# Find an isomorphism $f:\mathbb {K}^4/W\rightarrow \mathbb {K}$

Let

$$W :\ x_2+x_3-x_4=0$$

Are $$\mathbb {K}^4/W$$ and $$\mathbb {K}$$ isomorphic? If they are, find an isomorphism $$f:\mathbb {K}^4/W\rightarrow \mathbb {K}$$

First I found a basis of $$W$$, for example $$B=\left\{\left(\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix}\right), \left(\begin{matrix} 0 \\ 1 \\ 0 \\ 1 \end{matrix}\right), \left(\begin{matrix} 0 \\ -1 \\ 1 \\ 0 \end{matrix}\right)\right\}$$

Then $$W$$ has dimension $$3$$ and the dimension of $$\mathbb {K^4}/W$$ is $$4-3=1$$, so $$\mathbb {K}^4/W$$ and $$\mathbb {K}$$ have dimension $$1$$ and then they are isomorphic.

A basis of $$\mathbb {K}^4/W$$ is $$\left(\begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix}\right)+W$$

Now how can I make an isomorphism $$f:\mathbb {K}^4/W\rightarrow \mathbb {K}$$ ?

• Pick any $y$ such that $l(y) = y_2+y_3-y_4 = 1$ then for any $x \in K^4,l( x - l(x)y) = l(x)-l(x)l(y) = 0$ thus $x-l(x)y\in W$, $x \in W+ K y$ and the decomposition is unique. The isomorphism is $f : K^4/W \to K, f(x+W) = l(x)$ – reuns Jun 12 at 15:33

Complete the found basis to a basis of $$\mathbb{K}^4$$; it's sufficient to find a basis of the null space of $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & -1 & 1 & 0 \end{pmatrix}$$ which is the transpose of $$\begin{pmatrix} v_1 & v_2 & v_3 \end{pmatrix}$$ (with the vectors being the members of $$B$$).

A simple row reduction yields $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & -1 & 1 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \end{pmatrix}$$ so the required vector is $$v_4=\begin{pmatrix} 0 \\ -1 \\ -1 \\ 1 \end{pmatrix}$$ Now define $$g\colon\mathbb{K}^4\to \mathbb{K}$$ by $$g(v_1)=0,\quad g(v_2)=0,\quad g(v_3)=0,\quad g(v_4)=1$$ The kernel of this map contains $$W$$ and the image is $$\mathbb{K}$$. By counting dimensions, the kernel equals $$W$$, so $$g$$ induces an isomorphism $$f\colon\mathbb{K}^4/W\to\mathbb{K}$$ by the homomorphism theorem.

Start with the map $$F:\mathbb K^4\to \mathbb K$$ as $$F(x_1,x_2,x_3,x_4)^T=x_2+x_3-x_4.$$ Note that $$F$$ is linear.

Then define $$f:\mathbb K^4/W\to\mathbb K$$ as $$f(v+W)=F(v)$$.

This $$f$$ is well-defined because if $$v_1+W=v_2+W$$ then $$v_1-v_2\in W$$ and thus $$F(v_1-v_2)=0$$ and hence, since $$F$$ is linear, $$F(v_1)=F(v_2).$$

Claim: $$f$$ is one-to-one.

Proof: If $$f(v_1+W)=f(v_2+W)$$ then $$F(v_1)=F(v_2)$$ and hence $$F(v_1-v_2)=0$$ and thus $$v_1-v_2\in W$$ and $$v_1+W=v_2+W.$$ So $$f$$ is one-to-one.

Now show that $$f$$ is onto.