Question

Prove that R^(n)/R^(m) ~ R^(n-m)
as groups, where n, m∈ N, n ≥ m.

Answer 1

Answer on Question #81171 – Math – Abstract Algebra

Question

Prove that $\mathrm{R}^{\wedge}(\mathfrak{n}) / \mathrm{R}^{\wedge}(\mathfrak{m}) \sim \mathrm{R}^{\wedge}(\mathfrak{n} - \mathfrak{m})$ as groups, where $\mathfrak{n}, \mathfrak{m} \in \mathbb{N}, \mathfrak{n} \geq \mathfrak{m}$.

Solution

Consider $\varphi \colon R^n \to R^{n - m}$:

$$\varphi \big ( ( v _ { 1 } , v _ { 2 } , \ldots , v _ { n } ) \big ) = ( v _ { 1 } , v _ { 2 } , \ldots , v _ { n - m } ) \text{ for } ( v _ { 1 } , v _ { 2 } , \ldots , v _ { n } ) \in R ^ { n } .$$

We check that $\varphi$ is a homomorphism:

$$\begin{array}{l} \varphi \left( \left( v _ { 1 } , v _ { 2 } , \ldots , v _ { n } \right) + \left( u _ { 1 } , u _ { 2 } , \ldots , u _ { n } \right) \right) = \varphi \left( \left( v _ { 1 } + u _ { 1 } , v _ { 2 } + u _ { 2 } , \ldots , v _ { n } + u _ { n } \right) \right) \\ = \left( v _ { 1 } + u _ { 1 } , v _ { 2 } + u _ { 2 } , \ldots , v _ { n - m } + u _ { n - m } \right) \\ = \left( v _ { 1 } , v _ { 2 } , \ldots , v _ { n - m } \right) + \left( u _ { 1 } , u _ { 2 } , \ldots , u _ { n - m } \right) \\ = \varphi \left( \left( v _ { 1 } , v _ { 2 } , \ldots , v _ { n } \right) \right) + \varphi \left( \left( u _ { 1 } , u _ { 2 } , \ldots , u _ { n } \right) \right) \end{array}$$

for $(v_{1},v_{2},\ldots ,v_{n}),(u_{1},u_{2},\ldots ,u_{n})\in R^{n}$

Remark that $\ker (\varphi) = \{(0,\ldots ,0,v_{n - m},\ldots ,v_n)\}$ for all $(v_{n - m},\dots,v_{n})\in R^{m} \Rightarrow \ker (\varphi)\triangleq R^{m}$

$\varphi$ is surjective: if $(w_{1},w_{2},\ldots ,w_{n - m})\in R^{n - m}$ then $\varphi \big((w_1,w_2,\dots ,w_{n - m},0,\dots ,0)\big)$

$$= \left(w _ {1}, w _ {2}, \dots , w _ {n - m}\right) =>$$ $$= > \operatorname {im} (\varphi) = R ^ {n - m}$$

By the first isomorphism theorem, $R^n / \ker(\varphi) \triangleq \operatorname{im}(\varphi) \iff R^n / R^m \triangleq R^{n - m}$.

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