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I find difficulty to understand the proof of this theorem :

Theorem : Let $Y\subseteq\mathbb A^n$ be an affine variety. Let $Ρ\in Y$ be a point. Then $Y$ is nonsingular at $Ρ$ if and only if the local ring $\mathcal O_P$ is a regular local ring.

Let $P$ be the point $(a_1,\dots,a_n)$ in $\mathbb A^n,$ and let $\frak {a}$$=(x_1-a_1,\dots,x_n-a_n)$ be the corresponding maximal ideal in $A = k[x_1,\dots,x_n].$ We define a $\theta : A\longrightarrow k^n$ by $$\theta(f)=\left(\frac{\partial f}{\partial x_1},\dots,\frac{\partial f}{\partial x_n}\right).$$ So $\theta$ induce an isomorphism $\theta':\frak{a}/\frak{a}^2\longrightarrow$$k.$

Now let $\frak b$ be the ideal of $Y$ in $A,$ and let $f_1,\dots,f_n,$ be a set of generators of $\frak b.$ Then the rank of the Jacobian matrix $\left(\frac{\partial f_i}{\partial x_j}\right)_{i,j}$ is just the dimension of $\theta(\frak b)$ as a subspace of $k^n.$ Using the isomorphism $\theta'$ this is the same as the dimension of the subspace $\frak a^2+b/a^2$ of $\frak a/a^2,$ if $\frak m$ is the maximal ideal of $\mathcal{O}_Р,$ we have $$\frak m/m^2\cong a/(b+a^2).$$

I don't understand how I can use $\theta'$ to get $\dim_k \theta(\frak b)$$=\dim_k(\frak a^2+b/a^2)$ and why $$\frak m/m^2\cong a/(b+a^2).$$

Thank you for any help.

Med
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1 Answers1

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To the first point: you have $k$-linear maps $\theta\colon A \to k^n$ and $\theta'\colon {\frak a}/{\frak a^2} \to k^n$ and the second an isomorphism, so to compute the dimension of $\theta(\mathfrak{b})$ you could instead compute the dimension of $\theta'^{-1}(\theta(\mathfrak{b}))$. The latter is indeed $(\mathfrak{b} + \mathfrak{a}^2)/\mathfrak{a}^2$ since that ideal has the right image.

For the last bit: in general, if $R$ is a ring with maximal ideal $M$ then $M/M^2 \simeq (MR_M)/(MR_M)^2$. Here's how I think about this: $(MR_M)/(MR_M)^2 \simeq (M/M^2)_M$ but $M/M^2$ is a vector space over $R/M$, so the elements of $R - M$ already act invertibly and there is no need to localize. In our situation $R = A/\mathfrak{b}$, $M = \mathfrak{a}/\mathfrak{b}$, and $MR_M = \mathfrak{m}$.

You also want to use the fact that, in the notation of this Wikipedia article, $I^eJ^e = (IJ)^e$. In our case, we use this to show that $M^2 = (\mathfrak{a}/\mathfrak{b})^2 = (\mathfrak{a}^2 + \mathfrak{b})/\mathfrak{b}$. Then $M/M^2 \simeq \mathfrak{a}/(\mathfrak{a}^2 + \mathfrak{b})$ by one of the isomorphism theorems.

Hoot
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