Let $M$ denote the additive group of $ \mathbb{C} $. Why is $ \mathbb{C}$, as a ring, not isomorphic to $\mathrm{End}(M)$, where addition is defined pointwise, and multiplication as endomorphisms composition?


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    $\begingroup$ End(M) is huge. Really huge. Try to find a large class of elements it contains. $\endgroup$ Feb 6, 2011 at 21:07
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    $\begingroup$ Why would you expect it to be isomorphic? $\endgroup$ Feb 6, 2011 at 21:38
  • $\begingroup$ @Mariano - Well, $\mathbb{Z}$ is isomorphic to $End(\mathbb{Z},+)$, and a similar result holds for $\mathbb{Q}$. $\endgroup$
    – Pandora
    Feb 6, 2011 at 22:11
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    $\begingroup$ @Mariano: I see good sense in Pandora's expectation. Namely, for any commutative ring $R$, $R$ is isomorphic to the ring of $R$-module endomorphisms of $R$. When $R = \mathbb{Z}$, an $R$-module endomorphism is the same as an abelian group endomorphism, by definition. The same happens to hold for $\mathbb{Q}$. For most other rings it doesn't hold, but it's a good question... $\endgroup$ Feb 6, 2011 at 23:42
  • $\begingroup$ @Pete @Pandora: I did not say that it is a bad question, nor do I think so! It is a great working hypothesis to try on examples---and, as Pete says, for most rings it won't work. $\endgroup$ Feb 7, 2011 at 2:40

2 Answers 2


Method 1: $\operatorname{End}(M)$ is not an integral domain.

Method 2: Count idempotents. There are only 2 idempotent elements of $\mathbb{C}$, but lots in $\operatorname{End}(M)$, including for example real and imaginary projections along with the identity and zero maps.

Method 3: Cardinality. Assuming a Hamel basis for $\mathbb{C}$ as a $\mathbb{Q}$ vector space, there are $2^\mathfrak{c}$ $\mathbb{Q}$ vector vector space endomorphisms of $\mathbb{C}$, but only $\mathfrak{c}$ elements of $\mathbb{C}$.

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    $\begingroup$ Method 0: $\mathrm{End}(M)$ is not commutative. $\endgroup$
    – user26857
    Apr 19, 2016 at 20:41

The endomorphism ring of $M$ is its endomorphism group as a vector space over $\mathbb{Q}$ (since any additive function on $\mathbb{C}$ is in fact $\mathbb{Q}$-linear, and any $\mathbb{Q}$-linear map is necessarily additive).

How many endomorphisms does a vector space over $\mathbb{Q}$ of dimension $\kappa\geq\aleph_0$ have? Turns out it has $2^{\kappa}$ endomorphisms. Any endomorphism is completely determined by the image of a basis. Fix a basis $\beta$ for $\mathbf{V}$ over $\mathbb{Q}$, which is of cardinality $\kappa$.

Each endomorphism corresponds to a function $\beta\to\mathop{\oplus}\limits_{b\in\beta}\mathbb{Q} = \mathbb{Q}^{(\beta)}$. So the cardinality of the endomorphism ring is the cardinality of $(\mathbb{Q}^{(\beta)})^{\beta}$.

The cardinality of $\mathbb{Q}^{(\beta)}$ is $|\beta|=\kappa$, so the entire thing has cardinality $$|\mathrm{end}(M)| = \left|\mathbb{Q}^{(\beta)}\right|^{|\beta|} = \kappa^{\kappa} = 2^{\kappa}.$$ (For a proof that if $\kappa$ is infinite and $2\leq\lambda\leq 2^{\kappa}$, then $\lambda^{\kappa} = 2^{\kappa}{}{}$, see this previous answer).

In the case of $M$ and $\mathbb{C}$, since $M$ is of dimension $\mathfrak{c}=2^{\aleph_0}$ as a $\mathbb{Q}$-vector space, you have that $|\mathrm{End}(M)| = 2^{\mathfrak{c}}\gt \mathfrak{c}=|\mathbb{C}|$. So there aren't enough elements in $\mathbb{C}$ to give you all the endomorphisms.

(The elements of $\mathbb{C}$ correspond to the $\mathbb{C}$-linear maps, of course, by simple multiplication; but here you want a lot more maps).


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