$\bigcap\limits_{\varphi\in E^*}\ker(\varphi)$ and the Axiom of Choise

Question

Context.

Give a nonzero $K$-vector space $E$, it is known that $\displaystyle \bigcap_{\varphi\in E^*}\ker(\varphi)=0$ under AC.

It is also known that, without AC, there are models of ZF in which some non zero vector space do not have nonzero linear forms. For such an $E$, $\displaystyle \bigcap_{\varphi\in E^*}\ker(\varphi)=E.$

So my question is:

Question. Are there models of ZF for which the dimension of $\displaystyle \bigcap_{\varphi\in E^*}\ker(\varphi)$ can have any prescribed cardinality $\kappa$? (the model may depend on $\kappa$, even if it would be supernice to have a single model which works for any cardinality).

Or, do we have necessarily the alternative $\displaystyle \bigcap_{\varphi\in E^*}\ker(\varphi)=0$ or $E$?

I am also interested in the weaker question: can $\displaystyle\bigcap_{\varphi\in E^*}\ker(\varphi)$ be a nonzero proper subspace of $E$?

Side remark. There is no precise reason why I am interested in this, just plain curiosity. These questions popped out because I realized that I needed the existence of some complement of a given line of $E$ to prove that the intersection above is zero.

Answer 1

This is a rather boring/trivial answer to an interesting question. Maybe it can help you reformulate your question to one that admits a more interesting answer!

Let $E$ be a vector space with no non-zero linear form. Now consider the vector space $V=E\oplus K$. The linear form $\varphi(e,c)=c$ has kernel $E$, and conversely any linear form $\psi$ on $V$ restricts to a linear form on $E$, which must be zero, so $E\subseteq \ker(\psi)$. Thus $\bigcap_{\varphi\in V^*} \ker(\varphi)=E$.

More generally, for any set $I$, if we set $V=E\oplus \bigoplus_{i\in I} K$, we have $\bigcap_{\varphi\in V^*} \ker(\varphi)=E$.

This answers your second question, showing that the intersection of the kernels of linear forms can be a proper (and in fact a relatively small) subspace of $V$. For the first, you ask about the dimension of this subspace. I don't think this makes sense: any vector space with a basis has non-zero linear forms. In the absence of a basis, it's not clear what "dimension" means.