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Martin Brandenburg
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You can define $\mathfrak{m}_R := \{x ~|~ \forall y : 1 - xy \in R^*\}$. Then a homomorphism $R \to S$ of local rings is a map which is compatible with the ring structure and maps $\mathfrak{m}_R$ maps into $\mathfrak{m}_S$. However, this is not equivalent to the usual condition that images of non-units are non-units: In general it is not true that $R = R^* \cup \mathfrak{m}_R$. This is proven by Thierry Coquand in a remark about the theory of local rings. The counterexample is as follows: Consider the Zariski topos $C$ over $\mathbb{Z}$ and the structure sheaf $\mathcal{O}$ of $\mathrm{Spec}(\mathbb{Z})$. Then $\mathcal{O}$ is a local ring in $C$, one verifies that $\mathfrak{m}=\{0\}$ on global sections, so that in particular $2 \in \mathcal{O}^* \vee 2 \in \mathfrak{m}$ is not satisfied.

You can define $\mathfrak{m}_R := \{x ~|~ \forall y : 1 - xy \in R^*\}$. Then a homomorphism $R \to S$ of local rings is a map which is compatible with the ring structure and maps $\mathfrak{m}_R$ maps into $\mathfrak{m}_S$. However, this is not equivalent to the usual condition that images of non-units are non-units: In general it is not true that $R = R^* \cup \mathfrak{m}_R$. This is proven by Thierry Coquand in a remark about the theory of local rings.

You can define $\mathfrak{m}_R := \{x ~|~ \forall y : 1 - xy \in R^*\}$. Then a homomorphism $R \to S$ of local rings is a map which is compatible with the ring structure and maps $\mathfrak{m}_R$ maps into $\mathfrak{m}_S$. However, this is not equivalent to the usual condition that images of non-units are non-units: In general it is not true that $R = R^* \cup \mathfrak{m}_R$. This is proven by Thierry Coquand in a remark about the theory of local rings. The counterexample is as follows: Consider the Zariski topos $C$ over $\mathbb{Z}$ and the structure sheaf $\mathcal{O}$ of $\mathrm{Spec}(\mathbb{Z})$. Then $\mathcal{O}$ is a local ring in $C$, one verifies that $\mathfrak{m}=\{0\}$ on global sections, so that in particular $2 \in \mathcal{O}^* \vee 2 \in \mathfrak{m}$ is not satisfied.

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Martin Brandenburg
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ThereYou can define $\mathfrak{m}_R := \{x ~|~ \forall y : 1 - xy \in R^*\}$. Then a homomorphism $R \to S$ of local rings is no such axiomatizationa map which is compatible with the ring structure and maps $\mathfrak{m}_R$ maps into $\mathfrak{m}_S$. However, this is not equivalent to the usual condition that images of non-units are non-units: In general it is not true that $R = R^* \cup \mathfrak{m}_R$. This is proven by Thierry Coquand in a remark about the theory of local rings.

There is no such axiomatization. This is proven by Thierry Coquand in a remark about the theory of local rings.

You can define $\mathfrak{m}_R := \{x ~|~ \forall y : 1 - xy \in R^*\}$. Then a homomorphism $R \to S$ of local rings is a map which is compatible with the ring structure and maps $\mathfrak{m}_R$ maps into $\mathfrak{m}_S$. However, this is not equivalent to the usual condition that images of non-units are non-units: In general it is not true that $R = R^* \cup \mathfrak{m}_R$. This is proven by Thierry Coquand in a remark about the theory of local rings.

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Martin Brandenburg
  • 63.1k
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  • 207
  • 424

There is no such axiomatization. This is proven by Thierry Coquand in a remark about the theory of local rings.