I was totally confused about the elements of this field extension. The quotient sign made me think of cosets. To be able to comprehend what goes on, I tried to study on examples. If I let $n$ to be $3$ , I thought $(X^4 +X^3+ 2X^2+2)$ in $\mathbb R (X)$ would become $(X+1+ 2X^2+2)$ or $(2X^2+2)$ in $\mathbb R (X) /\mathbb R(X^3) $ . I was not sure which would be correct (?). I see that none is correct now, after the answers.

Thanks to all.

  • $\begingroup$ $x^4+x^3+2x^2+2$ is not an element of $\Bbb R(x^3)$ so it does not "become" anything since it was never in the base field to begin with. $\endgroup$ – Adam Hughes Apr 3 '17 at 13:51
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    $\begingroup$ $x^3+2x^2+2$ is also not in $\Bbb R(x^3)$ so it too does not "become" anything in the field extension. In fact, in general nothing "becomes" anything in a field extension, an extension adds more elements, it doesn't alter the ones that are already there. $\endgroup$ – Adam Hughes Apr 3 '17 at 13:57
  • $\begingroup$ Are you confusing extensions with quotients? The notation is the same but just that. See math.stackexchange.com/questions/2175482/…. $\endgroup$ – lhf Apr 3 '17 at 13:58
  • $\begingroup$ @Adam Hughes What are the elements not in $ \mathbb R(X^3) $ and in $\mathbb R(X)$? What are added? Yes, I am totally confused... $\endgroup$ – Tedebbur Apr 3 '17 at 14:41
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    $\begingroup$ @Serpenche some examples: $x$, $x+1$, $x^3+x$, anything which has powers of $x$ which are not multiples of $3$. $\endgroup$ – Adam Hughes Apr 3 '17 at 14:45

I think the idea you have in your head is that, given any finite field extension $E/F$ in characteristic zero* you can express $E$ as being generated over $F$ by adjoining some element: i.e. $E = F(\alpha)$.

Then, one might be interested in writing elements of $F(\alpha)$ in the form of a polynomial in $\alpha$ with coefficients from $F$.

This is a very common thing to do — to the point that you may have gotten the erroneous idea in your head that it must be done, and that's partly to blame for your confusion. But that's simply not so; it is an optional thing that one may choose to do.

But in any case, should one wish to do that here, the fully reduced form (meaning that the polynomial has degree less than that of the field extension) would be

$$X^4 +X^3+ 2X^2+2 = (2) X^2 + (X^3) X + (2 + X^3) $$

(once again, note that writing in fully reduced form is optional)

However! In situations like this I generally advise that one introduce a new variable; trying to work entirely with $X$'s like this is extremely error prone.

So, I would define $Y = X^3$, so that the original field extension becomes

$$ \mathbb{R}(X) / \mathbb{R}(Y) $$

Note, if needed, that $X$ is a root of an irreducible polynomial over $\mathbb{R}(Y)$: specifically, $f(t) = t^3 - Y$.

Then, going back to the original question, the fully reduced form of the given element would be

$$X^4 +X^3+ 2X^2+2 = 2 X^2 + Y X + (2 + Y) $$

*: I'm being overly specific for simplicity.


As discussed in the comments, there are two notions here that use the same notation. I'll describe each

A field extension is a way to add new elements to a field. For example, we might take $\mathbb Q$ and extend it by adding the element $\sqrt 2$. This produces the new field $\mathbb Q(\sqrt 2)$, which is a field extension of $\mathbb Q$. This is written $\mathbb Q(\sqrt 2)/\mathbb Q$ and is equal to the set $\{a+b\sqrt{2}:a,b\in\mathbb Q\}$.

