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An algebraically closed field of nonzero characteristic has no finite-index subfields.

Let \(F\) be a ordered field, and suppose \(a\in F\) is non-negative (and not a square). Then there is a field ordering on \(F(\sqrt{a})\) making \(F(\sqrt{a})/F\) ordered.

The only algebraic extensions of \(R\) are \(R\) itself and \(R(i)\).

\(\bar{R}=R(i)\).

A real closed field has no nontrivial field automorphisms.

Let \(R\) be a real field with no nontrivial real algebraic extensions. Then \(R\) is real closed.

Let \(F\) be an ordered field. Then \(F\) has a unique field ordering if and only if every non-negative element is a sum of squares.

There is a unique field ordering on \(\mathbb {Q}\).

A real closed field \(F\) is a real field such that

1. for all \(x\) in \(F\), either \(x\) or \(-x\) is a square in \(F\), and

2. every odd-degree polynomial over \(F\) has a root in \(F\).

Let \(F\) be an ordered field. A real closure of \(F\) is a real closed ordered algebraic extension of \(F\).

An algebraically closed field of characteristic 0 has an index-2 real closed subfield.

There is no nontrivial odd-degree finite extension of \(R\).

Let \(F\) be an ordered field with a real closure \(R\), and let \(K/F\) be a finite ordered extension. Then there is a unique order-preserving \(F\)-homomorphism \(K\to R\).

Let \(F\) be an ordered field with a real closure \(R\), and let \(K/F\) be a finite ordered extension. Then there is an \(F\)-homomorphism \(K\to R\).

The field \(R(i)\) is the unique quadratic extension of \(R\).

Let \(F\) be an ordered field, and let \(K/F\) be a field extension. Suppose that there is an \(F\)-linear functional \(\pi :K\to F\) such that, for all \(x\in K\), \(\pi (x^2)\geq 0\). Then there is a field ordering on \(K\) making \(K/F\) ordered.

Let \(F\) be an ordered field, and let \(K/F\) be an odd-degree extension. Then there is a field ordering on \(K\) making \(K/F\) ordered.

Suppose \(R\) is a field with unique nontrivial finite extension \(R(i)\). Then \(R\) is real closed.

Suppose \(R\) is a field with nontrivial algebraic closure \(R(i)\). Then \(R\) is real closed.

The monic irreducible polynomials over \(R[X]\) have form \(X-c\) for some \(c\in R\) or \((X-a)^2+b^2\) for some \(a,b\in R\) with \(b\neq 0\).

Polynomials over \(R\) satisfy the intermediate value property (with respect to the unique order).

Let \(F\) be an ordered field with real closure \(R\), and let \(K/F\) be algebraic. Then field orderings on \(K\) making \(K/F\) ordered correspond to \(F\)-homomorphisms \(K\to R\) via the order obtained by restriction from \(R\).

Let \(F\) be an ordered field, and let \(a\in F\). Then there is a unique ordering on the function field \(F(X)\) making \(F(X)/F\) ordered such that \(X{\gt}a\) but \(b{\gt}X\) for \(b{\gt}a\), and a unique one such that \(X{\lt}a\) but \(b{\lt}X\) for \(b{\lt}a\).

\(R\) has no nontrivial real algebraic extensions.

\(R\) has no nontrivial ordered algebraic extensions (with respect to the unique order).

An ordered field is real closed if

1. every non-negative element is a square, and

2. every odd-degree polynomial over \(F\) has a root in \(F\).

\(R\) has a unique ordering making it an ordered field. In this ordering, the non-negative elements are exactly the squares.

Every sum of squares in \(R\) is a square in \(R\).

Let \(F\) be an ordered field. Then \(F\) has a real closure.

There is no quadratic extension of \(R(i)\).

Let \(F\) be a field in which \(-1\) is not a square, and suppose every element of \(F(i)\) is a square. Then sums of squares in \(F\) are squares in \(F\).

Let \(F\) be a real field. There is a unique ordering on \(F\) if and only if, for each \(a\in F\), either \(a\) or \(-a\) is a sum of squares. In this ordering, the non-negative elements are precisely the squares in \(F\).

Let \(R\) be a field, and suppose \(\bar{R}\) is a finite extension of \(R\). Then \(R\) is real closed.

Let \(R\) be a field with \(\operatorname{char}K\neq 2\), and suppose \([\bar{R}:R]=2\). Then \(R\) is real closed.

Let \(F\) be an ordered field, and let \(K/F\) be a field extension. Then there is an ordering on \(K\) making the extension \(K/F\) ordered if and only if \(\sum _i x_i\alpha _i^2\neq -1\) for all choices of \(\alpha _i\in K\) and \(x_i\in F_{\geq 0}\).

The only finite extensions of \(R\) are \(R\) itself and \(R(i)\).

Let \(R\) be an ordered field whose polynomials satisfy the intermediate value property. Then \(R\) is real closed.

A field can be ordered if and only if it is real (that is, \(-1\) is not a sum of squares).

Let \(R\) be an ordered field with no nontrivial ordered algebraic extensions. Then \(R\) is real closed.

The finite-index subfields of \(\bar{\mathbb {Q}}\) are isomorphic copies of \(\mathbb {Q}_{\text{alg}}=\bar{\mathbb {Q}}\cap \mathbb {R}\) indexed by \(\operatorname{Gal}(\mathbb {Q}_{\text{alg}}/\mathbb {Q}(i))\).

Let \(R\) be a field. TFAE:

1. \(R\) is real closed.

2. \(\bar{R}=R(i)\) (and \(R(i)\neq R\)).

3. \(R\) is real, but has no nontrivial real algebraic extensions.

Let \(R\) be an ordered field. TFAE:

1. \(R\) is real closed.

2. \(R\) is maximal with respect to ordered algebraic extensions.

3. Polynomials over \(R\) satisfy the intermediate value property.

Let \(F\) be an ordered field. Then the real closure of \(F\) is unique up to unique \(F\)-isomorphism.

Let \(F\) be an ordered field, and let \(f\) be a polynomial over \(F\). Then \(f\) has the same number of roots in any real closure of \(F\).