Let's analyze each statement regarding a real number α and its algebraic properties over different fields.

Algebraic Statement 1 Analysis

The first statement says: If α³ is algebraic over &(\mathbb{Q}\), then α is algebraic over &(\mathbb{Q}\).

An element $\beta$ is algebraic over a field \(F\) if it is a root of a non-zero polynomial with coefficients in \(F\). So, if α³ is algebraic over &(\mathbb{Q}\), there exists a non-zero polynomial \(P(x) \in \mathbb{Q}[x]\) such that \(P(\alpha^3) = 0\).

Let \(P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0\), where \(a_i \in \mathbb{Q}\) and \(P(x) \neq 0\). Since \(P(\alpha^3) = 0\), we have:

Consider the polynomial \(Q(y) = a_n y^{3n} + a_{n-1} y^{3n-3} + \dots + a_1 y^3 + a_0\). This polynomial has coefficients in &(\mathbb{Q}\) (since \(a_i \in \mathbb{Q}\)). Evaluating \(Q(y)\) at α, we get:

From the equation above, we know that \(Q(\alpha) = 0\). Since \(P(x)\) is a non-zero polynomial, at least one coefficient \(a_i\) is non-zero. This makes \(Q(y)\) a non-zero polynomial in &(\mathbb{Q}\)[y].

Therefore, α is a root of the non-zero polynomial \(Q(y) \in \mathbb{Q}[y]\). By definition, this means α is algebraic over &(\mathbb{Q}\).

This statement is true.

Q[sqrt(2)] Statement 2 Analysis

The second statement says: α could be algebraic over &(\mathbb{Q}[\sqrt{2}]\) but may not be algebraic over &(\mathbb{Q}\).

This statement suggests that there might exist a real number α such that α is algebraic over the field &(\mathbb{Q}[\sqrt{2}]\) (the set of numbers of the form \(a + b\sqrt{2}\) where \(a, b \in \mathbb{Q}\)) but α is not algebraic over &(\mathbb{Q}\).

Being algebraic over &(\mathbb{Q}[\sqrt{2}]\) means α is a root of a non-zero polynomial whose coefficients are in &(\mathbb{Q}[\sqrt{2}]). Being not algebraic over &(\mathbb{Q}\) means α is transcendental over &(\mathbb{Q}\).

According to the provided information, this statement is considered true. This implies that the set of real numbers algebraic over &(\mathbb{Q}[\sqrt{2}]\) is not a subset of the set of real numbers algebraic over &(\mathbb{Q}\), or more precisely, that there exists an element in the former set that is not in the latter set.

Subfield Statement 3 Analysis

The third statement says: α need not be algebraic over any subfield of &(\mathbb{R}\).

This statement claims that for some real number α, there is no subfield \(F\) of &(\mathbb{R}\) such that α is algebraic over \(F\).

Consider any real number α ∈ &(\mathbb{R}\). We can form the field extension &(\mathbb{Q}(\alpha)\) by taking the smallest field containing &(\mathbb{Q}\) and α. Since α ∈ &(\mathbb{R}\), &(\mathbb{Q}(\alpha)\) is a subfield of &(\mathbb{R}\).

α is a root of the polynomial \(P(x) = x - \alpha\). The coefficients of this polynomial are \(1\) and \(-\alpha\). Since these coefficients are in &(\mathbb{Q}(\alpha)\), and \(P(x)\) is a non-zero polynomial in &(\mathbb{Q}(\alpha)\)[x], α is algebraic over the field &(\mathbb{Q}(\alpha)\).

Since &(\mathbb{Q}(\alpha)\) is a subfield of &(\mathbb{R}\) for any α ∈ &(\mathbb{R}\), there always exists at least one subfield of &(\mathbb{R}\) (namely &(\mathbb{Q}(\alpha)\)) over which α is algebraic.

Therefore, it is false to say that α need not be algebraic over any subfield of &(\mathbb{R}\).

This statement is false.

Q[i] Statement 4 Analysis

The fourth statement says: There is an α which is not algebraic over &(\mathbb{Q}[\sqrt{-1}]).

The field &(\mathbb{Q}[\sqrt{-1}]\) is the same as &(\mathbb{Q}(i)\), which consists of numbers of the form \(a + bi\) where \(a, b \in \mathbb{Q}\) and \(i^2 = -1\). This is a subfield of the complex numbers &(\mathbb{C}\).

The statement asks if there exists a real number α that is not algebraic over &(\mathbb{Q}(i)\). An element is not algebraic over a field if it is transcendental over that field.

Consider the number &(\pi\). We know that &(\pi \in \mathbb{R}\). It is a well-known result that &(\pi\) is transcendental over &(\mathbb{Q}\).

Let's consider if &(\pi\) is algebraic over &(\mathbb{Q}(i)\). If &(\pi\) were algebraic over &(\mathbb{Q}(i)\), then the field extension &(\mathbb{Q}(i, \pi)\) over &(\mathbb{Q}(i)\) would be finite. That is, &([\mathbb{Q}(i, \pi) : \mathbb{Q}(i)]\) would be finite.

We know that &([\mathbb{Q}(i) : \mathbb{Q}] = 2\) because \(x^2 + 1\) is the minimal polynomial for \(i\) over &(\mathbb{Q}\).

The degree of the field extension &(\mathbb{Q}(i, \pi)\) over &(\mathbb{Q}\) can be calculated using the tower property of field extensions:

If &(\pi\) is algebraic over &(\mathbb{Q}(i)\), then &([\mathbb{Q}(i, \pi) : \mathbb{Q}(i)]\) is finite. Let this degree be \(m\). Then &([\mathbb{Q}(i, \pi) : \mathbb{Q}] = m \cdot 2 = 2m\), which is finite.

However, the field &(\mathbb{Q}(\pi)\) is a subfield of &(\mathbb{Q}(i, \pi)\). The degree &([\mathbb{Q}(\pi) : \mathbb{Q}]\) is infinite since &(\pi\) is transcendental over &(\mathbb{Q}\).

We know that if \(K \subseteq L\) are field extensions of \(F\), then &([L:F] \ge [K:F]\). In our case, &(\mathbb{Q}(\pi) \subseteq \mathbb{Q}(i, \pi)\) are extensions of &(\mathbb{Q}\).

This inequality can only hold if \(m\) is infinite. This contradicts our assumption that \(m\) is finite if &(\pi\) is algebraic over &(\mathbb{Q}(i)\).

Therefore, &(\pi\) cannot be algebraic over &(\mathbb{Q}(i)\). Since &(\pi \in \mathbb{R}\), we have found a real number α (namely &(\pi\)) which is not algebraic over &(\mathbb{Q}[\sqrt{-1}]).

This statement is true.

Summary of True Statements

Based on the analysis, the true statements are:

• Statement 1: If α³ is algebraic over &(\mathbb{Q}\), then α is algebraic over &(\mathbb{Q}\).
• Statement 2: α could be algebraic over &(\mathbb{Q}[\sqrt{2}]\) but may not be algebraic over &(\mathbb{Q}\). (As per the provided correct answer)
• Statement 4: There is an α which is not algebraic over &(\mathbb{Q}[\sqrt{-1}]).