Algebraic subset or not ?
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Algebraic subset or not ?
Is the set \(\displaystyle{\left(0,1\right)}\) an algebraic subset of \(\displaystyle{\mathbb{A}^{1}_{\mathbb{R}}}\) ?
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- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Re: Algebraic subset or not ?
The answer is negative.
Proof
Suppose that the set \(\displaystyle{\left(0,1\right)}\) is an algebraic subset of \(\displaystyle{\mathbb{A}_{\mathbb{R}}^{1}}\) .
Then, \(\displaystyle{\left(0,1\right)=V(f(x))}\), where \(\displaystyle{f(x)\in\mathbb{R}[x]}\).
But, according to the Fundamental Theorem of Algebra, \(\displaystyle{V(f(x))}\) is finite, a contradiction,
since, \(\displaystyle{\left(0,1\right)}\) is infinite.
Proof
Suppose that the set \(\displaystyle{\left(0,1\right)}\) is an algebraic subset of \(\displaystyle{\mathbb{A}_{\mathbb{R}}^{1}}\) .
Then, \(\displaystyle{\left(0,1\right)=V(f(x))}\), where \(\displaystyle{f(x)\in\mathbb{R}[x]}\).
But, according to the Fundamental Theorem of Algebra, \(\displaystyle{V(f(x))}\) is finite, a contradiction,
since, \(\displaystyle{\left(0,1\right)}\) is infinite.
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