Let $k$ be a real number, and let $A$ denote the ring $\mathbb{R}[x]/(x^2+k)$. Find an $\mathbb{R}$-linear basis for $A$ and describe the multiplication law in terms of this basis.

I am not quite sure about the meaning of $\mathbb{R}$-linear basis. For example, if $k=1$, then $\mathbb{R}[x]/(x^2+k)$ is isomorphic to $\mathbb{C}$. So in this case, the $\mathbb{R}$-linear basis is simply a real scalar or just 1. Right?

In addition, what would be the general form of the basis for any $k$, please?

Also what does it mean by "describing the multiplication law"?

Extension Question:

As a variation to the above question, what would change if we change $\mathbb{R}$ to $\mathbb{Z}$, please? To be more specific, what is the basis for $\mathbb{Z}[x]/(x^2-2)$ this time, please? Thank you!

  • $\begingroup$ Nothing is changed: $\Bbb Z[x]/(x^2-2)$ is a free $\Bbb Z$-module of rank 2 and a basis is also $\{1,x\}$. (The division with remainder of polynomials in $\Bbb Z[x]$ by $x^2-2$ still holds.) $\endgroup$ – user89712 Nov 19 '13 at 8:38
  • $\begingroup$ I understand that $\Bbb R[x]/(x^2-1)$ is isomorphic to $\mathbb{R}\times \mathbb{R}$. Then what is $\Bbb Z[x]/(x^2-2)$ isomorphic to? $\endgroup$ – LaTeXFan Nov 19 '13 at 9:30
  • $\begingroup$ To $\Bbb Z\times\Bbb Z$ (as $\Bbb Z$-modules, or as abelian groups if you like, but not as rings!). $\endgroup$ – user89712 Nov 19 '13 at 9:32

If $A=\Bbb R[x]/(x^2+k)$, then let's look at the elements of this factor ring: take $f(x)\in\Bbb R[x]$ and consider $f(x)\pmod{(x^2+k)}$. In $\Bbb R[x]$ we have $f(x)=(x^2+k)g(x)+r(x)$ with $\deg r\le 1$, so $r(x)=ax+b$ and $f(x)\pmod{(x^2+k)}=r(x)\pmod{(x^2+k)}$. This shows us that the residue classes of polynomials modulo $(x^2+k)$ are in fact residue classes of polynomials of degree at most one. In conclusion, $A=\{ax+b:a,b\in\Bbb R\}$ and then $\{1,x\}$ is an $\Bbb R$-basis of $A$. (Actually, in this context, $x$ is in fact the residue class of $x$ modulo $(x^2+k)$. If you like, you can denote it by other letter.)

Now let's describe the multiplication law in terms of basis: $$(ax+b)(cx+d)=(ad+bc)x+bd-kac.$$ (Here I have used that $x^2=-k$ in $A$.)

  • $\begingroup$ Thank you for your answer. It is very helpful. $\endgroup$ – LaTeXFan Nov 18 '13 at 23:12

You have to remember polynomial division. If $f(x),g(x)\in\mathbb{R}[x]$, with $g\ne0$, there exist unique polynomials $q(x)$ and $r(x)$ such that $f(x)=g(x)q(x)+r(x)$ and the degree of $r$ is strictly less than the degree of $g$.

In your case, for any $f(x)\in\mathbb{R}[x]$ there are $q(x)$ and $a,b\in\mathbb{R}$ such that $$ f(x)=(x^2+k)q(x)+ax+b $$ because $x^2+k$ has degree $2$, so the remainder $r(x)=ax+b$ for some real $a$ and $b$.

Can you see the dimension of $A$ as a vector space over $\mathbb{R}$, now? Do you have any restriction about what $a$ and $b$ can be? Can you deduce that any element of $A$ can be represented as the equivalence class of a linear polynomial? How do you multiply those linear polynomials?

(Hint: the dimension is $2$.)

  • $\begingroup$ Thank you for your answer. It is quite clear. $\endgroup$ – LaTeXFan Nov 18 '13 at 23:10

By modding out $\mathbb{R}[x]$ by $(x^2+k)$, all you are doing is "extending" $\mathbb{R}$ by adjoining the roots of the polynomial $x^2+k=0$. In particular, as you pointed out, letting $k=1$ gives you $\mathbb{R}[x]/(x^2+1)\cong \mathbb{R}[i]\cong \mathbb{C}$. In this case, a $\mathbb{R}$-basis would be $\{1,i\}$, or if you were to express the basis as elements of $\mathbb{R}[x]/(x^2+1)$, you could say $\{1,x\}$ with multiplication law $x^2=-1$.

I hope this helps!

  • $\begingroup$ Thank you for your help. It makes sense to me now. $\endgroup$ – LaTeXFan Nov 17 '13 at 6:51
  • $\begingroup$ I'm glad I could help. $\endgroup$ – userCaltech Nov 17 '13 at 7:13

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