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In his answer to this question: Category of Field has no initial object, Arturo Madigin indicated that the field of rational numbers is the initial object in the category of fields of characteristic $ 0 $.

(There is also an interesting discussion trying to characterize such fields here: Examples of fields of characteristic $ 0 $.)

Does the category of fields of characteristic $ 0 $ have a final object? Somehow it would be great if it were the real numbers, but because of my limited background, I can’t imagine showing either hom existence or uniqueness.

Any ideas?

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  • $\begingroup$ I can't imagine why it would be the reals. There's always $\Bbb C$, for example. $\endgroup$
    – Cameron Buie
    Feb 26, 2013 at 4:56
  • $\begingroup$ Yeah but just because $\Bbb C$ contains $\Bbb R$ shouldn't matter. For example, the final object in sets is the singleton, not two, an infinite set or anything else. What's your rationale? $\endgroup$
    – alancalvitti
    Feb 26, 2013 at 4:58
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    $\begingroup$ No field containing an element squaring to $-1$ admits a morphism to $\mathbb{R}$ (such as $\mathbb{C}$). Neither does any field bigger than $\mathbb{R}$ (such as $\mathbb{R}$ adjoin $|2^{\mathbb{R}}|$ new variables). $\endgroup$
    – Qiaochu Yuan
    Feb 26, 2013 at 4:58
  • $\begingroup$ Qiaochu (in his answer) has hit upon a fundamental difference between morphisms of fields and those of sets (which I ought to have mentioned, honestly), which should tell you why it matters that $\Bbb C$ contains $\Bbb R$. $\endgroup$
    – Cameron Buie
    Feb 26, 2013 at 5:00

2 Answers 2

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No. Any morphism between fields is injective, and there are fields of characteristic $0$ of arbitrarily large cardinality.

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    $\begingroup$ Can you explain why you mentioned arbitrary large cardinality? Afaik, sets have a final object even if there are sets of arbitrarily large cardinality. Also, any good references as to why morphisms are required to be injective? $\endgroup$
    – alancalvitti
    Feb 26, 2013 at 5:04
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    $\begingroup$ @alancalvitti: Any morphism of fields is injective for the following reason: any morphism of unital rings $f:R\to S$ is required to take $1_R$ to $1_S$. Now consider what the kernel can be if $R$ is a field. $\endgroup$
    – Zev Chonoles
    Feb 26, 2013 at 5:08
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    $\begingroup$ @alancalvitti: As to your first question: can any given field $F$ be a final object if there are fields $L$ of larger cardinality than $F$ and any field morphism must be injective? $\endgroup$
    – Zev Chonoles
    Feb 26, 2013 at 5:09
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    $\begingroup$ @alan: if the cardinality remark were the only important step then I wouldn't have mentioned the fact that morphisms of fields are injective. Both ingredients are necessary. If $F$ is a field and $L$ a field larger than $F$, then there aren't any morphisms $L \to F$ (because morphisms of fields are injective, because fields have no nontrivial ideals, because nontrivial elements of nontrivial ideals are not invertible). Hence $F$ cannot be the terminal object. $\endgroup$
    – Qiaochu Yuan
    Feb 26, 2013 at 5:26
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    $\begingroup$ @alancalvitti: For comparison, the category of sets and injections doesn't have a finial object either. $\endgroup$
    – user14972
    Feb 26, 2013 at 5:47
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Given a category where all maps are monomorphisms, suppose that there are two different parallel maps $f,g: A\to B$. Then any given map $h: B\to T$ will produce two different maps $h\circ f, h\circ g: A\to T$. In particular, there cannot be a final object. Also, there cannot be any product $A\leftarrow P \rightarrow B$ because the two maps $(id_A,f),(id_A,g): A \to P$ would need to be equal and different at the same time (*).

For fields of characteristic $0$, take $A=B=\mathbb{C}$ (the complex numbers), the identity map for $f$, and conjugation for $g$.

Actually, that category has pullbacks. So, if you already know that it does not have products, you also know that it does not have a terminal object.

(*) I have seen this argument in some exercise, but I do not remember where.

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