I want to prove that in the theory $T$ of ordered fields, multiplication is no definible by addition, i.e, there´s no formula $\phi(x,y,z)$ in $\lbrace 0,1,+ \rbrace$ such that $T\vDash\phi(x,y,z)\Leftrightarrow x\cdot y=z.$
I want to use quantifier elimination or model completness but I don´t arrive to anything, if some one can help me I'll apreciate a lot.
Thanks.
The simplest way to do this is to use automorphisms: if there's an automorphism of $(\mathbb{R};0,1,+)$ which isn't an automorphism of $(\mathbb{R}; 0,1,+,\times)$, then $\times$ isn't definable in $(\mathbb{R}; 0,1,+)$.
Now there's a useful fact here: the structure $(\mathbb{R};0,1,+,\times)$ has no nontrivial automorphisms at all (if you haven't seen this before, it's a good exercise). So, you'll be done if you can find a single nontrivial automorphism of $(\mathbb{R};0,1,+)$. Do you see how to do this?
HINT: Think of $\mathbb{R}$ as a $\mathbb{Q}$-vector space ...