I am new to ring theory and so some of its concepts are still not clear to me. One of them being the quotient rings.
Let's say that I needed to construct this ring: $$\mathbb R[x] / (x^2 - 1)$$ The ring of polynomials with real coefficients "quotiented" by the ideal generated by the polynomial $(x^2 - 1)$.

Let $C$ be the set of cosets of the ideal. Then: $$C = \{p + (x^2-1) : p \in \mathbb R[x] \}$$ Now, consider this coset: $$D = x^3 + (x^2-1) = x^3 + kx^2 - k $$ And this $$E = x^5 + (x^2-1) = x^5+kx^2-k$$ $$k \in \mathbb R$$ Therefore $$D+E = x^5 + x^4 +kx^2-k $$ Now, I know that the original ring only contains linear polynomials, and so my expression should somehow simplify. but I don't know how to do it. I can imagine why the multiples of $(x^2-1)$ will be sent to zero - they form the coset containing the natural element.
But in my case the expression I got cannot be factored so that I can cancel anything but $(kx^2 - k)$ out.
How do I simplify this?
Although it may be inappropriate, please, could you give an answer in simple terms? I am not taking any course in abstract algebra, just some leisurely reading.

  • 1
    $\begingroup$ As an addition to the (correct) answer, in a quotient ring of a polynomial it's usually easiest to just write in representatives, so that $D = x^3$ and $E = x^5$, while keeping in mind that $x$ is no longer a "free variable"; it now obeys the rule $x^2 = 1$ (or equivalently $x^2 - 1 = 0$). If you stick to this way of writing/thinking about things it's obvious that $E = x^5 = x^2 \times x^2 \times x = x$. $\endgroup$ – Mees de Vries May 30 '18 at 9:06

The set of cosets of the ideal $(x^2-1)$ is the set$$\{p(x)+(x^2-1)\,|\,p(x)\in\mathbb{R}[x]\}.$$Note that $p(x)$ can be polynomial with any degree, not just a linear one.

In particular, if $p(x)=x^3$, then the coset is$$\{x^3+(x^2-1)\,|\,p(x)\in\mathbb{R}[x]\}=\{p(x)+q(x)\times(x^2-1)\,|\,q(x)\in\mathbb{R}[x]\}.$$And here $q(x)$ can be any polynomial.

And if $p^\star(x)=x^5$, you can use the fact that$$x^5+x^3=(x^3+2x)(x^2-1)+2x$$to deduce that, in $\mathbb{R}[x]/(x^2-1)$,$$\bigl(p(x)+(x^2-1)\bigr)+\bigl(p^\star(x)+(x^2-1)\bigr)=2x+(x^2-1).$$


Since in this quitient ring you "divided out" the polynomial $x^2-1$, you set this polynomial equal to zero, that is

$$ x^2-1=0\Rightarrow x^2=1 $$

this means

$$ x^3=x $$


$$ x^5=x. $$

Hope this helps.

  • $\begingroup$ So this means that my target ring will have two divisors of zero: $(x-1)$ and $(x+1)$? $\endgroup$ – Aemilius May 30 '18 at 9:06
  • $\begingroup$ Since you can factor $x^2-1$ as $(x-1)(x+1)$ the answe is yes. $\endgroup$ – Vinyl_cape_jawa May 30 '18 at 9:10

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