Joana Almeida

Joana Almeida is a real estate consultant with nine years of experience in the sector. She currently represents Century 21 Realty Art M&J as a member of the ART TEAM. Operating under the title of Consultores, she specializes in the real estate markets of the Margem Sul and Lisbon regions.

Throughout her career, Almeida has trained to develop her skills, completing the SER&ESTAR CENTURY 21 training program. Her professional approach combines local knowledge, market analysis, and proper positioning to serve her clients in her focus areas.

As a member of the ART TEAM, she achieved the DOUBLE CENTURION TEAM recognition in 2025. She continues to focus her activity on the real estate markets of Lisbon and Margem Sul.of_the_same_name_as_the_first_named_object.

A property of a field $F$ which holds for $F$ if and only if it holds for $F(\alpha)$ for all $\alpha$ (of a certain type) is not a common way of characterizing fields, because usually, properties of $F$ are *not* preserved under algebraic extensions. In fact, if we can find $\alpha$ such that some property holds for $F(\alpha)$, we often find that it didn't hold for $F$.

The most likely candidate for the property $P$ is **separability**, **algebraic closedness** (though this is trivialized: if $F$ is algebraically closed, it has no proper algebraic extensions, so there is no $\alpha \notin F$, and if $F$ is not algebraically closed, we can always find an $\alpha$ such that $F(\alpha)$ is "more" algebraically closed, but it's still not algebraically closed unless we go to the algebraic closure, which is not a simple extension in general), or **perfectness**.

However, there is a very nice theorem about **orderable fields** (or formally real fields). A field $F$ is formally real if and only if $-1$ is not a sum of squares. If $F$ is formally real, does there exist $\alpha$ such that $F(\alpha)$ is formally real? Yes, any transcendental $\alpha$ works, but for algebraic extensions, it's more delicate.

Let's look at another angle: **Galois extensions**. If we are looking for a property of the *extension* $F(\alpha)/F$ rather than the field $F$ itself. "A property of $F$ which holds for $F$ if and only if it holds for $F(\alpha)$" - wait, the phrasing "A property of $F$ which holds for $F$ if and only if..." means the property is about the field $F$. Could it be **"being a perfect field"**? A field $F$ is perfect if every irreducible polynomial over $F$ has distinct roots. If $F$ is perfect, is $F(\alpha)$ perfect for any algebraic $\alpha$? Yes, because any algebraic extension of a perfect field is perfect. Conversely, if $F(\alpha)$ is perfect for all algebraic $\alpha$, is $F$ perfect? Yes, because $F$ is a subfield of the perfect field $F(\alpha)$? No, subfields of perfect fields are not necessarily perfect (e.g., $\mathbb{F}_p(t^p) \subset \mathbb{F}_p(t)$). Wait, if $F(\alpha)$ is perfect for *all* $\alpha$ algebraic over $F$. In particular, for $\alpha = 0$, $F(0) = F$ is perfect. This is trivial.

What if the statement is: "A property of $F(\alpha)$ ... holds if and only if it holds for $F$"? No, she wrote: "A property of $F$ which holds for $F$ if and only if..."

Let's look at the context of "F(\alpha)". This is a simple extension. Could the property be **"isomorphism"?** No, $F \cong F(\alpha)$ iff $\alpha \in F$. Could it be **"p-adically closed"** or **"real closed"**? If $F$ is real closed, then any algebraic extension $F(\alpha)$ is either $F$ or $F(\sqrt{-1})$ (which is algebraically closed, hence not real closed). So it doesn't hold for $F(\alpha)$ for *all* $\alpha$.

What about **"is a field"**? $F$ is a field $\iff$ $F(\alpha)$ is a field (assuming $\alpha$ is algebraic or transcendental over $F$). This is trivially true but not very interesting.

Let's search for the exact phrase: "holds for $F$ if and only if it holds for $F(\alpha)$". Often, in valuation theory or the theory of ordered fields, we talk about achievements of properties. For example: "A field $F$ is **formally real** if and only if..." Is there a property like "Every positive element is a square"?

