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?

That’s what we mathematicians do. In the context of the original post it is absolutely clear that x+3 is a polynomial. There is no other reasonable interpretation.

When you write things like:

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.

it gives the impression that you don’t know what a polynomial is. The second sentence I quoted is not true. (EDIT: see note below, my interpretation of what was written was wrong.)

Of course if you change context then different interpretations arise. Which of course is the whole point of my original post. Like all spoken languages mathematical language is nuanced. Things must be interpreted in context.

When presented with the equation x+3=-1 x+3 is a polynomial. -1 is a polynomial.

I gather you do not think x^2 - x + 1 = 0 is a polynomial equation. Is x^3+4x a polynomial? Is there any other reasonable interpretation using accepted mathematical conventions? Perhaps you don’t think 2/(x+3) is a member of R(x). What is it a member of then?

Edit:

x+3 is a polynomial that defines a natural map from R to R. To solve the equation x+3=-1 is asking for the pre-image of -1 of this map. This is what it means to solve this equation. It’s solution set is an algebraic variety. I see no other reasonable interpretation. The whole branch of algebraic geometry is about precisely this. Studying zero sets of polynomial equations.

That we teach people rules they can apply to find the answer does not detract that what is really going is as I’ve described and as you did describe with the second quoted text.

I assume you are refering to the sentence:

> If we wanted to consider x+3 a polynomial, we would be asking for the t value such that (x+3)[t] = (-1)[t]

Bearing in mind that the example I have in mind is the equation "x+3=-1" with the solution of "2", in what sense in the above sentence not true?

>When presented with the equation x+3=-1 x+3 is a polynomial. -1 is a polynomial.

Fair enough. In that case, I assume you would consider the equation "x+3=-1" to be false, as it is clear that (3,1,0,0,...) != (-1,0,0,...). Unless of course you are asking, as I had suggested, that you are looking for the particular element at which polynomial evalutation yields an equal result on both sides. If this is the case then, as far as I can tell, you are introducing the machinery of formal polynomials for the sole purposes of overloading the "=" symbol in a confusing way.

>I gather you do not think x^2 - x + 1 = 0 is a polynomial equation.

Define "polynomial equation" If you are asking if I would consider that an equation taking place in R[x], then (absent some other context) the answer is no. Even with other context I would say that "x^2 - x + 1 = 0" is false as a polynomial equation. You might be able to get me to call equations done in the quotient ring R[x]/<x^2-x+1> polynomial equations, in which case "x^2 - x + 1 = 0" would be both a polynomial equation and true at the same time.

If you are asking if I would call "x^2 - x + 1 = 0" in an informal setting, then the answer is yes. However, I do not see how this is relevant, as the whole point of this comment chain was the formal notion of polynomials.

>x^3+4x a polynomial?

Informally, yes. Formally, it depends on context. However, absent some context, I would not consider "x^3 +4x" to be a formal polynomial.

>Is there any other reasonable interpretation using accepted mathematical conventions?

Yes, x^3 +4x is the member of the base ring corresponding to "(x * x * x) + (4 * x)", where x is some other member of the ring.

>Perhaps you don’t think 2/(x+3) is a member of R(x). What is it a member of then?

I am glad you asked. I believe my above answer regarding x^3+4x still applies. However, let me ask you: Is x^3+4x a member of R(x)?

>What is it a member of then?

Again, depending on context. Without context, I would consider 2/(x+3) to be a member of R.

I know of no mathematician who thinks x^2-x+1=0 is anything other than a polynomial equation. Specifically it’s shorthand notation for the variety of the ideal generated by x^2-x+1. And in general you don’t look associate this variety with the quotient ring of the ideal generated by the polynomial. You look at the quotient of the radical of that ideal.

Without any further information the only reasonable interpretation of x+3 is that it is a polynomial. Without any further context in an algebraic equation x is a variable and is not assumed to be an element of the base ring.

In the context of function spaces like C(R) it’s a different matter. And viewing x+3 as an element of C(R) the only reasonable interpretation of x+3=1 is that we are finding the pre-image of 1. And to do this for a complicated function means solving an equation. And solving an equation by hand, the context of my original comment, means reducing the equation to a simpler one. In the case I gave this means reducing x+3=1 to the simpler equation x=-2 whose solution is obtained by inspection. That’s the goal of all the algebraic manipulations we bore beginning algebra students with. Reduce complicated equation to simpler equation. One whose solution is obtained by inspection.

Your view of how to interpret x^3+4x is too simplistic because the only to way to algebraically manipulate that object is by considering it as an R[x] or R(x). You have to view the x as an indeterminate in some larger ring than the base ring.

For this to be the case, there would need to be a statement in the first place. And that statement would involve the =, so you necessarily still have the = symbol representing something other than questioness. I would further say that the equation itself is still just a statement, and any "question" interpretation is based entirely on the context where the equation is presented.

