Several questions actually.

All rings and algebras are supposed to be commutative and with $1$ here.

(1) Let $k$ be a field, and let $A$ and $B$ be two $k$-algebras. I need a proof that if $A$ and $B$ are reduced (i. e., the only nilpotents are $0$) and $\operatorname{char}k=0$, then $A\otimes_k B$ is reduced as well.

The condition $\operatorname{char}k=0$ can be replaced by "$k$ is perfect", but I already know a proof for the $\operatorname{char}k>0$ case (the main idea is that every nilpotent $x$ satisfies $x^{p^n}=0$ for some $n$, where $p=\operatorname{char}k$), so I am only interested in the $\operatorname{char}k=0$ case.

Please don't use too much algebraic geometry - what I am looking for is a constructive proof, and while most ZFC proofs can be made constructive using Coquand's dynamic techniques, the more complicated and geometric the proof, the more work this will mean.

BTW the reason why I am so sure the above holds is that some algebraist I have spoken with has told me that he has a proof using minimal prime ideals, but I haven't ever seen him afterwards.

Ah, and I know that this is proven in J. S. Milne's Algebraic Geometry (version 6.01, Proposition 5.17 (a)) for the case $k$ algebraically closed.

(2) What if $k$ is not a field anymore, but a ring with certain properties? $\mathbb{Z}$, for instance? Can we still say something? (Probably only to be thought about once (1) is solved.)

(3) Now assume that $k$ is algebraically closed. Can we replace reduced by connected (which means that the only idempotents are $0$ and $1$, or, equivalently, that the spectre of the ring is connected)? In fact, this even seems easier due to the geometric definition of connectedness, but I don't know the relation between $\operatorname{Spec}\left(A\otimes_k B\right)$ and $\operatorname{Spec}A$ and $\operatorname{Spec}B$. (I know that $\operatorname{Spm}\left(A\otimes_k B\right)=\operatorname{Spm}A\times\operatorname{Spm}B$ however, but this doesn't help me.)

PS. All algebras are finitely generated if necessary.

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    $\begingroup$ Tensor product of rings corresponds to pull-backs of affine schemes. So in question (3) you're asking if the product of connected affine schemes is connected (which is true). $\endgroup$ Dec 20, 2009 at 19:04
  • $\begingroup$ Thanks, but how do you prove this true fact? $\endgroup$ Dec 20, 2009 at 19:08
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    $\begingroup$ Since we're talking about affine schemes, my assertion above is equivalent to showing that the tensor product of two rings is a push-out in the category of commutative rings. So you just need to show that, given any ring $R$ and maps $A \to R$ and $B \to R$ which agree on $k$, this induces a map $A \otimes_k B \to R$ making the obvious diagram commute. This latter fact is a pretty straightforward exercise in ring theory. $\endgroup$ Dec 20, 2009 at 19:25
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    $\begingroup$ It is NOT true that the product of two connected affine schemes is connected. For example the affine scheme X=Spec(Q(i)) is connected since it has only one point, but the product of X with itself is equal to the sum (=coproduct) of two copies of X. Topologically it is a discrete space with two ponts, hence not connected. $\endgroup$ Dec 20, 2009 at 19:50
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    $\begingroup$ @HeinrichD: no. All I've learned from this thread is that commutative algebra is hard... $\endgroup$ Sep 18, 2016 at 8:12

5 Answers 5


In (1), A (resp. B) injects into the product of the residue fields at the minimal primes, so we can reduce to the case where A and B are fields which should be sufficiently well-known. Details are in Bourbaki, Algebra, ch.5 §15 no.2

(3) is answered in EGA IV, 4.5: the product of a connected scheme and a geometrically connected scheme is connected.

  • $\begingroup$ Thanks. I'll look into these sources (EGA IV is a bit high for me, bur Bourbaki seems just right, and anyway I don't care for (3) as much as I care for (1)). $\endgroup$ Dec 20, 2009 at 20:39

Over $\mathbb Z$, there are several possibilities.

  • If both $A, B$ are torsion-free over $\mathbb Z$, then $A\otimes_{\mathbb Z} B$ is connected if and only if its generic fiber $(A\otimes_{\mathbb Z} \mathbb Q)\otimes_{\mathbb Q} (B\otimes_{\mathbb Z} \mathbb Q)$ is connected, and you are reduced to the case of algebras over a field.

