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Carryless multiplication treats binary digits as coefficients of a polynomial over the field with two elements. Products are formed by polynomial multiplication, with exclusive or combining equal-degree terms instead of carrying to the next binary position.

No. Let $k={\mathbb{F}}_2$ be the field with two elements, put $A=k[\epsilon ]/({\epsilon }^2)$, and take the closed immersioninduced by the ring homomorphism $q:A\twoheadrightarrow k$ with $q(\epsilon )=0$.

For every field $K$, a ring homomorphism $\alpha :A\to K$ must send $\epsilon$ to zero: $\alpha (\epsilon {)}^2=0$, and a field has no nonzero nilpotent elements. Consequently $\alpha$ factors uniquely through $q$. Passing from ring homomorphisms to morphisms of schemes between affine schemes gives a bijectionThis concerns all morphisms of schemes, without requiring them to be over $k$: if $K$ does not have characteristic two, both sets are empty; otherwise each contains exactly one element.

Nevertheless, $f$ is not an isomorphism of schemes. Its map on global sections is $q$, whose kernel of a ring homomorphism contains the nonzero element $\epsilon$, so $q$ is not a ring isomorphism. The target is the nonreduced scheme defined by the dual numbers, whereas the source is a reduced scheme. This illustrates why field-valued points do not detect nilpotents.