Direct sum - exists?

Question

I have a simple question: does the direct sum exist in the category of 1-dimensional vector spaces?

This is my argument: In the category of 1-dimensional vector spaces, the direct sum of two vector spaces does not exist.

In this category, every vector space is isomorphic to $\mathbb{R}$ or $\mathbb{C}$, and any two non-isomorphic vector spaces are not related by a direct sum. If we try to define the direct sum of two 1-dimensional vector spaces, the resulting space would either be isomorphic to $\mathbb{R}^2$ or $\mathbb{C}^2$, which are not 1-dimensional vector spaces.

Answer 1

I assume the following setup: we have some field $\mathbb K$ and consider the full subcategory ${\cal C}$ of $Vect_{\mathbb K}$ that consists of one-dimensional vector spaces.

This category does not have coproducts.

(1) every morphism in ${\cal C}$ is either a zero map or an isomorphism.

(2) assume that two objects $V$ and $W$ have a coproduct in ${\cal C}$. This means that there is a coproduct diagram

$$ V \stackrel{i}{\longrightarrow} X \stackrel{j}{\longleftarrow} W $$

such that for any pair of morphism $f\colon V\to Y$ and $g\colon W\to Y$ there is a unique morphism $h=(f\mid g)$ with $f = h\circ i$ and $g=h\circ j$.

(3) Take $Y=V$, $f=id_V$ and $g=0$. This forces $i\neq 0$, $(id_V\mid 0)$ isomorphism and hence $j=0$.

(4) Take $Y=W$, $f=0$ and $g=id_W$. This forces $j\neq 0$, $(0\mid id_W)$ isomorphism and hence $i=0$.

Now, (3) and (4) contradict each other.

Answer 2

Your argument is incomplete.

Implicitly, you're relying on the fact that the category of vector spaces is enriched in commutative monoids: each set of morphisms between two objects has the structure of a commutative monoid, and composition of morphisms is a homomorphism of such in each variable.

In this situation, the direct sum of two objects $A_1$ and $A_2$ is a biproduct: an object $B$ equipped with pairs of morphisms $i_j\colon A_j\to B$ and $\pi_j\colon A_j\to B$ satisfying the equations

1. $\pi_j\circ i_k=1_{A_j}=1_{A_k}$ if $j=k$ and $0\colon A_k\to A_j$ otherwise.
2. $i_1\circ\pi_1+i_2\circ\pi_2=1_B$.

Indeed, given a pair of coproduct coprojections $i_j\colon A_j\to B$, the existence part of the universal property of the coproduct gives unique morphisms $\pi_j\colon B\to A_j$ satisfying the first set of equations. But the first set of equations implies $(i_1\circ\pi_1+i_2\circ\pi_2)\circ i_j=i_j$, whence the uniqueness part of the universal property of the coproduct implies the second equation. Conversely, $(f_1\circ\pi_1+f_2\circ\pi_2)\circ i_j=f_j$, and $f\circ i_1\circ\pi_1+f\circ i_2\circ\pi_2=f\circ 1_B=f$ show that $i_j\colon A_j\to B$ are coproduct coprojections.

In particular, since the biproduct is given by equations alone, the direct sum is absolute, so the structure of a direct sum in a subcategory is also a structure of a direct sum in the category. Thus, if the subcategory of $1$-dimensional subspaces had direct sums, then the direct sum of $1$-dimensional spaces would be $1$-dimensional in the category of all vector spaces.

If the direct sum of $1$-dimensional spaces were $1$-dimensional, then we can take $A_1=A_2=B$ (since all $1$-dimensional spaces are isomorphic) and consider the above equations as ones involving endomorphisms of the same object. In particular, such an object is its own biproduct if and only if we have

1. $\pi_ji_k=1$ if $j=k$ and $0$ otherwise
2. $i_1\pi_1+i_2\pi_2=1$.

for elements $\pi_j,i_j$ of the semi-ring $R$ of endomorphisms of the object.

Note that for the object being a $1$-dimensional vector space over a field, or more generally a free module of rank $1$ over a semi-ring, then $R$ is that field or semi-ring.

We see that such equations cannot hold for a non-trivial commutative ring $R$, as then $1=i_1\circ\pi_1+i_2\circ\pi_2=\pi_1\circ i_1+\pi_2\circ i_2=2$. They also cannot hold for a non-trivial semi-ring in which every non-zero element has a right inverse as then $\pi_1i_2=0=\pi_2i_1$ implies $\pi_1=\pi_2=0$ and hence $0=1$.

In other words, free modules of rank $1$ over commutative rings and semi-rings in which every non-zero element has an inverse are never also of rank $2$.

More generally, finite ranks of free modules are well-defined if there are no matrices $A$ and $B$ satisfying $AB=1_n$ and $BA=1_m$ where $1_n$ and $1_m$ are identity matrices of different sizes. The above equations have $A=\begin{pmatrix}\pi_1\\\pi_2\end{pmatrix}$ and $B=\begin{pmatrix}i_1&i_2\end{pmatrix}$. Since the condition is equational, it follows that a semi-ring has well-defined finite ranks of free modules if it admits a homomorphism into such a non-trivial semi-ring.

Answer 3

The following deserves no votes. It is nothing more than a "concrete" version of @MarcOlschok 's answer.

In the category of one-dimensional vector spaces over a fixed field $K,$

• every object is isomorphic to $K;$
• every morphism $f$ from $K$ to itself is determined by the scalar $f(1).$

Therefore, proving that it has no coproduct is equivalent to proving that there is no $(a,b)\in K^2$ (determining two morphisms from $K$ to $K$) such that $$\forall(c,d)\in K^2\quad\exists!e\in K\quad c=ea\text{ and }d=eb.$$ If such a pair $(a,b)$ existed, for $(c,d)=(1,0)$ we should have $ea=1$ hence $b=ad=0.$ But for $d=1$ we should have $eb=1$ hence $b\ne0,$ a contradiction.