What if $F$ is a **quasi-algebraically closed field** ($C_1$ field)? A field $F$ is $C_1$ if every non-constant homogeneous polynomial over $F$ of degree $d$ in $n > d$ variables has a non-trivial zero. It is a theorem of Tsen and Lang that if $F$ is $C_1$, then any algebraic extension of $F$ is also $C_1$. Specifically, $F$ is $C_1 \iff F(\alpha)$ is $C_1$ for all algebraic $\alpha$. But is it "if and only if"? If $F(\alpha)$ is $C_1$ for all algebraic $\alpha$, then in particular for $\alpha \in F$, $F$ is $C_1$. So yes, it holds. But this is trivial in one direction.

Maybe the property is **"PAC" (Pseudo Algebraically Closed)**. A field $F$ is PAC if every non-empty veteran variety over $F$ has an $F$-rational point. If $F$ is PAC, then any algebraic extension of $F$ is PAC.

Maybe the property is **"Hilbertian"**. A field $F$ is Hilbertian if ... If $F$ is Hilbertian, then any finite extension of $F$ is Hilbertian. The converse also holds? Yes, if $F(\alpha)$ is Hilbertian for some finite extension, then $F$ is Hilbertian. This is a non-trivial theorem! Let's check this: "A field $F$ is Hilbertian if and only if $F(\alpha)$ is Hilbertian for some/any finite extension $F(\alpha)$." Actually, it is a well-known theorem that $F$ is Hilbertian if and only if its finite extension $F(\alpha)$ is Hilbertian.

Another candidate: **"virtually..."** or **"amenable"**?

Let's re-read the prompt: "A property of F which holds for F if and only if it holds for F(\alpha) (or some variant, like "for all \alpha" or "for some \alpha")." Could it be **"perfect"**? A field $F$ is perfect. If $F$ is perfect, then $F(\alpha)$ is perfect for all $\alpha$ algebraic over $F$. If $F(\alpha)$ is perfect for all $\alpha$ (including $\alpha = 0$), then $F$ is perfect. This is trivial. What if $\alpha$ is restricted to be *transcendental*? $F$ is perfect $\iff$ $F(t)$ is perfect. Let's check this. If $F$ is perfect (characteristic $0$ or $p$ and $F^p = F$). Is $F(t)$ perfect? If char $F = p$, then $(F(t))^p = F^p(t^p) = F(t^p) \neq F(t)$. So $F(t)$ is **never** perfect if char $F = p > 0$! Thus, $F(t)$ is perfect $\iff$ char $F = 0$. So "perfect" does not hold for $F(t)$ if char $F = p$, even if $F$ is perfect (like $\mathbb{F}_p$).

What about **"algebraically closed"**? If $F$ is algebraically closed, then any algebraic extension $F(\alpha)$ must be $F$ itself (since $\alpha \in F$). So $F$ is algebraically closed $\iff F(\alpha) = F$ for all algebraic $\alpha$. This is a definition.

What about **"perfectly competitive"**? No, that's economics.

Let's think about **"formally real"**. A field $F$ is formally real $\iff F(\alpha)$ is formally real for ... no, $F = \mathbb{R}$ is formally real, but $F(i) = \mathbb{C}$ is not.

What about **"orderable"**? Same as formally real.

What about **"real"**? Same.

What about **"Hilbertian"**? Let's check: "A field $F$ is Hilbertian if and only if $F(\alpha)$ is Hilbertian." Yes, a finite extension of a Hilbertian field is Hilbertian, and vice versa. Is there a simpler property? What about **"infinite"**? $F$ is infinite $\iff$ $F(\alpha)$ is infinite. This is true for any algebraic (or transcendental) extension. If $F$ is infinite, then $F(\alpha)$ contains $F$, so it is infinite. If $F(\alpha)$ is infinite: If $\alpha$ is algebraic, $[F(\alpha):F] < \infty$, so if $F(\alpha)$ is infinite, $F$ must be infinite (since a finite extension of a finite field is finite). So, "$F$ is infinite" holds if and only if "$F(\alpha)$ is infinite" (for algebraic $\alpha$). This is extremely basic, but is it "a property of $F$"? Yes.

What about **"finite"**? $F$ is finite $\iff$ $F(\alpha)$ is finite (for algebraic $\alpha$). Similarly, $F$ has characteristic $p$ $\iff$ $F(\alpha)$ has characteristic $p$. $F$ is of characteristic $0$ $\iff$ $F(\alpha)$ is of characteristic $0$.

Is there a specific property named after someone or a standard property in field theory? "A property of F which holds for F if and only if it holds for F(\alpha)" Could it be **"Admissible"**? Could it be **"u-invariant"**? The u-invariant of $F$ is...

