x is actually just, in the language of computer science, syntactic sugar. In reality x+3 is really the infinite tuple
(3, 1, 0, 0, .....)
I think this is incorrect.
Lets continue to assume we are working over Q. Without further context I would take "x+3=-1" to mean that x,3, and -1 are all elements of Q. 3 and -1 being the obvious elements; and x being an a-priori unknown elements which we can easility derive to be 2.
Notably, x+3 is not a polynomial in the technical sense. If we wanted to consider x+3 a polynomial, we would be asking for the t value such that (x+3)[t] = (-1)[t]. Where (-1) is also a polynomial, and (g)[t] is the map Q[x] X Q -> Q given by standard polynomial evaluation.
Sure, this question is equivalent, but I see nothing in the original equation "x+3=-1" to suggest any involvement of formal polynomials.
(3, 1, 0, 0, ....)
A polynomial ring in one variable is an infinite direct sum of the base ring with addition component wise and multiplication defined in a certain way. The expression x+3 meets the definition of a polynomial.
What I am questioning is the necessity to interpret "x+3" as an element of Q[x].
>The expression x+3 meets the definition of a polynomial.
Only if you take x=(0,1,0,...) and adopt the convention that any member q of the base field Q is assumed to represent qx^0 = (q,0,0,...).
That is to say, x+3 only meets the definition of a polynomial because you insist on interpenetrating it as such.
However, we can also handle "x+3=-1" without ever defining the notion of a polynomial.
Eg, we can say, suppose x \in Q such that "x+3=-1". From this premise, we can derive presisly what specific element of Q x must be.
In a more general setting, we might only be able to derive a set of potential values that x could have, or derive that x cannot possibly exist.
As I mentioned in my prior comment, I see no reason to intererperet the "x+3" in "x+3=-1" as a polynomial. If we were to do so, the question would be asking: find t \in Q such that (x+3)[t]=(-1)[t]. Where (g)[t] is polynomial evaluation.
Applying the definition of polynomial evaluation, we would get that the above equation implies: t+3=-1.
Are you now going to insist that "t+3" is a polynomial. Bearing in mind that we have defined t to be an element of Q, which was necessary to apply it as the second argument of polynomial evaluation; and we only got "t+3" as the output of polynomial evaluation, which is defined to result in an element of the base field.
We could modify are notion of polynomial evaltuation to instead be of the form R[x] X R[x] -> R[x], which also gives us (for free) the ability to apply polynomials to other polynomials. But if we were to do this, then when we say that the solution to "x+3=-1", is -4, we are taking "-4" itself to be a polynomial.
In practice this is fine (we identify the base field with the subring of degree 0 polynomials all the time). However, this entire approach breaks down when you start working with functions that do not fit within the framework of polynomial rings.
For instance, suppose I said that "(x+3)! = 120". Are you still going to insist that "x+3" is a polynomial?
What if I define a function id: Q -> Q. In the equation "id(x+3) = 2, are you still going to insist that "x+3" is a polynomial?
