How many homomorphisms are there from $K$ to $\mathbb{R}$, where $K$ is some finite extension of $\mathbb{Q}$?

I dug around the web and understand that if the range is $\mathbb{C}$, we can use primitive element theorem and argument the number of homomorphism is exactly the degree of the minimal polynomial of the "primitive element".

But now that I "shrink" the space, how does the argument change?

Any help or insight is deeply appreciated.

  • 2
    $\begingroup$ It depends on $K$. If $K$ has degree $n$, then there are $n$ embeddings $K \to \mathbb{C}$, of which $r_1$ of them lie in $\mathbb{R}$ and $2r_2$ of them don't (this number is always even because every complex embedding has a conjugate complex embedding). These numbers show up in various results in number theory. $\endgroup$ – Qiaochu Yuan Sep 8 '17 at 1:56

There's no simple answer. In general, if $\alpha$ is a primitive element of $K$ with minimal polynomial $f$ over $\mathbb{Q}$, then there is a homomorphism $K\to \mathbb{R}$ for each root of $f$ in $\mathbb{R}$. There are at most $\deg f$ roots in $\mathbb{R}$, but there could be less since $\mathbb{R}$ is not algebraically closed. The number of homomorphisms $K\to\mathbb{R}$ is commonly written $r_1$, and $[K:\mathbb{Q}]-r_1$ is commonly written $2r_2$ (this difference is always even, since nonreal roots of $f$ in $\mathbb{C}$ come in complex conjugate pairs).


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