Answer to Question #104926 in Abstract Algebra for Sourav Mondal

Question

Prove that R^5/R=R⁴ as rings.

Answer 1

Consider the mapping $f\colon\mathbb R^5\to\mathbb R^4$, where $f(a,b,c,d,e)=(a,b,c,d)$. It is a ring homomorpism, because $f((a_1,b_1,c_1,d_1,e_1)+(a_1,b_1,c_1,d_1,e_1))=$

$=f(a_1+a_2,b_1+b_2,c_1+c_2,d_1+d_2,e_1+e_2)=$

$=(a_1+a_2,b_1+b_2,c_1+c_2,d_1+d_2)=$

$=(a_1,b_1,c_1,d_1)+(a_2,b_2,c_2,d_2)=$

$=f(a_1,b_1,c_1,d_1,e_1)+(a_2,b_2,c_2,d_2,e_2)$ and

$f((a_1,b_1,c_1,d_1,e_1)\cdot(a_1,b_1,c_1,d_1,e_1))=$

$=f(a_1a_2,b_1b_2,c_1c_2,d_1d_2,e_1e_2)=$

$=(a_1a_2,b_1b_2,c_1c_2,d_1d_2)=$

$=(a_1,b_1,c_1,d_1)\cdot(a_2,b_2,c_2,d_2)=$

$=f(a_1,b_1,c_1,d_1,e_1)\cdot(a_2,b_2,c_2,d_2,e_2)$

So $\mathbb R^5/\ker f\approx Im f$

$Im f=\mathbb R^4$ , because for every $(a,b,c,d)\in\mathbb R^4$ we have $(a,b,c,d)=f(a,b,c,d,0)$ .

Find $\ker f$ : if $(a,b,c,d,e)\in\ker f$, then $(0,0,0,0)=f(a,b,c,d,e)=(a,b,c,d)$ , that is $a=b=c=d=0$ , so $\ker f=\{0\}^4\times\mathbb R\approx\mathbb R$

We obtain $\mathbb R^5/\mathbb R\approx\mathbb R^4$