When $A, B$ are $K$-modules, then $A \times B$ is the same as $A \oplus B$.

Let $A, B$ be two $K$-algebras, where $K$ is a field. Is $A \times B$ the same as $A \oplus B$?

Thank you very much.

Edit: $A \times B$ is direct product and $A \oplus B$ is direct sum.

Edit: I am asking this question because I am confused with Lemma 1.6 on page 46 of the book Elements of representations of associative algebras. It is said that $\{e_1, \ldots, e_n\}$ is a complete set of primitive orthogonal idempotents. This means that $A=e_1A\oplus \cdots \oplus e_nA$. But it is said that $A$ is not nessesarily connected (connected means $A$ is a product of two non-trivial algebras, the definition of connected is on page 18 of the book, above example 4.1).

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    $\begingroup$ What do these symbols mean? If this is coproduct and product, the answer is most likely no. For commutative $k$-algebras the coproduct is $\otimes_k$ and the normal product is product. $\endgroup$ Commented Mar 29, 2013 at 7:54
  • $\begingroup$ A related fact you may be interested in, the product and coproduct of spaces are not necessarily the same, but if there is some finiteness property, they are. Sorry, don't recall, maybe someone else can expand on this. $\endgroup$ Commented Mar 29, 2013 at 7:55
  • $\begingroup$ See en.wikipedia.org/wiki/Biproduct and possibly qchu.wordpress.com/2012/09/14/… . $\endgroup$ Commented Mar 29, 2013 at 8:08
  • $\begingroup$ @AlexYoucis Since he is asking about finite direct sums of modules, I guess the answer should be yes. $\endgroup$
    – Hui Yu
    Commented Mar 31, 2013 at 14:06

2 Answers 2


To my very regret, old-fashioned terminologies are all over the place in mathematics and still prevent us from using the universal benefit of (the deep idea of) category theory, for example the unification of various scattered notions in mathematics. So let me answer what $A \times B$ and $A \oplus B$ should denote (although most books have not adopted this yet).

In any category, $\times$ denotes the product, and $\oplus$ or $\sqcup$ denotes the coproduct. Even in the category of $K$-modules, they are not the same. Only the underlying objects are the same. A (co)product is more than just an object, it is a a special (co)limit (co)cone. Thus, a product comes equipped with projections, and a coproduct comes equipped with inclusions. Of course there is this usual abuse of notation. But in this case, in my opinion, it is important to realize that (even the underlying objects) $A \times B$ and $A \oplus B$ really have a different flavor, and that it really is a coincidence that they agree. Instead of being surprised that they don't agree for, say, groups, one should be surprised that they agree for abelian groups. Unfortunately, the current literature and courses for beginners do not support this. The only reason why one can write $A \times B = A \oplus B$ safely is that there is a canonical isomorphism between the underlying objects, and in fact there is the notion of a biproduct which fuses product and coproduct in arbitrary linear categories.

Now let $k$ be a commutative ring. For me, rings are always unital unless otherwise specified. The product in the category of $k$-algebras is as usual: We just take the cartesian product of the underlying sets, and then declare addition, scalar multiplication, multiplication and unit pointwise. This works for every algebraic category. So the elements of $A \times B$ are easy to understand and easy to manipulate. The coproduct $A \oplus B$ is more complicated. The underlying $k$-module is some quotient of $k \oplus A \oplus B \oplus A \otimes_k B \oplus B \otimes_k A \oplus A \otimes_k A \otimes_k B \oplus \dotsc$, where we basically identify $x$ with $x \otimes 1$ and $1 \otimes x$. The multiplication is defined in an obvious way. Of course, in the category of commutative $k$-algebras this simplifies a lot, namely we may reduce to $A \otimes_k B$.

