What ring is isomorphic to factor-ring?

Question

$R=\mathbb{Z}[x]$ - polynomials with integer coefficients

$I = (x^2+x-1)\mathbb{Z}[x]$

In this case we have that classes of factor-ring $R/I$ represented in the next form: $K_{a,b}=ax+b$.

What ring is isomorphic to factor-ring $R/I$? I have suggestion that it will be $\mathbb{R}$, but I'm not sure and can't prove this.

All answers are highly appreciated. Thank you in advance.

This will not be $ℝ$ – there are only countably many polynomials in $R = ℤ[X]$, so there are only countably many elements in $R/I$. — k.stm, Dec 11 '14 at 20:06

Timbuc · Accepted Answer · 2014-12-11 20:09:36Z

3

Hint:

Put $\;\phi:=\frac{1+\sqrt5}2\;$ , and define

$$f:\;\Bbb Z[x]\to\Bbb Z[\phi]\;,\;\;f(p(x)):=p(\phi)$$

Show the above is an epimorphism of rings. What's $\;\ker f\;$ ? How in the world did I come up with that particular $\;\phi\;$ ?!

answered Dec 11 '14 at 20:09

Timbuc

32.1k2 gold badges24 silver badges46 bronze badges

$\begingroup$ $\;\phi:=\frac{1+\sqrt5}2\;$ because one root of $x^2+x-1$ is $x_0=\frac{-1-\sqrt5}2\;$ $\endgroup$ – Nikita Dec 11 '14 at 20:22
$\begingroup$ @Nikita Yes indeed. $\endgroup$ – Timbuc Dec 11 '14 at 20:30
$\begingroup$ Since $K_{a,b}=ax+b$ and $x_0=\frac{-1-\sqrt5}2\;$. We have $a(\frac{1+\sqrt5}2)+b = b+\frac{a}{2}+\frac{a\sqrt5}{2}$. What next? Am I thinking in the right way or not? $\endgroup$ – Nikita Dec 11 '14 at 22:31
$\begingroup$ The elements in $\;\Bbb Z[\phi]\;$ are of the form $\;f(\phi)\;,\;\;f(x)\in\Bbb Z[x]\;$ . You may want to represent these elements as $\;ax +b+I\;,\;\;I:=\langle x^2+x-1\rangle\;,\;\;a,b\in\Bbb Z\;$ , since in the quotient ring every coset has as representative a linear polynomial. I don't know what else to tell you... $\endgroup$ – Timbuc Dec 12 '14 at 5:12
$\begingroup$ As a result i got that $R/I$ is isomorphic to $\mathbb{Q}(\sqrt5) = \left\{a+b\sqrt5\middle\lvert a,b\in\mathbb{Q}\right\}$ $\endgroup$ – Nikita Dec 12 '14 at 12:08

| show 1 more comment

1 Answer 1