Solve the problem involving Rings

chwala · Jul 26, 2023

Its a bit clear; i can follow just to pick another polynomial say

##(x+1)^3## are we then going to have ##(2x-2)+ x+3##?

or it has to be a polynomial with

##x^2+1## being evident? cheers...

fresh_42 · Jul 26, 2023

Let's see ...
\begin{align*}
(x+1)^3&=(x+1)^2\cdot (x+1)=2x\cdot (x+1)=2x^2+2x=2\cdot (x^2+1) -2+2x=2x-2=2\cdot (x-1)\\
&\text{crosscheck}\\
(x+1)^3&=x^3+3x^2+3x+1=x\cdot (-1)+3\cdot (-1)+3x+1=2x-2=2\cdot(x-1)
\end{align*}
We identify ##x^2=-1.##

chwala · Jul 26, 2023

...I was getting a bit lost on this line:

$$2x^2+2x=2\cdot (x^2+1) -2+2x=2x-2=2\cdot (x-1)$$

...but i now i got it by letting,

##x^2+x=x^2+1+m##

therefore

##m=x-1##

other steps follow well.

chwala · Jul 26, 2023

...which means that

##(x+1)^4=(x+1)^2 (x+1)^2 =2x ⋅2x=4x^2=4(-1)=-4##

Cross check:

##(x+1)^4=x^4+4x^3+6x^2+4x+1=1-4x-6+4x+1=-4##

chwala · Jul 26, 2023

Lastly on this just to get the drift of things, if we have a polynomial say;

##x^3+4##

then this becomes,

##x(x^2)+4=x(-1)+4=4-x##

...

fresh_42 · Jul 26, 2023

chwala said:

Lastly on this just to get the drift of things, if we have a polynomial say;

##x^3+4##

then this becomes,

##x(x^2)+4=x(-1)+4=4-x##

...

Right. The ring ##\mathbb{R}[x] ## is an integral domain, the ideal ##\langle x^2+1 \rangle## is prime and maximal. This makes ##\mathbb{R}[x]/\langle x^2+1 \rangle ## a field, the complex numbers. So ##x= \mathrm{i}.##

Here we have ##x^3+4= \mathrm{i}^3+4=-\mathrm{i} +4 =4-x.##

Solve the problem involving Rings

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