Simple extension fields If I am correct simple extension fields are extensions generated by one element. I have learned that this means that elements of a simple extension can be written as powers of that element as long as the power is greater than the degree of that element. Also this means that all elements of the simple ext. Can be written in terms of decending powers of the generating element where the highest power is the degree of the generator minus 1. This concept seems to work for generating elements whose irreductable in a certain field is a descending degree polynomial. Looking at the construction of the complex numbers by starting with $\Bbb R$ (and doing $\Bbb R [x]/\langle x^2+1\rangle$), it is said that all elements are of the form $a+b\cdot \text{generator}$. But when the generator is evaluated for the irreductable, $x^2+1$, we see... $(\text{generator})^2=-1$. Thus there is no first power of gen. Can someone explain where all elements of the extension just explained are of the form $a+b\cdot \text{generator}$ comes from or explain this situation for anytime where the irreduc ( gen, F)  is not a descending degree polynomial. I hope my question is clear.
Thank you
 A: If $k$ is any field whatsoever and $K$ is an extension of $k$, then to say that $K$ is a simple extension is (by definition) to say that there is an element $\alpha\in K$ such that $K=k(\alpha)$, where the notation ``$k(\alpha)$" means (by definition) the smallest subfield of $K$ containing both $k$ and $\alpha$. 
There are two different things that can happen here. First, $\alpha$ could be transcendental: this means that there is no non-zero polynomial $p(X)\in k[X]$ such that $p(\alpha)=0$. If this is the case, then the ``evaluation map" $k[X]\to K=k(\alpha)$ is injective (that's just a reformulation of the definition of transcendental I've just given). Because $k[X]$ is an integral domain, $K$ is a field, and this map is injective, the universal property of the field of fractions of $k[X]$, which is denoted $k(X)$, yields an extension of this map to a (necessarily injective) $k$-algebra map $k(X)\to K$ which we know sends $X$ to $\alpha$. The image of this map is a subfield of $K$ which contains the image of $X$, namely $\alpha$, so the image contains $K=k(\alpha)$, i.e., the map is surjective, hence an isomorphism: $k(X)\simeq K$. 
If $\alpha$ is not transcendental, then by definition there is a non-zero polynomial $p(X)\in k[X]$ with $p(\alpha)=0$ (we say $\alpha$ is algebraic over $k$). This means that the $k$-algebra map $k[X]\to K$ sending $X$ to $\alpha$ is not injective ($p(X)$ is a non-zero element in the kernel). This $k[X]$ is a principal ideal domain, the kerne of this map is generated by $g(X)$ for some (monic) non-zero, non-constant polynomial (it's non-constant because the map $k[X]\to K$ is not the zero homomorphism). So we get an induced injection of $k$-algebras $k[X]/(g(X))\to K$ sending $X+(g(X))$ to $\alpha$. Now, because $K$ is a field and $k[X]/(g(X))$ is isomorphic to a subring of $K$, $k[X]/(g(X))$ is a domain. Therefore $g(X)$ is irreducible. But then $k[X]/(g(X))$ is actually itself a field. So the image of the $k$-algebra injection $k[X]/(g(X))\to K$ is a subfield containing $k$ and $\alpha$, and hence is all of $K$, so we have a $k$-algebra isomorphism $k[X]/(g(X))\simeq K$. You can check directly that the elements $1+(g(X)),X+(g(X)),\ldots,X^{n-1}+(g(X))$ ($n=\deg(g(X))$) form a $k$-basis for $k[X]/(g(X))$, so their images, which are $1,\alpha,\ldots,\alpha^{n-1}$, form a $k$-basis for $K$. This means that each $\beta\in K$ can be written uniquely as a sum $a_0+a_1\alpha+\cdots+a_{n-1}\alpha^{n-1}$ with $a_i\in k$, i.e., as a polynomial in $\alpha$ of degree $\leq n-1$ with coefficients in $k$.
Doing the construction in reverse, for any irreducible polynomial $g(X)\in k[X]$, the ring $K=k[X]/(g(X))$ is a field containing a canonical copy of $k$, and if $\alpha=X+(g(X))$ is the image of $X$ in this ring, then $g(\alpha)=0$, and $K=k(\alpha)$ (where again the notation $k(\alpha)$ is not intrinsic to $k$, it is intrinsic to $K$, by definition the smallest subfield of $K$ containing both $k$ and $\alpha$). A specific example is $k=\mathbf{R}$ and the irreducible polynomial $X^2+1$ (the square of any real number is positive, so this polynomial has no roots, hence is irreducible since it's quadratic). The resulting simple field extension $K=\mathbf{R}[X]/(X^2+1)$ is then what we call the field of complex numbers $\mathbf{C}$. 
A: You might not have that a given generator is of finite degree, take
$$\Bbb Q(x)/\Bbb Q$$
which is transcendental of infinite degree, so that no finite set of powers can generate it. Hence "descending" powers doesn't make any sense, since there is no "largest" power of the generating element, i.e. $x$ beyond which all powers are algebraically dependent upon lower powers.
For the case you mention let's just use the base field $\Bbb Q$, since your notation $R$ might already contain $i$, and is a bit imprecise as-is. Then $\Bbb Q(i)=\Bbb Q[x]/(x^2+1)$ and the first power of the generator does show up in the quotient, in fact they are exactly elements of the form $a+bi$ with $a,b\in\Bbb Q$, it's only beyond that that you can reduce to lower degree element sums because every time you hit $ax^n$ with $n\ge 2$ you can reduce the degree to be $<2$ by iterating the procedure $ax^n=-ax^{n-2}$.
