show/hide this revision's text 3 DLS --> David Savitt.

Another way to think of a place of a number field is in terms of equivalence classes of absolute values. You should work out how this definition is the same as the one given by DLSDavid Savitt.(I'm sorry for not being able to refer to him by his real name.)

Let $k$ be any field. An absolute value on $k$ is a homomorphism $|\ |:k^\times\to\mathbb{R}^{\times\circ}$ from the multiplicative group of $k$ into the ordered group of stritctly positive reals which is not trivial and such that the triangular inequality $$ |x+y|\le |x|+|y| $$ is satisfied for all $x,y\in k$, with the convention that $|0|=0$.

An absolute value $|\ |$ gives rise to a distance $d(x,y)=|x-y|$ on $k$, making it into a metric space. It can thus be completed to a field $k_{|\ |}$ in which $k$ is a dense subfield. Two absolute values on $k$ are equivalent if they induce the same topology on $k$, and thus give rise to the same completion.

Ostrowski (1918) determined all absolute values on a number field, and Artin (1932) gave a beautifully simple proof of his theorem. You can read all about it in many places, including my online notes arXiv:0903.2615
show/hide this revision's text 2 added 2 characters in body; added 43 characters in body

Another way to think of a place of a number field is in terms of equivalence classes of absolute values. You should work out how this definition is the same as the one given by DLS. (I'm sorry for not being able to refer to him by his real name.)

Let $k$ be any field. An absolute value on $k$ is a homomorphism $|\ |:k^\times\to\mathbb{R}^{\times\circ}$ from the multiplicative group of $k$ to into the ordered group of stritctly positive reals which is not trivial and such that the triangular inequality $$ |x+y|\le |x|+|y| $$ is satisfied for all $x,y\in k$, with the convention that $|0|=0$.

An absolute value $|\ |$ gives rise to a distance $d(x,y)=|x-y|$ on $k$, making it into a metric space. It can thus be completed to a field $k_{|\ |}$ in which $k$ is a dense subfield. Two absolute values on $k$ are equivalent if they induce the same topology on $k$.k$, and thus give rise to the same completion.

Ostrowski (1918) determined all absolute values on a number field, and Artin (1932) gave a beautifully simple proof of his theorem. You can read all about it in many places, including my online notes arXiv:0903.2615
show/hide this revision's text 1

Another way to think of a place of a number field is in terms of equivalence classes of absolute values. You should work out how this definition is the same as the one given by DLS. (I'm sorry for not being able to refer to him by his real name.)

Let $k$ be any field. An absolute value on $k$ is a homomorphism $|\ |:k^\times\to\mathbb{R}^{\times\circ}$ from the multiplicative group of $k$ to the ordered group of stritctly positive reals which is not trivial and such that the triangular inequality $$ |x+y|\le |x|+|y| $$ is satisfied for all $x,y\in k$, with the convention that $|0|=0$.

An absolute value $|\ |$ gives rise to a distance $d(x,y)=|x-y|$ on $k$, making it into a metric space. It can thus be completed to a field $k_{|\ |}$ in which $k$ is a dense subfield. Two absolute values on $k$ are equivalent if they induce the same topology on $k$.

Ostrowski (1918) determined all absolute values on a number field, and Artin (1932) gave a beautifully simple proof of his theorem. You can read all about it in many places, including my online notes arXiv:0903.2615