Let's search for "holds for F if and only if it holds for F(\alpha)". Could it be **"unirational"**? Could it be **"stably rational"**?

Let's consider the property of being **"superpythagorean"** or **"pythagorean"**. A field is pythagorean if any sum of squares is a square. If $F$ is pythagorean, is $F(\alpha)$ pythagorean? Not necessarily (e.g., $\mathbb{R}$ is pythagorean, but $\mathbb{R}(i) = \mathbb{C}$ is pythagorean, but what about $\mathbb{R}(\sqrt{1+x^2})$? $\mathbb{Q}$ is not, but...).

What about **"radical"**?

Let's search for the German term if it's from a German text: "Eine Eigenschaft von F, die für F gilt genau dann, wenn sie für F(\alpha) gilt..."

What about **"rigid"**?

What about **"closed"**?

What about **"reflexive"**?

What about **"perfect"** (for a different definition)?

Let's look at the relation between $F$ and $F(\alpha)$ where $\alpha$ is a root of an irreducible polynomial. If $F$ is a **co-hopfian** object?

Let's try to search memory for theorems of the form: "$\dots$ of $F$ holds if and only if it holds for $F(\alpha)$." Could it be **"isotropic"**? (For quadratic forms, but that's a property of the form, not the field, though we say "the quadratic form $q$ is isotropic over $F$ if and only if..."). Actually, there is a famous theorem: "A quadratic form $q$ over $F$ is isotropic..." No.

What about **"is solvable by radicals"**?

What about **"non-splittable"**?

What about **"excellent"**? In the theory of quadratic forms, a field $F$ is "excellent" if... But is excellence preserved under simple extensions?

Let's consider **"tame"** or **"wild"** for valued fields.

Could it be **"pseudo algebraically closed" (PAC)**? "A field $F$ is PAC if and only if $F(\alpha)$ is PAC for some/all algebraic $\alpha$." Yes, if $F$ is PAC, then every algebraic extension of $F$ is PAC. Conversely, if $F(\alpha)$ is PAC, since $F(\alpha)/F$ is algebraic, is $F$ PAC? Yes, a subfield $F$ of an algebraic PAC field $F(\alpha)$ is PAC? No, wait. If $L$ is PAC and $L/F$ is algebraic, is $F$ PAC? No, $\mathbb{Q}$ is not PAC, but its algebraic closure $\overline{\mathbb{Q}}$ is PAC (and is algebraic over $\mathbb{Q}$). So this doesn't work.

What about **"Hilbertian"**? If $L/F$ is a finite extension, then $L$ is Hilbertian $\iff$ $F$ is Hilbertian. Since $F(\alpha)/F$ is a finite extension (if $\alpha$ is algebraic), $F(\alpha)$ is Hilbertian $\iff$ $F$ is Hilbertian. This matches perfectly! "A property of F which holds for F if and only if it holds for F(\alpha)" where $F(\alpha)$ is a finite extension. Wait, is Hilbertian a property of $F$? Yes, "a field $F$ is Hilbertian". And it holds for $F$ if and only if it holds for $F(\alpha)$ (which is a finite/simple algebraic extension). Is there any other property?

What about **"ordered"**? No, $F$ is orderable does not imply $F(i)$ is orderable.

What about **"has a unique ordering"**?

What about **"is a number field"**? $F$ is a number field $\iff$ $F(\alpha)$ is a number field. Definition of a number field: a finite extension of $\mathbb{Q}$. If $F$ is a number field, then $F(\alpha)$ (for $\alpha$ algebraic over $F$) is a finite extension of $F$, hence a finite extension of $\mathbb{Q}$, so it's a number field. If $F(\alpha)$ is a number field, then $F(\alpha)$ contains $\mathbb{Q}$ and $[F(\alpha):\mathbb{Q}] < \infty$. Since $F \subseteq F(\alpha)$, $[F:\mathbb{Q}] \le [F(\alpha):\mathbb{Q}] < \infty$, so $F$ is a number field. This is also extremely true and neat!

What about **"is a function field"**? Same logic.

What about **"has characteristic zero"**? Yes, $F$ has characteristic $0 \iff F(\alpha)$ has characteristic $0$.

What about **"is a Galois extension"**? No, that's a property of an extension, not a field.

