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Let $L/K$ a finite field extension. We can endow the tensor product $L \otimes_K L$ with structure of $L$-algebra via $L \to L \otimes_K L,\ l \mapsto l \otimes 1_L$. Therfore $L \otimes_K L$ carries also structure of a $L$-module via $L$-multiplication in first factor $l \cdot (a \otimes b) := la \otimes b$.

Let $d: L \otimes_K L \to L, a \otimes b \mapsto ab$ be the canonical diagonal map and denote it's kernel by $I:= \operatorname{ker}(d)$.
I have once already asked in Tensor product $L \otimes_K L$ has no nilpotent elements iff $I/I^2=0$ about the proof of the statement that

$L \otimes_K L$ reduced (has no nilpotent elements) is equivalent to $I= I^2$, ie $I/I^2=0$

I obtained there a completely satisfying answer but nevertheless I'm still badly curious if it is also possible to give a 'constructive' proof in the sense that if $L \otimes_K L$ is not reduced, so it contains a nilpotent non zero $a \in L \otimes_K L$, then this $a$ helps me to to construct an explicit element $i(a) \in I \backslash I^2$ and vice versa a $b \in I \backslash I^2$ helps to construct a nilpotent $a(b) \in \sqrt{(0)} \subset L \otimes_K L$.

Question: Are such direct constructions here feasible?

Note: we can use the fact that $I/I^2$ is generated as $L$-module by $a \otimes 1_L -1_L \otimes a, a \in L$.

user267839
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  • Examining the other proof, it seems the only "non constructive" part of the proof is the chinese remainder lemma and Thm 8.7 in Atiyah-Macdonald. Do you know a "constructive proof" of these results? If you find such proofs you may use the existing proof. You find some information here: https://mathoverflow.net/questions/65332/matrix-decomposition-the-other-way/65338#65338 – hm2020 Sep 12 '21 at 12:46
  • It seems the main theme in the proof of AM.Thm 8.7 is the chinese remainder lemma and this lemma has a constructive proof. – hm2020 Sep 16 '21 at 09:48

1 Answers1

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Question: "I obtained there a completely satisfying answer but nevertheless I'm still badly curious if it is also possible to give a 'constructive' proof."

Answer: It seems the main theme in the proof of the assertaion is the chinese remainder lemma and this lemma has a (it has several I believe) constructive proof.

Tensor product $L \otimes_K L$ has no nilpotent elements iff $I/I^2=0$

hm2020
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  • I not see how you can use the proof you gave in the linked question + Thm 8.7 in Atiyah-Macdonald extract following two explicite constructions: (I) assume $L \otimes_K L$ is not reduced, then $\sqrt{(0)} \neq 0$. How we can explicitely use non-zero elements from $\sqrt{(0)} \neq 0$ to construct an element $a \in I \backslash I^2$ and (II) if we have $I \neq I^2$, how we can use elements from $I \backslash I^2$ to construct explicitely a non-zero nilpotent element $n \in \sqrt{(0)} \neq 0$? – user267839 Sep 17 '21 at 21:19
  • @Isakthe XI - I agree that it is not immediate, but the chinese remainder lemma and its constructive proof may be the first a place to look. – hm2020 Sep 18 '21 at 16:26
  • Chinese remainder lemma solves my problem at least in case $L= K(a) =K[X]/f$ ($f$ minimal polynomial of $a \in L$): Then we have in $L[X]$ decomposition in prime factors $f= (X-a)^{n_a} \cdot \prod f_i^{n_i}$ and therefore $L \otimes_K L= L[X]/(X-a)^{n_a} \times \prod_i L[X]/(f_i)^{n_i}$.

    Then the diagonal map $d_L: L \otimes L \to L$ is given by projection $L[X]/(X-a)^{n_a} \times \prod_i L[X]/(f_i)^{n_i} \to L[X]/(X-a)=L, X \mapsto a $. Then $I = ker(d_L)= (X-a) \times \prod_i L[X]/(f_i)^{n_i}$.

     – user267839 Sep 18 '21 at 23:08
  • Now if $\sqrt{(0)} \neq 0$, then $f$ is not separable and there exist a seprable $g \in K[X]$ with $f= g(X^{p_m})$ for $p= char(K), m > 0$, a known fact from algebra. Since $f(a)=g(a^{p_m})=0$ we conclude $n_a= p^m \ge 2$. Choose $x_a \in (X-a) \backslash (X-a)^2$. Then $(x_a, 0, 0, ..., 0) \in I \backslash I^2$.