Further, lets take seriously the notion that "x+1" is a polynomial in the formal sense. What does it mean to find the set of values for which x+1=2 is true? Normally, I would say that we are looking for the set of x values which makes that equation true. However, we are insisting that the LHS is a polynomial (again, in the formal sense). This means that there is no variable. The "x" in the LHS is literally a value. It makes no sense to ask what values of (0,1,0,0,...) make that equation true. The fact that we give (0,1,0,...) a standard name of x does not suddenly make the question sensical. Nor does the fact x is often used to represent variables.

>I know of no mathematician who thinks x^2-x+1=0 is anything other than a polynomial equation.

To be clear, outside of very particular contexts I would still call x^2-x+1=0 a polynomial equation, because it is extremly useful to talk about polynomials without invoking all of the machinery of formal polynomials.

>And viewing x+3 as an element of C(R) the only reasonable interpretation of x+3=1 is that we are finding the pre-image of 1.

I disagree. Viewing x+3 as an element of C(R), the only reasonable interpretation of x+3=1 is the statement (x↦x+3)=(x↦1).

Viewing x+3 as an element of R would allow us to treat the equation x+3=1 in the "obvious" way. We can prove that the statement x+3=1 implies that x=-2.

Further, I would agree with you that, absent other context, when given an equation which contains an "x" in it, there is some implication that we are supposed to solve for x.

>Your view of how to interpret x^3+4x is too simplistic because the only to way to algebraically manipulate that object is by considering it as an R[x] or R(x).

Why? Suppose I don't know what R[x] or R(x) is. We certainly don't teach highschoolers what either of those are, and they seem to be able to do "algebra" just fine.

Here is a simple approach to dealing with x^3+4x without considering it a member of R[x]. For concreteness, I want to solve x^3+4x=0.

Suppose x \in R such that x^3 +4x = 0.

By the distributive property, this equation is true iff x(x^2 +4x)=0.

By direct calculation, we can verify that x=0 is consistent with this equation, and therefore consistent with the original equation.

Consider the case where x != 0.

Note that the function f(n) = n/x is a bijection. Therefore, we have x(x^2 + 4)=0 iff f(x(x^2+4)) = f(0).

By direct computation, we get that this is true iff x^2+4 = 0.

We know that x^2 >=0, and 4>0.

Therefore, x^2+4 > 0.

This is a contradiction, which means that the case where x!=0 is impossible.

This means that we have proven that x=0.

Now, suppose we were working over C.

Continuing from x^2 + 4 = 0, we can show that:

x^2+4 = 0 iff

(x+2i)(x-2i) = 0 iff

x+2i = 0 OR x-2i = 0 iff

x=2i OR x=-2i

Since the cases x=0 and x!=0 are exhaustive, we have proven the statement x \in {0, 2i, -2i}.

I solved this using the method we teach school children and without invoking any notion of polynomials.

..x=0 is consistent..

You don't see the problem with this? Saying x=0 and that x is an element base ring means that x is the element 0. You can't later in your problem write "...x=2i..." if you are going to persist in your view that x is an element of the base ring. What you have shown is that the variety of the ideal generated by x^3+4x is the same as the variety of the ideal generated by x, x-2i, x+2i if we are talking about C as the base ring. If the base ring is Q then a different thing is shown.

You can't logically say, in a consistent manner, that x is in R and x^3+4x = 0 and that x is 3 different values. An element of a ring is not three different elements. An element of a ring is itself. If you want to vary the object x then you need to enlarge your ring to an algebraic structure that admits x so that it behaves the way you wish to view it. Your view of what is really happening when solving an equation is not rigorously sound. The proper way to view this is in the context of algebraic geometry.

I never said that x=0. I said that x could be 0. More specifically, I stated that the statement x^3+4x=0 AND x=0 is not inconsistent.

All I have shown in is that (assuming we are working in C), the statement x^3 + x=0 implies that x \in {0, 2i, -2i}.

I suppose you could complain that I have not defined a sense in which {0, 2i, -2i} is correct while {0, 2i, -2i, 7} is incorrect, as it is still a true statement that x^3+4x=0 implies x \in {0, 2i, -2i, 7}.

However, you can easily make this intuition rigourous by saying that the question is to compute the set {x | x^3+4x=0}. Sure, this is invoking machinery not explicitly present in the statement x^3+4x=0. I will even concede that we do not make this machinery explicit when teaching highschool students. However, it is far less machinery than your approach.

I am not claiming that the algebraic approach is not rigourously sound; merely that it is not the only rigourously sound approach.

As far as I can tell, you are claiming that it is the only rigourously sound way of stateing the question.

>You can't logically say, in a consistent manner, that x is in R and x^3+4x = 0 and that x is 3 different values.

I believe I have made this point clear, but I never claimed x is 3 different values. The claim I made was that x is a member of the set {0, 2i, -2i}

This means that we have proven that x=0.

