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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 $f:\mathbb {K}^4/W\rightarrow \mathbb {K}$ ?

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    $\begingroup$ 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 $f : K^4/W \to K, f(x+W) = l(x)$ $\endgroup$ – reuns Jun 12 at 15:33
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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.

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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.

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