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We define the maps $f_+$ and $f_-$ from $\mathbb{Q}[X]/(X^2 -2) \mathbb{Q}[X]$ to $\mathbb{Q} + \mathbb{Q} \cdot \sqrt2$ in the following manner: For any residue class $g+ (X^2-2) \mathbb{Q}[X]$, we write:

$$ f_+(g+ (X^2-2) \mathbb{Q}[X]) = g \left( \sqrt2 \right)$$

$$ f_-(g+ (X^2-2) \mathbb{Q}[X]) = g \left(- (\sqrt2) \right)$$

I want to show that these maps are well-defined. Because the degree of our polynomial $X^2-1$ is two, every representative in the residue class will have a degree less than this, upon division. We thus expect polynomials in the domain of the form $aX+b$ with $a,b \in \mathbb{Q}$; the elements in the image will have the form $p+ q \sqrt2$, with $p,q \in \mathbb{Q}$.

I have a hard time figuring out what this map precisely does to elements in the domain. If I would understand this, it might be easier to prove it is well-defined ( it does not matter which representative we pick).

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    $\begingroup$ No need to write $\;(x^2-2)\Bbb Q[x]\;$ , enough with $\;(x^2-2)\;$ to denote the polynomial ideal, otherwise both the writing and the reading become cumbersome $\endgroup$ – DonAntonio Oct 18 '18 at 17:13
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    $\begingroup$ There are infinitely many rational numbers. The same holds for the extension field present in this question. Therefore the tag finite-fields was inappropriate. $\endgroup$ – Jyrki Lahtonen Oct 19 '18 at 4:33
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    $\begingroup$ To get started on the question. What are the potential sources of non-well-definedness here? You should notice that on the right hand side a polynomial $g$ is evaluated at a point. Yet, look at the left hand side, we seek to define the image of a coset of an ideal of polynomials. In other words, the polynomial $g$ is simply a representative of its coset. What could go wrong, if we use some other polynomial to represent the coset? $\endgroup$ – Jyrki Lahtonen Oct 19 '18 at 4:38
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    $\begingroup$ Compare: the "definition" of a function $f:\Bbb{Z}/5\Bbb{Z}\to\Bbb{Z}/4\Bbb{Z}$ by the recipe $f(\overline{n})=\overline{n}$ does not work. It is not well-defined, because the result of the function depends on the choice of integer $n$ representing the coset $\overline{n}\in\Bbb{Z}/5\Bbb{Z}$. For example, $0$ and $5$ belong to the same coset (or residue class) of integers modulo five. In other words $\overline{0}=\overline{5}$ in the domain. If $f$ is to be a function we need $f(\overline{0})=f(\overline{5})$. $\endgroup$ – Jyrki Lahtonen Oct 19 '18 at 4:42
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    $\begingroup$ (cont'd) But in the codomain $\overline{0}\neq\overline{5}$ because $0$ and $5$ are not congruent to each other modulo four. Therefore this recipe does not give a well-defined function. A point of this exercise is to study what happens to polynomials. Remember that the cosets $g+(x^2-2)$ and $g'+(x^2-2)$ belong to the same coset of the ideal $(x^2-2)$ if their difference $g-g'$ is a (polynomial) multiple of $x^2-2$. The key point is whether this matters at all in the definition of $f_+$ or $f_-$? $\endgroup$ – Jyrki Lahtonen Oct 19 '18 at 4:48
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The tip in the comments answered my question. The function is well-defined because any equivalent elements $g$ and $g'$ have images $g(\pm \sqrt{2})$ and $ g'(\pm \sqrt{2})$.

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    $\begingroup$ Correct. Well done! $\endgroup$ – Jyrki Lahtonen Oct 20 '18 at 20:15

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