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Let $k$ be a field. Assume that you already know that the category $\mathrm{Alg}(k)$ of $k$-algebras (everything here is commutative and unital) has a coproduct $\sqcup$. But you don't know that this actually comes from the tensor product of vector spaces over $k$. You just know the universal property of $A \sqcup B$ (aka $A \otimes_k B$). From this you can deduce:

  • $\sqcup$ is commutative and associative up to natural isomorphisms
  • $(A/I) \sqcup B \cong (A \sqcup B) / \langle i_A(I) \rangle$
  • $A \sqcup k[x_1,\dotsc,x_n] \cong A[x_1,\dotsc,x_n]$
  • $A \sqcup -$ commutes with colimits

In particular, we can compute the tensor product of arbitrary algebras using presentations: $$k[\{x_i\}]/I \sqcup k[\{y_j\}]/J \cong k[\{x_i\},\{y_j\}]/\langle I,J \rangle$$

Question. How can we prove that for every injective homomorphism $\phi : A \to B$ of $k$-algebras the induced homomorphism $\phi \sqcup \mathrm{id} : A \sqcup C \to B \sqcup C$ is also injective for every $k$-algebra $C$?

For example, this is clear when $C$ is a polynomial algebra over $k$. In general, $C$ is free as a module over $k$, but we cannot use the isomorphism $C \cong k^{(I)}$ since this leaves the category of $k$-algebras.

Background: I assist a lecture where the students have just learned what the tensor product of algebras is, without knowing the tensor product of modules. Now they have to believe somehow some of the well-known properties, because they are usually proven with the help of the tensor product of modules. But perhaps we can do it with algebras alone. Since the lecture is about elementary algebraic geometry, you may assume that $k$ is algebraically closed and some basic results about affine varieties (but not about their fiber products ;)).

Appendix: Further properties which follow from the universal property:

1) Let us denote the coproduct inclusions by $i_A : A \to A \sqcup B$ and $i_B : B \to A \sqcup B$. It is easy to see that $\otimes : A \times B \to A \sqcup B, (a,b) \mapsto i_A(a) \cdot i_B(b)$ is $k$-bilinear and that the span of the image generates $A \sqcup B$ (since the image satisfies the same universal property).

2) When $A,B \neq 0$, then we also have $A \otimes_k B \neq 0$. Geometrically: The fiber product of two non-empty schemes is non-empty.

Proof: Since the zero algebra only maps to the zero algebra, we may replace $A,B$ by quotients. In particular, we may assume that $A,B$ are field extensions of $k$. Let $P$ be a transzendence basis of $A/k$ and $Q$ one of $B/k$. Let $M$ be a set containing disjoint copies of $P$ and $Q$ and let $C$ be the algebraic closure of $k(M)$. Then there are maps $A \to C$ and $B \to C$, which induce by the universal property a map $A \sqcup B \to C$. Since $C \neq 0$, we also have $A \sqcup B \neq 0$.

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You probably want \amalg or \sqcup instead of \coprod. –  Zhen Lin May 22 '12 at 7:22
    
The theory of tensor products of vector spaces is somewhat simpler than tensor products of modules. Is it good enough to quickly develop enough of the former to use? –  Hurkyl May 22 '12 at 8:45
    
@Hurkyl: Of course this is possible, but this is not the question. [at]Zhen: Ok I've used sqcup now. –  Martin Brandenburg May 22 '12 at 18:05
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The real question is, how did those students get to know what the tensor product of algebras is before knowing what the tensor product of vector spaces is, and why oh why?! –  Mariano Suárez-Alvarez May 22 '12 at 19:07
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At Qiaochu: Yes, first line in my question. At Mariano: This is not my question. And there is not "the reason". I know what you can read in textbooks. At Hurkyl: When you are not interested, you don't have to care for this question. –  Martin Brandenburg May 22 '12 at 19:32

2 Answers 2

up vote 2 down vote accepted

Here is a geometric argument that will prove something slightly weaker. I assume that $k$ is algebraically closed. I will use letters $A,B, C$ to denote $k$-algebras, and $X,Y,Z$ to denote corresponding algebraic sets.

