p$, which has degree $<\deg(p)$. So that, every polynomial is represented by its remainder mod $p$, and that's it (and $p$ is $0$). Check the operation. Also check that two remainders cannot be equal modulo $p$.

A standard easy example is $\Bbb C\cong \Bbb R[x]/(x^2+1)$. In this quotient ring, we have $x^2+1=0$, that is, $x^2=-1$, and the elements are the remainders mod $(x^2+1)$, that is, polynomials of degree at most $1$, i.e. of the form $ax+b$. When multiplying them, for example, we get that $$(ax+b)(ux+v)=aux^2+(av+bu)x+bv \equiv au\cdot(-1)+(av+bu)x+bv $$ modulo $x^2+1$. Now if we rename the variable $x$ to $i$, all gets more familiar...

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If I may add another example to the perfect answer by @Berci, think of how you compute with expressions like $$ a + b \alpha + c \alpha^{2},\tag{1} $$ where $\alpha = \sqrt[3]{2}$, and the coefficients are rational, say.

Say you want to compute the product of two expressions of this kind. You first expand the product as a linear combination of powers of $\alpha$, as it were a polynomial in $\alpha$.

Then you substitute $\alpha^{3} = 2$ everywhere to get again an expression of the form (1). The quotient ring business formalizes this second step, as clearly explained by @Berci.

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