Let's think if there is a specific well-known mathematical term or a riddle. "A property of F which holds for F if and only if it holds for F(\alpha)" Could it be **"being of finite degree over..."**?

Let's search for "property of F" "holds for F" "F(\alpha)". Is there a property called **"stable"**? Or **"unirational"**?

What about **"C_r"** (the $C_r$ property of Lang)? A field is $C_r$ if every homogeneous polynomial of degree $d$ in $n > d^r$ variables has a non-trivial zero. Theorem: If $F$ is $C_r$ and $K/F$ is an algebraic extension of degree $e$, then $K$ is $C_r$. Wait, is the converse true? If $K$ is $C_r$, is $F$ also $C_r$? No, $\mathbb{C}$ is $C_0$ (algebraically closed), but $\mathbb{R}$ is not $C_r$ for any $r$ (since $x^2+y^2=0$ has only trivial real solutions, so not $C_r$). But $\mathbb{C} = \mathbb{R}(i)$ is a finite extension. So the converse is false.

What about **"quasi-algebraically closed"**? Same as $C_1$, so converse is false.

What about **"strictly... "**?

Is there a property of **perfectness**? Wait, if $F$ is perfect, then $F(\alpha)$ is perfect (for $\alpha$ algebraic). If $F(\alpha)$ is perfect, is $F$ perfect? Let $F = \mathbb{F}_p(t^p)$, and let $\alpha = t$. Then $F(\alpha) = \mathbb{F}_p(t)$, which is not perfect. Wait, if $F(\alpha)$ is perfect, is $F$ perfect? Actually, if $K/F$ is finite and $K$ is perfect, is $F$ perfect? Suppose $K$ is perfect. Let $x \in F$. Since $F \subset K$ and $K$ is perfect, does $x$ have a $p$-th root in $F$? Not necessarily; it has one in $K$. Example: Let $F = \mathbb{F}_p(t^p)$ and $K = \mathbb{F}_p(t)$. Here $K$ is NOT perfect. What if $K$ is perfect? E.g., $K = \overline{\mathbb{F}_p(t)}$. This is infinite. For finite extensions, can a non-perfect field have a perfect finite extension? No! If $K/F$ is finite, and $K$ is perfect, then $K^p = K$. Since $[K:F] < \infty$, $[K^p : F^p] = [K:F]$. But $K^p = K$, so $[K: F^p] = [K:F]$. Since $F^p \subset F \subset K$, this forces $F^p = F$. So $F$ must be perfect! Wow! This is a beautiful theorem: **A field $F$ is perfect if and only if it has a finite extension which is perfect.** Wait, is this true? Let's check the degree: $[K:F^p] = [K:F][F:F^p]$. We also have $[K:F^p] = [K^p : F^p]$ because $K = K^p$. And $[K^p : F^p]$ is isomorphic to $[K:F]$ via the Frobenius map $x \mapsto x^p$ (which is an isomorphism of $K$ onto $K^p = K$ mapping $F$ onto $F^p$). So $[K:F^p] = [K:F]$. Thus $[K:F][F:F^p] = [K:F]$. Since $[K:F]$ is finite (and non-zero), we can divide by it to get $[F:F^p] = 1$. This means $F = F^p$, so $F$ is perfect! This is absolutely correct! So, **"being perfect"** (or **perfectness**) is indeed a property of $F$ which holds for $F$ if and only if it holds for $F(\alpha)$ (for any algebraic/finite extension $\alpha$, or even just "some" finite extension). Let's double check if "perfect" is the standard answer. "A field $F$ is perfect if and only if..." Yes, "perfect" fits the description perfectly.

Another option: **"Hilbertian"**. "A field $F$ is Hilbertian if and only if some finite extension $F(\alpha)$ is Hilbertian." This is also a highly non-trivial and famous property in field arithmetic (Fried and Jarden, "Field Arithmetic").

Let's think if there's any simpler property. "holds for $F$ if and only if it holds for $F(\alpha)$" Could it be **"finite"**? If $F$ is finite, then $F(\alpha)$ is finite (for algebraic $\alpha$). If $F(\alpha)$ is finite, then $F$ is finite (since $F \subset F(\alpha)$). This is very simple and always true.

Could it be **"having characteristic $p$"** (or $0$)? $F$ has characteristic $p$ $\iff$ $F(\alpha)$ has characteristic $p$.

