Naive set theory question on "=" So I picked up a couple of good undergraduate-level books over the weekend and have been working through them...
In Algebra: Chapter 0, the author of the text writes:

The prototype of the well-behaved relation is '=', which corresponds to 'the diagonal'
  $$\{ (a, b) \in S \times S \, | \, a = b \} = \{(a, a) \, | \, a \in S \} \subseteq S \times S $$

The problem I'm having is that the way I'm interpreting this expression makes it seem uninteresting, almost tautological, which I'm thinking can't be correct.  What I think this expression is saying is this:  Suppose I give you elements $a$ and $b$ in the set $S$ to compare for equality.  To do that, write $a = b$ as the ordered pair $(a,b)$, which is the same element as $(a, a)$, which by a previous definition we are told is true iff $\{b\} = \{a\}$. 
But when I try to bring this to a concrete level, by letting $S$ be the set of all polynomials with integer coefficients, say, I become confused.  Since $(0, 6x + (-6)x)$ is not the same ordered pair as $(0, 0)$, aren't we left to conclude that $(6x + (-6)x) \neq 0$?  That doesn't seem right. The only thing I can see this definition being useful for is writing tautologies like $A = A$. 
And then I wondered: What does it mean to write an expression like $x^2 = 2x$?  Am I correct in saying that the "$=$" there is not the same "$=$" as in the definition given above?  How would one write the definition of this new "$=$" in set-theoretic terms, with universal quantifiers?

Consider $S=\{1,2,3\}$. The diagonal is $\{ (1,1), (2,2), (3,3) \}$. There is nothing trivial or tautological about it.
BTW, $(0, 6x + (-6)x)$ is the same ordered pair as $(0, 0)$ because $6x + (-6)x=0$ as polynomials. You need to recall the definition of equality of polynomials.