  • Show that the set of (closed) points of the product is the product of the sets of closed points. (Easy from the universal property of $\otimes_k$.)

  • Show that if $X \to Y$ is surjective, then so is $X \times Z \to Y \times Z$. (Follows from the first point.)

  • Show that if $U \hookrightarrow X$ is an open immersion, say of a distinguished open $D(f)$, then the same is true of $U\times Z \hookrightarrow X \times Z$. (Should be easy enough.)

  • Show that if $A \to B$ is injective, then the map of varieties $X \to Y$ is dominant (i.e. contains a dense open subset of the image), and conversely, provided that $A$ and $B$ are reduced.

  • Now suppose that $X \to Y$ is dominant. We may find $V \hookrightarrow Y$ distinguished open, with preimage $U \hookrightarrow U$ distinguished open, so that $U \to V$ is surjective. Then $U \times Z \to V \times Z$ is surjective. Now it should be easy to deduce that $X \times Z \to Y \times Z$ is dominant.

  • Going back to $k$-algebras, we've shown that $A \to B$ injective implies $A \otimes_k C \to B \otimes_k C$ is injective, up to nilpotents.

Dealing with nilpotents will be trickier.


Added in response to the OP's request for more details on the 2nd last point:

Let $\varphi:X \to Y$ denote the given dominant map. This means that the image of $\varphi$ is dense, and so (e.g. by Chevalley's theorem) contains a dense open subset, and hence a distinguished dense open subset, of $Y$.

Choose a distinguished open $V:= D(f)$ in $Y$ contained in the image of $X$. Let $f' : = f \circ \varphi$ be the pull-back of $f$ to $X$ and let $U = D(f')$ (a distinguished open in $X$).

Now $x \in U$ iff $f'(x) \neq 0$ iff $(f \circ \varphi)(x) \neq 0$ iff $f(\varphi(x)) \neq 0$ iff $\varphi(x) \in V$, and so $U = \varphi^{-1}(V)$. As $V \subset \varphi(X)$, we certainly have that $\varphi(\varphi^{-1}(V)) = V$, and so $\varphi: U \to V$ is surjective, as required.

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Thank you! This is already a big step. Can you explain why in the second last point we can find $V \hookrightarrow Y$ distinguished open which lies in the image of $X \to Y$? –  Martin Brandenburg May 23 '12 at 20:33
    
@Martin: Dear Martin, Choose a distinguished open $U = D(f)$ contained in the image of $X$. If we pull-back $f$ to a function $f'$ on $X$, then $V = D(f')$ should do the trick (if I haven't blundered!). Best wishes, –  Matt E May 23 '12 at 23:30
    
Hm, you just repeat the claim in your first sentence. –  Martin Brandenburg May 24 '12 at 9:31
    
@Martin: Dear Martin, Sorry, this was meant to indicate why the statement is true. Extra argument added to the main body of the answer to explain. Regards, –  Matt E May 24 '12 at 11:27
    
@Martin: Dear Martin, Ah, I just looked at your comment again and realized I misunderstood. I was explaining why, given $V$, we could choose $U$ such that $U \to V$ is surjective (and I added more details on this point, but probably you don't need them; sorry). I realized now that you were asking about why we can find $V$ at all. The answer is that this a form of Chevalley's theorem, saying that the image of a map of varieties is constructible. You can find the theorem, with a sketch of the proof, stated as an exercise somewhere early in Hartshorne Chapter II (maybe section 3 or 4). ... –  Matt E May 24 '12 at 11:41

I come also here to bore you...

ANyway, you know that the achievement is true iff $k$ is a field, but your points are true also for a general commutative rings $k$. THen you need some "Extra" points:

1)You have a functor $U: Alg_k\to Vect_k$ that map any monomorphism is a section,

2) Exist a bifuctor $B: Vect_k\times Vect_k\to Vect_k$ such that $U\circ (A\coprod C)\cong B(U(A), B(C)$ (naturally in $A, C$)

from this you prove your claim.

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