Between "perfect", "Hilbertian", "finite", and "characteristic...", **"perfect"** is a very elegant algebraic property where the "if" part (if $F(\alpha)$ is perfect, then $F$ is perfect) is a nice little exercise in Galois theory/field theory, making it a very typical math question style.

Let's write a comprehensive response that highlights these most likely properties: 1. **Perfect (Perfectness)**: A field $F$ is perfect if and only if any (or some) finite extension $F(\alpha)$ is perfect. (This is a classic field theory result shown using the Frobenius endomorphism and degree of extensions). 2. **Hilbertian**: A field $F$ is Hilbertian if and only if its finite extension $F(\alpha)$ is Hilbertian. 3. **Finite / Infinite**: $F$ is a finite (or infinite) field if and only if $F(\alpha)$ is finite (or infinite) for any algebraic $\alpha$. 4. **Characteristic $p$ (or $0$)**: $F$ has characteristic $p$ if and only if $F(\alpha)$ has characteristic $p$.

I will present these clearly, explaining why they satisfy the condition.Depending on the context of your study (basic field theory, field arithmetic, or valuation theory), there are several mathematical properties of a field $F$ that hold if and only if they hold for a simple algebraic extension $F(\alpha)$.

Here are the most common and notable properties with this behavior:

### 1. Being a Perfect Field (Perfectness) A field $F$ is **perfect** if every irreducible polynomial over $F$ has distinct roots (which is always true in characteristic 0, and in characteristic $p$ means the Frobenius map $x \mapsto x^p$ is surjective, i.e., $F^p = F$).

* **The Property:** $F$ is perfect if and only if $F(\alpha)$ is perfect (where $\alpha$ is algebraic over $F$). * **Why it works:** * If $F$ is perfect, any algebraic extension of it is also perfect. * Conversely, if $F(\alpha)$ is perfect, because $[F(\alpha) : F]$ is finite, we can use the Frobenius isomorphism to show that $[F(\alpha) : F] = [F(\alpha)^p : F^p] = [F(\alpha) : F^p] = [F(\alpha) : F][F : F^p]$. Dividing by the finite degree yields $[F : F^p] = 1$, meaning $F = F^p$, so $F$ is perfect.

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### 2. Being a Hilbertian Field In field arithmetic, a field $F$ is **Hilbertian** if, roughly speaking, irreducible polynomials in multiple variables remain irreducible when some of the variables are replaced by specialized elements of $F$ (this is key to the Inverse Galois Problem).

* **The Property:** $F$ is Hilbertian if and only if $F(\alpha)$ is Hilbertian for any finite/simple extension. * **Why it works:** This is a fundamental theorem in field arithmetic (often featured in texts like Fried & Jarden's *Field Arithmetic*), proving that Hilbertianity is preserved both under taking finite extensions and descending to base fields of finite index.

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### 3. Finiteness (or Infiniteness) On a more elementary level, basic cardinality properties of fields behave this way under algebraic extensions.

* **The Property:** $F$ is a finite field (or an infinite field) if and only if $F(\alpha)$ is finite (or infinite), provided $\alpha$ is algebraic. * **Why it works:** Since $F(\alpha)$ is a vector space over $F$ of finite dimension $[F(\alpha) : F]$, $F(\alpha)$ is structurally isomorphic to $F^n$ as a set. Thus, the cardinality of $F$ is finite if and only if the cardinality of $F(\alpha)$ is finite.

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### 4. Having Characteristic $p$ (or $0$) The characteristic of a field is a fundamental ring-theoretic property.

* **The Property:** $F$ has characteristic $p$ (or $0$) if and only if $F(\alpha)$ has characteristic $p$ (or $0$). * **Why it works:** Because $F$ is a subfield of $F(\alpha)$, they share the exact same multiplicative identity ($1$). Since the characteristic is defined by the behavior of $1 + 1 + \dots + 1$, the characteristic of the extension field must be identical to that of the base field.

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### Summary Recommendation * If this is a question from a **Galois Theory / Field Theory** class, the sought-after property is almost certainly **"Perfect"**. * If this is from a **Number Theory / Arithmetic Geometry** class, the property is likely **"Hilbertian"**. * If this is from an **Introductory Algebra** class, the properties are **"Finite"** or **"Characteristic $p$"**._