Defining the existence of a non algebraic element in the language $L:= \{0,1,+,\cdot\}$

Question

I raise following question after reading this post.

Is it possible in the language $L:= \{0,1,+,\cdot\}$ to write sentences for which a model will necessarily contain a copy of $\mathbb Q$ and a non algebraic element?

If yes, what is this set of sentences? If not can you provide a proof?

Thanks.

I added the word necessarily to precise the question.

Answer 1

If I understand you correctly, you're asking for an $L$-theory $T$, presumably extending the theory of fields, such that every model of $T$ is of characteristic $0$ and hence contains $\mathbb{Q}$ (this is easy) and contains a transcendental element (this is harder).

There are such theories, but analyzing them requires a lot of algebraic work. As a concrete example, let $R$ be the real closure of $\mathbb{Q}$ (its elements are the real numbers which are algebraic over $\mathbb{Q}$). Now consider $R(t)$, the field obtained by adjoining one transcendental $t$. It's a fact that $R$ is definable in $R(t)$. That is, there is a first-order formula $\varphi_R(x)$ in the language of fields such that for all $a\in R(t)$, $R(t)\models\varphi_R(a)$ if and only if $a\in R$.

I can even write down the formula explicitly: $\exists y\, 1+x^4 = y^2$. For a proof that this works, see Proposition 3.3 in this book, noting that every real closed field is Pythagorean (every sum of two squares is non-negative, hence is a square) of characteristic $0$.

Now let $T = \text{Th}(R(t))$. $T$ expresses the following:

1. The set of all elements satisfying $\varphi_R$ is a subfield which is real closed.
2. There is some element not satisfying $\varphi_R$.
3. For every polynomial $p(x)$ of degree $d$ with coefficients satisfying $\varphi_R$, if $a$ does not satisfy $\varphi_R$, then $a$ is not a root of $p$ (this is because $R$ is relatively algebraically closed in $R(t)$).

So any model $K\models T$ has a subfield $\varphi_R(K)$ which is real closed, and hence contains $\mathbb{Q}$, but $K$ also contains elements which are not algebraic over $\varphi_R(K)$, and hence are not algebraic over $\mathbb{Q}$.

Answer 2

Here is another example, similar to what Alex gave.

Let $T$ be the theory of fields asserting that every nonconstant polynomial with coefficients in $\mathbb{Z}$ has a root, that the charactersitic is zero, and that not every element has a square root. This theory is finitely satisfiable: simply pick a finite field of large characteristic containing the roots of any given collection of polynomials with coefficients in $\mathbb{Z}$, making sure that the characteristic is also larger than the absolute values of these coefficients.

Every model of this theory will contain $\mathbb{Q}$, since it will be a field of characteristic zero. It will also contain the algebraic clousre of $\mathbb{Q}$, since every polynomial with integer coefficients has a root. However, the field must be bigger than that, since not every element has a square root.