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Let $k$ be a field. Then $(k,\text{id}_k)$ is a $k$-algebra. Let two further $k$-algebras $(A,\varphi_A)$ and $(B,\varphi_B)$, such that both are local as rings, and a $k$-algebra homomorphism $f:A\rightarrow B$, i.e. $\varphi_B=f\circ\varphi_A$, be given.

Assume that the induced ring homomorphism $q:k\rightarrow B/\mathfrak{m}_B$ is a bijection. Here $\mathfrak{m}_B$ denotes the maximal ideal of $B$.

Now I have to show that $f$ is actually a local ring homomorphism.

I am having trouble doing this! My idea was to show that $\mathfrak{m}_A$ will be mapped to the zero element in $B/\mathfrak{m}_B$ under the composition$A\rightarrow B\rightarrow M/\mathfrak{m}_B$, but I don't know how.

Any help/ hint/ advice is appreciated!

TwoStones
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    hint: consider the composition $$A\xrightarrow{f} B\to B\big/\mathfrak{m}_B\xrightarrow{q^{-1}}k.$$ this map is surjective (why?), so what can you conclude about its kernel? – Atticus Stonestrom Jun 04 '21 at 20:16
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    Thanks for the hint! The kernel is a maximal ideal in this case and therefore equal to $\mathfrak{m}_A$. Since $q$ is bijective the result follows, right? – TwoStones Jun 04 '21 at 20:27
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    dear @TwoStones, yes, that is absolutely right!! :) then you can complete your desired argument immediately. alternatively the elementwise argument of Mark (+1) of course also works – Atticus Stonestrom Jun 04 '21 at 20:31
  • $\def\ker{\operatorname{Ker}} \def\frm{\mathfrak{m}} $I don't know if this is the intended solution by @AtticusStonestrom, but here is my argument: denote $\pi:B\to B/\frm_B$ to the projection. Let $x\in A\setminus\ker(\pi\circ f)$. Since $B/\frm_B\cong k$, there is $\lambda\in k^*$ with $x-\lambda\in\ker(\pi\circ f)$. Thus, $x-\lambda$ is a non-unit of $A$, and due to the fact that the subset of non-units of a local ring is closed under sums, $x$ must be a unit. Thus $\ker(\pi\circ f)=f^{-1}(\ker\pi)=f^{-1}(\frm_B)$ is the maximal ideal of $A$. – Elías Guisado Villalgordo Feb 17 '23 at 10:16
  • I just found out that this is in EGA IV, 3rd part. Second paragraph of (10.10.3). – Elías Guisado Villalgordo Mar 04 '23 at 13:38
  • @AtticusStonestrom Why is it surjective? – Jeremy Lindsay Aug 03 '23 at 13:44
  • @JeremyLindsay fix any $\lambda\in k$. how does the composition act on $\varphi_A(\lambda)$? – Atticus Stonestrom Aug 04 '23 at 12:20

2 Answers2

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If I understand correctly, $q$ is the composition of $\varphi_B$ with the quotient map to $m_B$. To be honest, you should give the precise definitions in the question.

We want to show that $f(m_A)\subseteq m_B$. So let $x\in m_A$. By assumption there is some $y\in k$ such that $q(y)=\varphi_B(y)+m_B=f(x)+m_B$. Thus $\varphi_B(y)-f(x)\in m_B$. But note that:

$\varphi_B(y)-f(x)=f(\varphi_A(y))-f(x)=f(\varphi_A(y)-x)$

So we have $f(\varphi_A(y)-x)\in m_B$. This means that the element $f(\varphi_A(y)-x)$ is not invertible in $B$. Since homomorphisms map invertible elements to invertible elements we conclude that $\varphi_A(y)-x\in A$ is not invertible, i.e $\varphi_A(y)-x\in m_A$. Since $x\in m_A$ it follows that $\varphi_A(y)\in m_A$ as well, i.e $\varphi_A(y)$ is not invertible. Again, it follows that $y\in k$ is not invertible, but since $k$ is a field this means $y=0$. But then $\varphi_B(y)=0$, and so $f(x)=f(x)-\varphi_B(y)\in m_B$.

Mark
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  • Thank you very much! You are right about $q$ and I had a typo in the question regarding $q$, sorry for that! – TwoStones Jun 04 '21 at 20:31
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We have the following diagram:

We know all of the map are $k$-algebra homomorphism, therefore $\phi$ is $k$-algebra homomorphism from $k \to B/{\frak{m}}_{B}\cong k$. This map has to be identity map.

Therefore we have $A\to B/{\frak{m}}_B$ being surjective map, therefore by the first isomorphism theorem the kernel of this map has to be a maximal ideal, in the local ring is ${\frak{m}}_A$ therefore $f:A \to B$ maps ${\frak{m}}_A$ into ${\frak{m}}_B$

yi li
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