TITLE: Without excluded middle: simple example of field with nilpotents?
QUESTION [5 upvotes]: I would like to verify I'm not making a mistake.
Does the statement "fields do not have non-zero nilpotents" depend on the law of excluded middle?
All proofs of this fact I can conjure rely on the dichotomy $a=0\vee a\neq 0$ along with the implication (for fields) that $a\neq 0\implies a$ invertbile. Moreover, in algebraic geometry it's possible to show the generic ring of Zariski topoi is internally a field (has the implication $a\neq 0\implies a$ invertible, but not necessarily excluded middle).
Is there some super simple example of a (commutative unitary) ring which is internally a field and has nilpotents?
Added. What about the statement "in a field, the only nilpotent is zero"?
Remark. The implication ($\neg (a=0)\implies a$ is invertible) is not of my invention; it is taken as the definition of a "ring of fractions" in section 2 of this paper by Kock. Fields are defined analogously.

REPLY [2 votes]: I think the intuitionistic definition of field you assume has to be disputed: First, you have $$(\ddagger):\qquad\forall s: \neg(s=0)\Rightarrow (s\text{ invertible})\quad\Longleftrightarrow\quad \forall s: \neg(s=0)\Leftrightarrow (s\text{ invertible}).$$ There, since $\neg\neg\neg\Phi\Leftrightarrow\neg\Phi$ is intuitionistically valid for any proposition $\Phi$, you get that
$$(\star)\qquad\left[\neg(s=0)\Leftrightarrow (s\text{ invertible})\right]\quad\Longrightarrow\quad \left[\neg\neg(s\text{ invertible})\Leftrightarrow (s\text{ invertible})\right].$$
It's questionable you want to impose this restriction, because e.g. it does not hold when interpreting the domain of $s$ as the structure sheaf ${\mathscr O}_X$ of a reduced scheme $X$, which internally you might want to think of as a field.
The reason the R.H.S. of $(\star)$ does not hold in this case is that the truth value of $(s\text{ invertible})$ is the (scheme-theoretic) non-vanishing locus $\{x\in X\ |\ s(x)\neq 0\text{ in }k(x)\}$ of $s$, so $\neg\neg(s\text{ invertible})$ holds iff the latter is dense in $X$, which is the case iff $s$ is regular. 
On the other hand, the alternative implication/equivalence defining intuitionistic fields
$$(\dagger):\qquad\forall s: \neg(s\text{ invertible})\Rightarrow (s=0)\quad\Longleftrightarrow\quad \forall s: \neg(s\text{ invertible})\Leftrightarrow (s=0)$$
does hold for reduced schemes. Also, with this definition you can quickly prove that
$$s\text{ nilpotent}\quad\Longrightarrow\quad s=0$$
since $s\text{ nilpotent}\Rightarrow\neg(s\text{ invertible})$. 
Source: Ingo Blechschmidt's paper is a very interesting and pleasant read on all of this.