Caution: not every field extension has such an easily calculable general form. For example, if $\omega$ is a third root of unity then $$\mathbb Q(\omega)=\{a+b\omega+c\omega^2:a,b,c\in\mathbb Q\}\neq\{a+b\omega:a,b,c\in\mathbb Q\}$$

To elaborate a little on the notation, $\mathbb Q(\sqrt 2)$ is the name of the field. We write $\mathbb Q(\sqrt 2)/\mathbb Q$ when we want to highlight the fact that we are considering it as a field extension of $\mathbb Q$. Just like large composite numbers can be thought of as multiples of many different numbers, so can large fields be thought of as extensions of many different fields. Specifying the base field is important when we are interested in what elements have been added and what elements started off in the field. $F(\alpha)$ is defined to be the unique smallest field (under intersection) that contains all the elements of $F$ and also $\alpha$. We can make this notion formal by using Zorn's lemma and working within the universe of all fields containing $F$ and $\alpha$ that aren't "too big."

$\mathbb R(x)/\mathbb R(x^3)$ is a field extension. The base field is $\mathbb R(x^3)$ (which is itself a field extension of $\mathbb R$) and the produced field is $\mathbb R(x)$. In the base field we have the rational polynomials with exponents that are multiples of $3$, such as $\frac{1+x^3}{x^3}$ and $x^9-3x^6+3x^3-1$. In the case of $\mathbb{R}(x)$, we have all rational polynomials, such as $\frac{1}{x}$, $x^2+2x+1$, and $\frac{x^3}{1+x+x^2+x^3}$, none of which were in the base field.

This is mostly unrelated to the notion of quotients. $K/F$ as a quotient refers to what happens when we consider $\{kF:k\in K\}$. Here $kF$ denotes a left-coset of $F$, and is equal to the set $kF:=\{kf:f\in F\}$. When we look at the cosets, some turn out to produce the same sets. We then identify all the cosets that are the same set as the same object inside the set $K/F$. We often refer to a coset by a "coset-representative" which is simply an element in the coset used to stand for the entire coset. When possible, we often aim to choose the smallest or simplest coset representative.

Example: $\mathbb Z/(3\mathbb Z)$

This isn't a field, but is an easier example and works the same way. Here we have $F=3\mathbb Z =\{3x:x\in\mathbb Z\}=\{\ldots,-6,-3,0,3,6,\ldots\}$. When you look at the cosets, you get: $$1+\{3x:x\in\mathbb Z\}=\{3x+1:x\in\mathbb Z\}=\{3x+1:x\in\mathbb Z\}$$ $$2+\{3x:x\in\mathbb Z\}=\{3x+2:x\in\mathbb Z\}=\{3x+2:x\in\mathbb Z\}$$ $$3+\{3x:x\in\mathbb Z\}=\{3x+3:x\in\mathbb Z\}=\{3x:x\in\mathbb Z\}$$ $$4+\{3x:x\in\mathbb Z\}=\{3x+4:x\in\mathbb Z\}=\{3x+1:x\in\mathbb Z\}$$ $$5+\{3x:x\in\mathbb Z\}=\{3x+5:x\in\mathbb Z\}=\{3x+2:x\in\mathbb Z\}$$ $$6+\{3x:x\in\mathbb Z\}=\{3x+6:x\in\mathbb Z\}=\{3x:x\in\mathbb Z\}$$ $$\cdots$$ Here we see that only three distinct sets for, and that we can refer to them by the coset representatives $0,1,$ and $2$. To distinguish these coset representatives from the usual natural numbers, we will sometimes write $\mathbb Z/3\mathbb Z=\{\overline{0},\overline{1},\overline{2}\}$

The map that sends $h$ to its coset representative $hF$ is known as the "induced homomorphism" or the "modular homomorphism" or the "quotient homomorphism." computing it is quite simple, as the map $\varphi:K\to K/F$ is simply given by $\varphi(k)=kF$. Often times, after writing down the set defined by $\varphi(k)$ you'll want to find a small coset representative to use. The key here is that $h,k$ are in the same coset if and only if $h-k\in F$ because cosets are groups. The usual approach is to just use a little trial and error by looking at $h-f$ where $f\in F$ (this is equivalent to the statement in the previous sentence).


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