Re: Algebraic over field

On Sun, 18 Dec 2005, Jannick Asmus wrote:
> On 18.12.2005 12:19, William Elliot wrote:
> > On Sun, 18 Dec 2005, Tanja wrote:
> >
> >>We know that root of 2 in R is algebraic of degree 2 over Q and of
> >>degree 1 over R.
> >>Can vi tell that "3/2" is algebraic over Z (integers)?
> >>I know that (Z,+,*) is a ring, but not a field,...
> >>Can someone explain this to me very-very simple?
> >
> > 3/2 is the root of 2x - 3 in Z[x].
> > So it's algebraic of degree 1 over Z.
>
> What about googling the expression 'algebraic' first? I doubt your

Google sucks ever since it went commercial.

> result, since 3/2 is not integral over Z.

Let F be an extension field of K and u and element of F.
If there's some non-zero polynomial p(x) in K[x], then
u is said to be algebraic over K,

Since the question was about Z, which isn't a field,
the above definition needs be extended to rings to give
meaning to OP's question. For fields, it's moot question
whether p is monic or not.

> > By adjuncting 1/2 to Z, one finds 3/2 in Z[1/2].
> > Of course Z[1/2] = Z[3/2] and it's not a field.
> >
.

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