EDIT the above claim is incorrect. See a partial answer here.

  • If $A$ (or $B$) have positive characteristic, then $A_{\rm red}$ is an algebra over a product of finite fields, then so is $A\otimes_{\mathbb Z} B$. Again you are reduced to the case of algebras over a field (if at least two finite fields involve really, then the tensor product is not connected.

  • The remainning case: if $A$ is noetherian, then ${\rm Spec}(A)$ has a flat part and a ''vertical part'': take the ideal $I$ of $A$ consisting in torsion elements. Then $A/I$ is flat and $mI=0$ for some non-zero integer $m$, and we have ${\rm Spec A}=V(I)\cup V(mA)$.
    The same is true for $A\otimes B$. To have the connectedness, you want [EDIT: it is enough (but far to be necessary) that] the flat and vertical parts to be connected and meet each other. Sorry, I don't have a simple statement.

  • $\begingroup$ Thanks a lot, though much of this answer is over my head (I'm still not beyound Atiyah/Macdonald in commutative algebra). Anyway, the main reason why I asked the connectedness question has just collapsed: the product I was thinking of is not the tensor product. $\endgroup$ Feb 26, 2010 at 0:01

The answer to (2) is no. For example, for every prime $p$, $$\mathbb{Z}[\sqrt{p}] \otimes_{\mathbb{Z}} \mathbb{F}_p = \mathbb{Z}[x]/(x^2-p) \otimes_{\mathbb{Z}} \mathbb{F}_p = \mathbb{F}_p[x]/(x^2)$$ is not reduced.

  • $\begingroup$ What about $A\otimes_CB$ if $C\to A$ and $C\to B$ are assumed to be reduced at each fiber, that is to say, after base change to $\kappa(\mathfrak p)$ for all $\mathfrak p\in\operatorname{Spec}(C)$ (and I don't know how to describe this without referring to prime ideals)? $\endgroup$
    – user20948
    May 23, 2020 at 18:05

Ad (1): It is perfectly true that the tensor product of two reduced algebras over a perfect field is reduced. You can find a proof in Bourbaki's Algebra (Chapter V; §15; 4,5), but of course this venerable author is not especially known for his enthusiasm toward constructive mathematics .

Ad (2): I don't know.

Ad (3): The relation you are looking for is the simplest possible, namely $$Spec(A\otimes_k B)=Spec(A)\times_k Spec(B).$$ But that doesn't help because the product of connected schemes has no reason to be connected. I can only give you the following sufficient condition for connectedness of tensor products of algebras.

Suppose $k \subset K$ is a separable extension of fields such that $k$ is algebraically closed in $K$. Such an extension is called (rather unimaginatively) a REGULAR extension.With this definition, we can state the

Theorem: If $k \subset K$ is regular, then for every $k$-subalgebra $A$ of $K$ and every $k$-domain $B$ (not related at all to $K$), the tensor product $A\otimes_k B$ is a domain and in particular has connected spectrum.

Here are examples of regular extensions :

a)Every purely transcendantal extension of $k$ is regular.

b)If $k$ is algebraically closed, every extension of $k$ is regular.

PS: Separable extension above means universally reduced and, of course, does not imply that the extension is algebraic.


Let A and B finitely generated reduced k-algebras. Then A and B are coordinate rings of two affine closed sets X and Y: A=k[X] and B=k[Y]. It is true that A ⊗k B=k[X]⊗kk[Y]=k[X×Y] (See Shafarevich, Basic algebraic geometry, p. 25, example 4), and then A ⊗k B is finitely generated reduced k-algebra too.

  • $\begingroup$ I don't know what an affine closed set is when $k$ is not algebraically closed. $\endgroup$ Dec 22, 2009 at 8:20
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    $\begingroup$ Francisco, this is not true if k is not perfect. Take for k a field of characteristic p and let a be an element in some overfield such that a is not in k but a^p is in k. Then if A=B=k[a]=k(a), the k-algebra A \otimes B has the non-zero element (a \otimes 1 - 1 \otimes a) whose p-th power is zero.Hence the tensor product of A and B is NOT reduced. Shafarevich states explicitly that k is algebraically closed at the beginning of the paragraph you quote. $\endgroup$ Dec 22, 2009 at 9:37

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