    Conversely if $I \neq I^2$ then $n_a \neq 1$ since otherwise $I = 0 \times \prod_i L[X]/(f_i)^{n_i}$ would imply $I=I^2$. Therefore we can take a $x_a \in (X-a) \backslash (X-a)^2$. Then $0 \neq (x_a, 0, 0, ..., 0) \in \sqrt{(0)} $.

     – user267839 Sep 18 '21 at 23:09
  • What about general $L$? I not see a direct way to reduce the discussion to case above with $L= =K[X]/f$. Say we have for arbitrary $a \in L \backslash K$ an inclusion of fields $K \subset K(a) = K[X]/f \subset L$ which induces inclusion $K(a) \otimes_K K(a) \subset L \otimes L$. Denote by $d_L: L \otimes L \to L$ and $I_L= ker(d_L)$ and the restricted diagonal map $d_{K(a)}: K(a) \otimes_K K(a) \to K(a)$ and $I_{K(a)}=ker(d_{K(a)})$. – user267839 Sep 18 '21 at 23:11
  • Also let $\sqrt{(0)}L \subset L \otimes L$ the nilpotent ideal of $L \otimes L$ and $\sqrt{(0)}{K(a)} := \sqrt{(0)}L \cap K(a) \otimes_K K(a) \subset K(a) \otimes_K K(a)$. Now if we want to be able to reduce the problem for general $L$ to the case $K(a)$ then we have to make sure that following two points are true: if $\sqrt{(0)}_L \neq 0$, then $\sqrt{(0)}{K(a)} \neq 0$ and if $I_L \neq I_L^2$, then $I_{K(a)} \neq I_{K(a)}^2$. Do you see why these two points are true? – user267839 Sep 18 '21 at 23:11
  • @IsaktheXI - Why in the above comment can you assume that $n_a\geq 1$? – hm2020 Sep 19 '21 at 08:07
  • @Isakthe XI - if you look in Atiyah-Macdonald and the proof of Thm 8.7 you will find that the essential part of this proof is the CRL which "is constructive". – hm2020 Sep 19 '21 at 09:28
  • For 'Why $n_a$ can be assumed $n_a\geq 1$?' Above I was only dealing with case $L=K(a) = K[X]/f$ ($f \in K[X]$ minimal polynomial of $a$). Therefore $f$ splits in $L[X]$ ! in prime factors $f= (X-a)^{n_a} \cdot \prod f_i^{n_i}$ and since $f(a)=0$ by construction in $L[X]$ we have $(X-a) | f$ and that's why $n_a\geq 1$. – user267839 Sep 19 '21 at 17:52
  • back to general case. The Thm 8.7 from AM states that $A:= L \otimes_K L$, which is Artin ring since $A$ is a finite $L$-module, decomposes as $A= \prod_{j=1}^n A/(\mathfrak{m}_j)^{k_j}$ where $\mathfrak{m}_j$ are maximal ideals in $A$. Then $A$ is reduced iff $k_j=1$ for all $j$, so in non reduced case it's easy to construct the nilpotent elements.

    I think here is partial a solution for one direction: Every ideal $K$ in a product of rings $R_1 \times R_2$ has structure $K= K_1 \times K_2$ with $K_i$ ideal of $R_i$ (note $K_i$ can be R_i$).

