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I am reading a proof that if $(A,\mathfrak m)$ is a regular local ring, then $A$ is an integral domain. I put the major questions I'm worried about in bold, but there are a lot of little things I'm slightly unsure about and made note of. Please let me know if any of them are in error.

The proof proceeds by induction on the dimension of the ring. The base case is fairly simple. If $\dim A =0$, then by regularity $0=\dim \mathfrak m/\mathfrak m^2$, so $\mathfrak m = \mathfrak m^2$, which means $\mathfrak m=0$ by Nakayama's lemma.

When $d>0$, suppose the statement is true for $d-1$. Let $k=A/\mathfrak m$ and let $s$ be the closed point of $\operatorname{Spec} A$. Fix $f\in \mathfrak m/\mathfrak m^2$. We have, according to the proof,

$$d-1\le \dim A/fA\le \dim_k T_{V(f),s}=d-1.$$

I see why the first two inequalities hold. The first is a standard theorem in dimension theory, and the second is just the statement that the dimension of the tangent space is at least the dimension of the ring of global sections, where here the scheme is $V(f)$ embedded in $A$ (right?). It seems intuitively correct to me that the last equality should hold because we are killing off a member of a basis for $\mathfrak m/\mathfrak m^2$, but I can't prove it.

Question 1. Why does the last equality hold?

Let $\mathfrak p$ be a minimal prime for $A$ such that $\dim V(\mathfrak p)=d$. (We can make such a choice since if we have a chain of of primes of length $d$ as in the definition of dimension, we can just take the smallest one, right?) Let $B= A/\mathfrak p$, and $g$ be in the image of $f$ in $B$. Again, we should have

$$d-1\le B/gB\le \dim_k T_{V(g),s} \le \dim_k T_{V(f),s} = d-1.$$

Question 2. Why does the second to last inequality above, the one between the tangent spaces, hold?

The above shows that $A/(\mathfrak p +fA)=B/gB$ has dimension $d-1$ and is regular. Since $A/(\mathfrak p +fA)$ and $A/fA$ are integral domains of the same dimension, we must have $\mathfrak p + fA=fA$ (otherwise we could always make a chain in $ fA$ longer by adding $\mathfrak p$, right? So the dimension would not line up?) and $\mathfrak p\subset fA$.

Then, for $u\in \mathfrak p$, we can find $v\in A$ such that $u=fv$, and $f\notin \mathfrak p$ because $\dim V(f)< \dim V(\mathfrak p)$. (This holds because the first has dimension $d-1$ and the second dimension $d$, yes?). So we must have $v\in \mathfrak p$, and $\mathfrak p\subset f\mathfrak p\subset \mathfrak m\mathfrak p.$ So $\mathfrak p=0$ and $A$ is an integral domain. (This last claim is justified by Nakayama.)

(I have found other more easily understood proofs elsewhere, but I really want to nail down the details on this one.)

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The maximal ideal of $A/fA$ is given by $I=\mathfrak{m}/fA$. Then $I^2$ should be $\mathfrak{m}^2+fA/fA$. The second isomorphism theorem will then identify the cotangent space $I/I^2$ with $\mathfrak{m}/(\mathfrak{m}^2+fA)$. There is an evident surjection from $\mathfrak{m}/\mathfrak{m}^2$ to $\mathfrak{m}/(\mathfrak{m}^2+fA)$, whose kernel is the one dimensional space generated by $f$. So your intuition for question 1 seems to be correct. – Jeff Tolliver Jun 26 '13 at 3:14
$fA$ is not contained in $\mathfrak{m}^2$, so one would not usually talk about $\mathfrak{m}^2/fA$. But $\mathfrak{m}^2+fA/fA \cong \mathfrak{m}^2/\mathfrak{m}^2\cap fA$; In other words, it is the image of $\mathfrak{m}^2$ in $A/fA$. – Jeff Tolliver Jun 26 '13 at 4:02
I think the key to question 2 is that $V(g)=V(f)\cap \mathrm{Spec} B$, so it can be viewed as a closed subscheme of $V(f)$ (since $\mathrm{Spec}(B)$ is closed). The inequality just expresses the fact that the tangent space of a closed subscheme cannot be larger than the tangent space of the ambient scheme. – Jeff Tolliver Jun 26 '13 at 4:07
@JeffTolliver I think that's the right intuition. Liu hasn't developed much machinery for talking about tangent spaces geometrically yet, though. Could that statement be reframed in terms of the cotangent space, using an argument like you gave for my first question? – Potato Jun 26 '13 at 4:14
(Thank you, by the way.) – Potato Jun 26 '13 at 4:16

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