Let $I$ be the ideal generated by the polynomial $f(x) = x^2 +1 $. Hence, $I = \langle x^2 + 1 \rangle $.

Generally, how Can I compute the elements of the field $\mathbb{R}[x] / \langle x^2+1\rangle $ ?

I see how to compute things like $\,x^2 = -1\,\overset{\large \times\ x}\Longrightarrow\, x^3 = -x,\ $ but it is not immediately clear how to do general computations.

  • 1
    $\begingroup$ Use long division. Each element (an equivalent class of polynomials) has a representative of degree at most $1$: the remainder after division by $x^2+1$. $\endgroup$
    – Tom
    Feb 12, 2015 at 13:18
  • $\begingroup$ I added some context to your question to prevent it from being deleted by others (it is now linked by many other answers). Of course you can edit it as need be to more accurately reflect your specific situation. $\endgroup$ Jan 5, 2020 at 22:14

4 Answers 4


Well, there's a representative of each element of the field (each element is an equiv class of polynomials, where two are "the same" if they differ by a multiple of $x^2 + 1$). Here's a way to find the representative: take your polynomial, and replace every $x^2$ with $ -1$ (and hence $x^3$ with $-x$, etc.). When you're done, you'll have something with only an $x$ and a constant term. So your field elements are of the form $$ [ax + b] $$ In short: it looks "two dimensional".

What about products? Well, most are easy, but what's $[x] \cdot [x]$? It's $[x^2] = [-1]$. So $x$ is an element whose square is $-1$. This sounds a lot like the complex numbers, yes? In fact, your field is isomorphic to $\mathbb C$, under the map $$ [ax + b] \mapsto a\mathbf i + b. $$


Understanding such quotient rings is easier if one builds on intuition from analogous quotients. Modular polynomial arithmetic is analogous to modular integer arithmetic since, in both cases, we have available an (Euclidean) algorithm for Division with Smaller Remainder. This allows us to efficiently calculate canonical normal-form reps of cosets, viz. those of least size, i.e. remainders.

If $R$ is a ring and $\,f\in R[x]\,$ has lead coeff $\,1\,$ (or any invertible) then we can divide by $f$ with unique remainder: $ $ for any $\,g\in R[x]\,$ there are $\,q,r\in R[x]\,$ with $\,g = q f\! +\! r\,$ and $\,\deg r < \deg f.\,$ Said equivalently, $\,r\,$ is the least degree element of the coset $\,g + fR[x]\,$ in $\,S = R[x]/f.\, $ This implies that a complete system of reps for the quotient ring $S\,$ is all such remainders mod $f,\,$ which is clearly the set of all polynomials of smaller degree than $f.\,$ They are all incongruent if unequal, else $f\,$ divides some difference $\,d\neq 0,\,$ contra $\,\deg d < \deg f$ (this depends on the lead coef of $\,f\,$ not being a zero divisor, otherwise e.g. $\,(2x\!-\!3)(2\!-\!3x) = x\,$ in $\,\Bbb Z/6).$

Same as in $\,\Bbb Z/m,$ we can transport the ring structure to this set of normal reps, e.g. to calculate the product of two reps, we multiply them as elements of $\,R[x]$ then reduce the result $\,{\rm mod}\ f\,$ to obtain the least degree rep of the product, i.e. $\,g * h = (g h\,\ {\rm mod}\,\ f),\,$ e.g. in your ring $\,\Bbb R[x]/(x^2\!+1)\,$ our reps are all $\, a+bx,\,$ for all $\,a,b\in \Bbb R,\,$ with product $(a\!+\!bx) * (c\!+\!dx)\,$ being

$$\begin{align} {\rm mod}\,\ \color{#c00}{x^2\!+1}\!:\qquad &(a\!+\!bx)\ \ (c\!+\!dx) =\, ac + bd \,\color{#c00}{x^2}\!\! + (ad\!+\!bc)\,x \\ &\phantom{a\!+\!bx)\ (c\!+\!dx)}\ \ \,\equiv\ ac \color{c00}{- bd}\,\ +\:\!\ (ad\!+\!bc)\,x,\ \ {\rm by}\ \ \color{#c00}{x^2\equiv -1}\\[.2em] \Longrightarrow\qquad &\!\!\!\!\bbox[5px,border:1px solid #c00]{(a\!+\!bx)*(c\!+\!dx) =\, ac-\color{c00}{bd}\,\ +\ (ad\!+\!bc)\,x}&&\\ \end{align}$$

This is just the usual rule for multiplication of complex numbers, once we rename $\,x\,$ to $\:i.\,$

As above, we can eliminate the use of rings and replace it by equivalent congruence arithmetic, as in replacing arithmetic in $\,\Bbb Z/m\,$ by congruence arithmetic $\!\bmod m,\,$ e.g. see here for an explanation of congruences $\!\bmod (n,x^r-1),\,$ i.e. where $\,n\equiv 0,\, x^r\equiv 1,\,$ as used in the AKS primality test.

This normal form rewriting using the polynomial division algorithm generalizes to multivariate rings over nice coefficient rings (e.g. fields, Euclidean domains) - see the Gröbner basis algorithm.


Consider $\mathbb R[\beta] = \{f(\beta); f(x) \in \mathbb R [x]\}$. Then $\mathbb R[\beta]$ is a domain and $\mathbb R \subset \mathbb R[\beta] $.

If you take $J = \{f (x) \in K[x] ; f(\beta) = 0\}$ (maximal ideal) then you may conclude using the surjective homomorphism $\psi : \mathbb R[x] \to \mathbb R[\beta]$ defined as $\psi(f)(x) = f(\beta)$ and the Isomorphism Theorem that

$$\frac{\mathbb R[x]}{\langle x^2 + 1\rangle} \simeq \mathbb R [\beta] \simeq \mathbb R(\beta)$$

Now if $p(x) \in \mathbb R[x]$ is any polynomial then there exist $q(x), r(x) \in \mathbb R[x]$ such that

$$p(x) = q(x)f(x) + r(x)$$

with $r(x) = 0$ or $\partial r (x) < \partial f (x) = 2$. Notice that

$$\require{cancel} p (\beta) = q (\beta) \color{#A05}{\cancel{f (\beta)}^0} + r (\beta) = a\beta + b$$

Hence any element of $\mathbb R[\beta]$ is of the form $a\beta + b$, then we may write $$\mathbb R(\beta) = \{ \color{teal}{a \beta + b} ; a, b \in \mathbb R\}$$

You may also show that $$\begin{align}\phi : \mathbb R(\beta) &\to \mathbb C\\a\beta + b &\mapsto ai + b \end{align}$$

Is a isomorphism.


Question asked in Two exercises about Euclidean algorithm in Euclidean ring $\mathbb{Z}_2[x]$. , was stated as duplicate of above. and I don't have enough points to comment or reopen it. so, providing answer here.

  • if someone reopen and post below answer there, it would be helpful to @Vizik.

@Vizik you have a field with elements $${\{1,x,x^2,1+x,x+x^2,1+x+x^2,1+x^2\}}$$ generated by irreducible polynomial ${1+x+x^3}$ where ${1+x=x^3}$

  • Now it is easy to see that ${(1+x) \cdot (x+x^2) = 1 (mod(1+x+x^3))}$
  • Hence the inverse of ${(1+x)}$ is ${(x+x^2)}$

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