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Let $F_1$ be the standard axiomatic system defining the concept of an algebraic field. $F_1$ has two primitive binary operations: $+$ (addition) and $\cdot$ (multiplication).

Let $F_2$ be another axiomatic system defining the concept of an algebraic field that has, in addition to the two primitive binary operations mentioned above, also the primitive unary operations $-$ (additive inverse) and $^{-1}$ (multiplicative inverse).

Intuitively, I feel that $F_1$ and $F_2$ are equivalent in the sense that every statement that can be proved in $F_1$ can also be proved in $F_2$, and vice versa. However, how can this feeling be formalized?

One obstacle that I've met when trying to formalize the equivalence of $F_1$ and $F_2$ is that since the symbol $^{-1}$ is not used in $F_1$, it can be defined there to be an object other than the multiplicative inverse, and then there will be statements in $F_1$ that won't be provable in $F_2$ or that won't even make sense in $F_2$, as when $^{-1}$ is defined to be a field scalar rather than a function.

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The idea you are looking for is that of conservativity. Let $\mathcal{L}_1$ and $\mathcal{L}_2$ be first order languages with $\mathcal{L}_1 \subseteq \mathcal{L}_2$. Let $T_1$ be an $\mathcal{L}_1$ theory and $T_2$ an $\mathcal{L}_2$ theory. If, for any $\mathcal{L}_1$ statement $\phi$, $T_1 \vdash \phi \iff T_2 \vdash \phi$, then $T_2$ is conservative over $T_1$.

In this context, we can let $\mathcal{L}_1$ contain the symbols $+$ and $\times$ and $T_1$ be the theory of fields in this language. Let $\mathcal{L}_2$ be the language with $+$, $\times$, $-$ and $^{-1}$ and $T_2$ be the theory of fields in this language. It is not hard to prove that $T_1 \subseteq T_2$. It may take more work to show that if $\phi$ is any $\mathcal{L}_1$ statement such that $T_2 \vdash \phi$, then $T_1 \vdash \phi$, but there is a simple model-theoretic proof of this relying on completeness: any model of $T_1$ can be expanded to a model of $T_2$ just by interpreting the $-$ and $^{-1}$ symbols correctly, so if $T_2 \vdash \phi$, then every model $F \models T_1$ will also satisfy $\phi$, and therefore $T_1 \vdash \phi$.

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    Thank you! I'm a little rusty on Logic. Could you please recommend me a good textbook where all the concepts and results that you mentioned in your answer are clearly defined and proved? – 2017-02-02
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    This answer is absolutely correct, but I just want to add two comments. 1. Remember that function symbols (primitive operations) are usually required to be total. So you'll have to pick an ad hoc output for $^{-1}$ on $0$. A natural choice is to take $0^{-1} = 0$. 2. This is a special case of what's called a "definitional expansion". That is, the new symbols in the language $\mathcal{L}_2$ are all definable from the symbols in $\mathcal{L}_1$. – 2017-02-02
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    You can read about "definitional equivalence" of theories, which may be the notion you're looking for, in Section 2.6 of *A Shorter Model Theory* by Hodges. – 2017-02-02
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    @AlexKruckman: Thanks, Alex! – 2017-02-02
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    @AlexKruckman: Is there a connection between definitional equivalence and Skolemization? – 2017-02-02
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    Well, Skolemization is typically *not* a definitional expansion (unless the base theory has what are called "definable Skolem functions" or "definable choice functions"). – 2017-02-02
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    On the other hand, Morleyization *is* a definitional expansion, and there should be a precise way of making sense of the statement that two theories are definitionally equivalent if and only if their Morleyizations are the same up to renaming relation symbols. – 2017-02-02