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I am having difficulty in understanding exactly the elements of the set $\mathbb{R}[x]/(x^3)$. I'll explain my thought process.

The Quotient Ring is the set of additive cosets, so we have that $$\mathbb{R}[x]/(x^3) = \{f+(x^3) : f\in\mathbb{R}[x]\}.$$ So we have the relation $$f-g\equiv 0 \mbox{ mod } x^3,$$ hence $x^3|f-g$.

Now, right now I understand this as a whole bunch of notation taken strictly from the definition. But what exactly are the elements of this quotient ring?

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They are cosets, exactly as you say. I'm not sure what kind of answer you expect here. There are other ways of thinking of the elements, e.g. as equivalence classes of elements under the relation $f\sim g$ iff $x^3\mid f-g$. Is that the sort of thing you mean? – Matthew Pressland Feb 25 '13 at 13:57
Since powers $\ge3$ of $x$ are zero in the quotient, every element can be written as $a+bx+cx^2$, with obvious addition, and multiplication given explicitly by $$(a+bx+cx^2)(e+fx+gx^2)=ae+(af+be)x+(ag+bf+ce)x^2.$$ – anon Feb 25 '13 at 14:00
@MattPressland Well I am tasked with finding the zero divisors of this particular ring, so I was looking trying to find a classification, similar to anon above. – Domonic Mei Feb 25 '13 at 15:08
@anon I was going to ask why, but it's because $x^3$ obviously divides every power of x greater than three. Thanks! – Domonic Mei Feb 25 '13 at 15:10
up vote 9 down vote accepted

Let $p = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots + a_n x^n \in \mathbb{R}[x]$. Then the image of $p$ under the natural map $\mathbb{R}[x] \to \mathbb{R}[x]/\langle x^3 \rangle$ is $a_0 + a_1 x + a_2 x^2 + \langle x^3 \rangle$.

I.e., every element in $\mathbb{R}[x]/\langle x^3 \rangle$ is of the form $$(\text{polynomial of degree $2$}) + \langle x^3 \rangle$$

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Elements of $\mathbb{R}[x] / (x^3)$ are real polynomials... modulo $x^3$.

Sure, that's nearly an empty statement, but it is a very general and useful view on quotient objects of any sort -- the elements of the quotient are "named" by elements of the original object, and the relation (here, congruence modulo $x^3$) tells us when two names are equal.

However, for many quotient rings, you can choose canonical (or otherwise reduced in some sense) representatives. A familiar example is that each element of $\mathbb{Z} / (47)$ has a canonical representative as an integer in the interval $[0, 46]$. And it can be computed efficiently through division with remainder.

Univariate polynomial rings (over fields) are similar -- division with remainder gives us a way to select canonical representatives in its quotient rings. Since $x^3$ has degree three, every element of $\mathbb{R}[x] / (x^3)$ has a canonical representative of degree $\leq 2$.

In this specific case, in the way polynomials are usually represented, the representative is especially easy to compute: just truncate!

$$a + bx + cx^2 + dx^3 + \ldots \equiv a + bx + cx^2 \pmod{x^3}$$

For multivariate polynomial rings, you need the idea of a Gröbner basis to do similar things.

As a general principle, univariate polynomial rings over fields (especially finite fields) are extremely similar to the integers in terms of their algebraic properties; most notions you have about one sort can be translated to the other.

Possibly interestingly, another interpretation is available here. We can view elements of $\mathbb{R}[x] / (x^3)$ as second-order approximations to real power series (whether you mean formal power series, or the notion of convergent power series from calculus).

So, we can view its elements as analytic functions, where we allow ourselves to have errors in the third order.

This is sometimes rather useful. I've seen some calculations greatly simplified in $\mathbb{R}[x] / (x^2)$ by rewriting

$$ 1 + a x \equiv \exp(ax) \pmod{x^2} $$

and other similar sorts of things, although I don't recall any examples at the moment.

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You would imagine $\mathbb{R}[x]/(x^3)$ as an unitary ring generated by $y$ with a single defining relation $y^3=0$.

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Hint $\ $ Because one can divide with unique remainder by any monic polynomial $\rm\,f,\,$ the coset $\rm\: g +(f) \in R[x]/(f)\:$ may be uniquely represented by its least degree element, the remainder $\rm\:g\ mod\ f\, =\, g - hf.\:$ Therefore the polynomials of degree $\rm < deg\, f\,$ form a complete system of representatives of $\rm\,R[x]/(f).\,$ Thus we can represent the ring by these normal forms, and pullback the ring operations to the normal form reps (transport of structure), e.g. $\rm\: g * h := gh\ mod\ f.\:$

For example, Hamilton's presentation of $\Bbb C$ as pairs of reals is a special case $\:\Bbb R[i]\cong \rm\Bbb R[x]/(x^2\!+1).\:$ Here the normal forms are all linear polynomials ${\rm\:a + bx }\:$ with the transported multiplication

$$\rm\begin{eqnarray}\rm (a\! +\! bx)&&\rm(c\! +\! d x)\, &\rm\,\equiv\,&\rm (ac\!-\!bd) + (ad\!+\!bc)\, x\ \ \ (mod\ x^2\!+1)\\ \rm i.e.\quad (a\! +\! b{\it i}\,)&&\rm(c\! +\! d {\it i}\,)\, &\, =\,&\rm (ac\!-\!bd) + (ad\!+\!bc)\,{\it i}\end{eqnarray}$$

The same remainder representation works for any Euclidean domain with unique remainders, i.e. any domain with a division algorithm with unique smaller remainder, e.g. the familiar, case of of $\rm\,\Bbb Z/m = $ integers $\rm\,mod\ m,\,$ represented by $\,0\,$ and the least positive element of each nonzero coset $\rm\:k + m\,\Bbb Z\,\to\, k\ mod\ m,\:$ with transported multiplication $\rm\ j k\, :=\, jk\ mod\ m.$

There are multidimensional generalizations of the division algorithm (e.g. Grobner bases) which extend the above to certain multivariate polynomial rings $\rm\,R[x,y,z\ldots]/(f,g,h,\ldots).\:$

The above can be viewed as ring-theoretic special cases of very general methods in term rewriting systems for solving word problems in (quotient) equational algebras, e.g. the Knuth-Bendix completion algorithm. For more on the ring-theoretic perspective see George Bergman's classic paper: The diamond lemma for ring theory, 1978. and its errata and updates. Chasing links to this will locate recent literature on these topics (generalizations of Grobner bases, etc).

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@Downvoter If something is not clear then please feel welcome to ask questions and I will gladly elaborate. – Math Gems Feb 25 '13 at 18:37
Maybe the "highbrow" answer looks not useful to some? – Pedro Tamaroff Mar 8 '13 at 2:02

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