Later you have x=2i or x=-2i. Persisting with the idea that you can view x as an element in the base ring and at the same time allowing its value to change or vary indicates you don’t really understand these issues. If x is in R it is a single value. If you want to vary it you need to expand R to include x and add the approroate algebraic structure.

It’s shocking that you think x^3+4x is not a polynomial. The whole discussion I started originally was that math language, like all other human languages, is nuanced and there are lots of abuse of notations. This is ok because math is written for humans by humans. The standard interpretation of x^3+4x is that it’s a polynomial. This is not disputable.

To be clear, by the time I got to this statement, I had changed the question (from a base of Q to C)

Further, the complete statement I was making at that point was:

If ((x^3+4x=0) AND x != 0) then ((x=2i) OR (x=-2i))

I am not allowing x to vary at all here. Suppose, for the sake of arguement, we had x=2i. It would still be true that ((x=2i) OR (x=-2i)).

> Persisting with the idea that you can view x as an element in the base ring and at the same time allowing its value to change or vary indicates you don’t really understand these issues.

Persisting with the idea that I am doing this indicates that your are not reading what I am writing.

We don’t teach high schoolers what is really going on. We mask what is really going on because making it all precise is not effective or helpful at this stage of development. We teach rules to manipulate equations. We don’t use the language of algebraic geometry because that is too complicated. We don’t say to them the variety of the ideal generated by x+2 is the same as the ideal generated 4x+8 are the same so that x=-2 is equivalent to 3x+1=-x-7.

However, we are insisting that the LHS is a polynomial (again, in the formal sense). This means that there is no variable. The "x" in the LHS is literally a value. It makes no sense to ask what values of (0,1,0,0,...) make that equation true. The fact that we give (0,1,0,...) a standard name of x does not suddenly make the question sensical. Nor does the fact x is often used to represent variables.

You are clearly not an algebraist. The polynomial

(3, 1, 0, 0, 0, ....)

Induces a natural map from R to R that is generally called “evaluation”. It maps the number 5 to 8 for instance. We abuse notation and say to begininnng students, “replace x with 5”. We dumb things down. Instead of asking for the pre-image of 8 under this natural map we ask for what values of x do we get a value of 8. The shorthand way of writing this is to say solve the equation:

x+3=8

That you don’t know this is disconcerting since you’ve obscuoulsy had more than an elementary mathematical education. You are confusing the simplistic view of what is taught in basic courses with what is really going on.

Ask a million mathematicians, “is x^2+3x a polynomial” and without hesitation they’ll say yes. Because in standard uasage of that expression it is a polynomial. That’s the default interpretation.

Read the first chapter of any beginning algebraic geometry textbook. x^2+3x-1=0 is an algebraic variety. This is the standard interpretation of that equation.

I am well aware of this. If you look at my comments within this very chain, you will see that I made reference to polynomial evaluation, which I will call Ev(f, x). I Am not denying that the partial application given by f' = x -> f(f,x) is naturally induced by the polynomial f. Nor that this is so natural that it often makes sense to identify f with f' so that we would consider f=f', even though they are different types of objects.

>Ask a million mathematicians, “is x^2+3x a polynomial” and without hesitation they’ll say yes.

As will I, because the distinction between formal polynomials and expressions which can be naturally modeled as polynomials is so unimportant that it is almost never worth thinking about.

Put another way, how would you compute the following sets:

{ x \in C | x^3 - x = 8 }

{ x \in C | x^3 = x + 8 }

{ x \in C | log(x) = x^x }

{ x \in C | sin(x) = .7 }

{ x \in N | exists y \in N such that 5x + 3y = 1 }

Are you really claiming that these questions are ill-posed without stating them in terms of algebraic geometry?

You don’t understand the underlying algebraic theory. This is evidenced by your claim

The "x" in the LHS is literally a value. It makes no sense to ask what values of (0,1,0,0,...) make that equation true.

You can’t write such a statement if you really understand that “what values make x+3 the number 5” is a nicer way of conveying the question “under the natural evil map induced by x+3 what is the pre-image of 5”. The text of your I quoted is wrong.

Then you are not reading my comments.

> To be clear, outside of very particular contexts I would still call x^2-x+1=0 a polynomial equation, because it is extremely useful to talk about polynomials without invoking all of the machinery of formal polynomials.

>You don’t understand the underlying algebraic theory.

My claim is that the underlying algebraic theory is not necessary to rigorously state or solve the problems under discussion.

>You can’t write such a statement if you really understand that “what values make x+3 the number 5” is a nicer way of conveying the question “under the natural evil map induced by x+3 what is the pre-image of 5”.

My claim is that "what values make x+3 the number 5" translates directly into { x \in F | x+3=5}. I further claim that the "x+3" in this interpenetration is not a polynomial in the formal sense

I am not disputing that this is set is the same set as the pre-image of 5 under the natural map induced by x+3.