     – user267839 Sep 19 '21 at 17:52
  • For $d_L: A = L \otimes L \to L$ the ideal $I=ker(d_L)$ is maximal in $A$, and therefore if we write $I= \prod_j I_j \subset \prod_j A/(\mathfrak{m}j)^{k_j}$ there exist a unique $s$ with $I{s} = \mathfrak{m}{s} \subset A/(\mathfrak{m}{s})^{k_s}$ and $I_j= A/(\mathfrak{m}j)^{k_j}$ for other $j \neq s$. WLOG, $s=1$. Then $I = \mathfrak{m}_1 \times \prod{j=2}^n A/(\mathfrak{m}j)^{k_j}$. Then $I=I^2$ iff $k_1=1$. If $k_1 >1$ then in local Artin ring $A/(\mathfrak{m}{1})^{k_1}$ we have $\mathfrak{m}{1} \neq \mathfrak{m}{1}^2$ and – user267839 Sep 19 '21 at 17:54
  • we can easily construct a $r = (r_1, 0,....,0) \in I \backslash I^2$ for $r_1 \in \mathfrak{m}{1} \backslash \mathfrak{m}{1}^2$. – user267839 Sep 19 '21 at 17:54
  • The converse is not clear. Assume $A$ is not reduced, but $k_1=1$ where $I = \mathfrak{m}{1} \times \prod{j=2}^n A/(\mathfrak{m}j)^{k_j}$. That is there exist a $k_j >1$ with $j \neq 1$. But then $I=I^2$ since $\mathfrak{m}{1}= (0)$ in $A/(\mathfrak{m}1)$, that's strange. The problem is to show that if $A$ is not reduced, the kernel $I$ has the structure $\prod{j=1}^{s-1} A/(\mathfrak{m}j)^{k_j} \times \mathfrak{m}_s \times \prod{j=s+1}^{n} A/(\mathfrak{m}_j)^{k_j}$ with $k_s >1$. I don't know now to prove it. – user267839 Sep 19 '21 at 18:03
  • @IsaktheXI - If you are given a direct sum $A \cong A_1\oplus \cdots \oplus A_d$ it follows there is a 1-1 correspondence between prime ideals in $A$ and prime ideals in $A_i$ for $i=1,..,d$: Any prime ideal $\mathfrak{p} \subseteq A$ is on the form $\mathfrak{p}=A_1\oplus \cdots \oplus A_{i-1} \oplus \mathfrak{p}i \oplus A{i+1} \oplus \cdots \oplus A_d$ with $\mathfrak{p}_i \subseteq A_i$ a prime ideal. – hm2020 Sep 20 '21 at 09:42
  • The disjoint union of a finite number of affine schemes is affine: https://math.stackexchange.com/questions/4254117/proving-that-the-disjoint-union-of-two-affine-schemes-is-isomorphic-to-an-affine/4255251#4255251 – hm2020 Sep 20 '21 at 09:44
  • I think that what you wanted to say is that such decomposition of prime ideals hold for direct products of rings; there exist no direct sum of rings due to category theoretical reasons (see https://en.wikipedia.org/wiki/Direct_sum#Direct_sum_of_rings , or https://math.stackexchange.com/questions/1209200/why-isnt-there-necessarily-a-direct-sum-of-rings)

    Note that in previous comments I already used implicitely this fact for maximal ideals.

     – user267839 Sep 20 '21 at 20:13
  • The relation of your last link and this problem I not understand. – user267839 Sep 20 '21 at 20:13
  • @IsaktheXI - when you are considering a finite set of rings $A_1,..,A_d$, it follows the direct product equals the direct sum: $\oplus_i A_i = \prod_i A_i$. – hm2020 Sep 21 '21 at 08:35
  • I disagree. What is the $1$ in $\oplus_i A_i$? Note that we talk about the category of rings and not modules! I agree with you that if $R$ ring, and $M_1,..., M_n$ are $R$-modules then $\oplus_i M_i = \prod_i M_i$ as modules, but in case of rings I don't think that it's true. Have you already checked the discussion in the linked mse thread? – user267839 Sep 23 '21 at 22:02
  • @IsaktheXI - in this context you do not need to worry too much about category theoretical properties of direct sum/direct product. Your ring $A$ is a finite dimensional $k$-algebra with $k$ a field, and you need a constructive proof that $A \cong A_1\oplus \cdots \oplus A_d$ is a direct sum (or equivalently a direct product) of Artinian local rings $A_i$. I believe this should follow from the constructive "chinese remainder lemma", but as I say above: It is not immediate. The multiplicative unit $1$ is $1:=(1,1,...,1) \in A$ and the product is componentwise. – hm2020 Sep 24 '21 at 09:22
  • Again: When I write $A\cong A_1\oplus \cdots \oplus A_d$ i mean that $A$ is the set of $d$-tuples $(a_1,..,a_d)$ with $a_i \in A_i$ and where multiplication and addition is componentwise. If you are to be "strict" this definition does not satisfy the correct universal property: The embedding $A_i \rightarrow A$ mapping $a$ to $(0,..,a,..,0)$ is not unital. The projection map $A \rightarrow A_i$ is a well defined map of unital commutative rings. – hm2020 Sep 24 '21 at 11:46
  • But this universal property is not the issue here: What you want is a constructive proof of existence of such a "decomposition" - this componentwise structure is clear from the proof of Thm 8.7 in AM. – hm2020 Sep 24 '21 at 11:47