Many people still write $A \oplus B$ for what should be denoted by $A \times B$, because the elements of $A \times B$ may be written as $(a,0) + (0,b)$ and there is no harm in identifiying $(a,0)$ with $a$ etc. But meanwhile it should be well-known that morphisms are as important as elements, in the context of universal constructions even far more important than elements, and more importantly it glosses over the fact that $a \mapsto (a,0)$ is not a homomorphism of algebras, since it does not preserve the unit. So this construction usually denoted by $A \oplus B$ actually takes place in the category of non-unital $k$-algebras:

Let more generally $(A_i)_{i \in I}$ be a family of non-unital $k$-algebras. As a $k$-module, their coproduct $\oplus_{i \in I} A_i$ is again a large direct sum of tensor products (for example $A \oplus B = A \oplus B \oplus A \otimes_k B \oplus B \otimes_k A \oplus \dotsc$ with a suitable multiplication). But many people denote by $\oplus_{i \in I} A_i$ the subalgebra $C$ of $\prod_{i \in I} A_i$ consisting of those tuples with finite support (see Frank McGovern's answer). But this is not a coproduct and therefore I find it confusing to denote it as such, contradicting the usual use of this symbol (I will suggest a new symbol below). Nevertheless, it has an interesting universal property. In some sense, it is the universal coproduct with the additional requirement that $A_i \cdot A_j = 0$ for $i \neq j$. More precisely, there are inclusions $A_i \to C$ such that $A_i \otimes_k A_j \to C$ is zero for $i \neq j$, and for every family of homomorphisms of non-unital $k$-algebras $A_i \to B$ such that $A_i \otimes_k A_j \to B$ is zero for $i \neq j$, there is a homomorphism $C \to B$ making the obvious diagrams commute. Sometimes elements $x,y$ of a ring with $xy=0$ are called orthogonal (motivated by the case of projections on some space). Therefore, I would call $C$ the orthogonal coproduct and denote it by $\oplus_{i \in I}^{\perp} A_i$. What do you think?

See here for a quite similar construction for groups.

Notation should be consistent and work for all categories. It only gets confusing when for different categories the notation differs.

Edit: Surprisingly, my point of view also appears on Wikipedia: Direct sum of rings

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    $\begingroup$ I would comment that one inconsistent notation is the one you mention but don't note as inconsistent: $\oplus$ should mean biproduct and $\coprod$ the coproduct. Your last long equation for $A \oplus B$ seems to suffer from an ambiguity that would be resolved by doing this. $\endgroup$
    – Ryan Reich
    Commented Sep 18, 2013 at 15:35
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    $\begingroup$ When $\oplus$ is biproduct, why do people use this for rings (where there is no biproduct) ... $\endgroup$ Commented Nov 4, 2013 at 10:12
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    $\begingroup$ After reading this answer after some years, it really strikes me how upset I was back then about basically everything in math which was not in line with category theory ... $\endgroup$ Commented Jan 24, 2020 at 19:42

Yep! There's only a difference in the direct product and the direct sum in the infinite case.

For example, if $\mathbb N$ is our infinite indexing set and $\{A_n\}_{n\in \mathbb N}$ are $K$-algebras, then $\prod_{n\in \mathbb N}A_n$ is the set of tuples $(x_1,x_2,\ldots,x_i,\ldots)$ such that $x_n\in A_n$. However, we define $\bigoplus_{n\in N}A_n$ to be the set of tuples $(x_1,x_2,\ldots,x_i,\ldots)$ such that $x_n\in A_n$ and all but finitely many $x_n=0$. Thus $\bigoplus_{n\in\mathbb N}A_n\subsetneq \prod_{n\in \mathbb N}A_n$.

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    $\begingroup$ @MartinBrandenburg In my experience, if $A$ and $B$ are rings, and someone writes $A \oplus B$, they mean $A \times B$. It is true that this is not a coproduct. (In the category of unital commutative rings, the coproduct is $A \otimes B$. In the category of not necessarily unital commutative rings, I believe the coproduct is $A \times B \times (A \otimes B)$. I'm not sure about the noncommutative cases.) In any case, I have never seen $\oplus$ in a category of rings used to mean coproduct, even for finite sums. $\endgroup$ Commented Mar 29, 2013 at 13:49
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    $\begingroup$ See my answer for more details. $\endgroup$ Commented Mar 30, 2013 at 11:53

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