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So, I have to show that the kernel of a linear functional is a subspace of codimension one. Someone suggested to me that the following:

If $f:X\rightarrow \mathbb{K}$ and $f\not\equiv 0$, then $\ker(f)$ is a subspace of codimension one and a subspace of codimension one is kernel of non-zero linear functional.

To prove this, we can do the following:

$f$ induces an isomorphism between $X\backslash \ker(f)$ and $\mathbb{K}$. Conversely, if $Z$ is a hyperplane, let $g:X\rightarrow X\backslash Z$ be the natural map and let $T:X\backslash Z\rightarrow \mathbb{K}$ be an isomorphism. Then, $f=T\circ g$ is a linear functional on $X$ and $\ker(f)=Z$.

But my problem is: how do I establish those isomorphisms? Can somebody help me out here? Thanks in advance.

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The first isomorphism you are looking for is precisely the first isomorphism theorem for modules.

To check that $f=T\circ g$ is a linear functional you simply apply the definitions. Finally $x\in \ker f$ iff $T(g(x))=0$, but $T$ is an linear functional so this occurs iff $g(x)=0$, but $g(x)=0$ iff $x\in Z$ by the definition of the natural quotient map.

If you were wondering how to construct $T$: By axiom of choice there exists a basis $\{e_i\}$ for a $Z$. We can append to that an element $e$ so that $\{e\}\cup \{e_i\}$ is a basis for $Z$. Define $T':X\to \mathbb K$ by extending $\alpha e\mapsto\alpha$ by Hahn-Banach. Define $T:X\backslash Z\to \mathbb K$ in the natural way from $T'$. I leave it to you to show that $T$ is an isomorphism. This is actually the standard way to show the converse.

Note we are assuming that $\dim X>1$ here, else the result is trivial.

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