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Let $R$ be a (commutative unitary) ring object in a topos $\mathcal E$. Say $R$ is a Fermat ring if it satisfies $$\forall f:R\to R\;\exists !g:R^2\to R\;: \forall x,x^\prime \in R\;[fx^\prime -fx=g(x,x^\prime )\cdot (x^\prime -x)].$$ As Kock writes in his SDG book, this is an alternative synthetic foundation for calculus.

Suppose now $\mathcal E$ is a topos of $k$-algebras for $k$ a commutative unitary ring. $\mathcal E$ has the ring object $R=\operatorname{Spec}k[x]$. I want to prove that $$\mathcal E\models R\text{ is Fermat}.$$

A chain of isomorphisms quickly reminds me that $f:R\to R$ is just a univariate polynomial with coefficients in $k$. If an arrow $g:R^2\to R$ were just a bivariate polynomial, then it seems we can get a desired $g$ by looking at $f(Y)-f(X)$. Indeed, since $Y-X\mid Y^k-X^k$, then $Y-X\mid f(X)-f(Y)$ and any bivariate polynomial which gives $f(Y)-f(X)$ when multiplied by $Y-X$ should work.

My problem is that I can't seem to get a chain of isomorphisms leading to $k[x,y]$; all I get is

$$\mathsf{Hom}(\operatorname{Spec}k[x]\times \operatorname{Spec}k[x],\operatorname{Spec}k[x])\cong \mathsf{Hom}(k[x],k[x]\times k[x])\cong \mathsf{Hom}(\mathbf{2},Uk[x])$$i.e two univariate polynomials...

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    $\begingroup$ Your second step of isomorphisms is wrong; it's $\text{Hom}(k[x], k[x] \otimes k[x])$, which is indeed bivariate polynomials. $\endgroup$ – Qiaochu Yuan Sep 27 '16 at 6:37
  • $\begingroup$ Indeed something much more general is true in the big Zariski topos of a ring $k$: The canonical map $\mathcal{A}\to\mathcal{H}\mathrm{om}(\mathcal{H}\mathrm{om}_{\mathrm{Alg}(\mathbb{A}^1)}(\mathcal{A},\mathbb{A}^1),\mathbb{A}^1)$ is an isomorphism, for any quasicoherent $\mathbb{A}^1$-algebra $\mathcal{A}$. For instance, if $\mathcal{A}=\mathbb{A}^1[X_1,\ldots,X_n]/(f_1,\ldots,f_m)$, then this says that the canonical map $\mathcal{A}\to\mathcal{H}\mathrm{om}(X,\mathbb{A}^1)$ is an isomorphism, where $X=\{(x_1,\ldots,x_n):\mathbb{A}^1)^n|f_1(x_1,\ldots,x_n)=\dots=f_m(x_1,\ldots,x_n)=0\}$. $\endgroup$ – Ingo Blechschmidt Sep 28 '16 at 15:14
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Since $$\operatorname{Spec}k[x]\times \operatorname{Spec}k[x]\cong\operatorname{Spec} k[x]\otimes k[x]\cong\operatorname{Spec}k[x,y],$$ a map $\operatorname{Spec}k[x]\times \operatorname{Spec}k[x]\to\operatorname{Spec}k[x]$ corresponds to a map $k[x]\to k[x,y]$, not a map $k[x]\to k[x]\times k[x]$ as you wrote. Such a map is determined by where it sends $x$, i.e., a bivariate polynomial.

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  • $\begingroup$ Thanks. Silly mistake but oh well. $\endgroup$ – Arrow Sep 27 '16 at 7:03

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