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I learned how to see quotient groups intuitively when I learned of a group mod its commutator subgroup. If we take a group and mod out all the elements that do not commute, we get a quotient group which is abelian. Simple, nothing technical.

Now I am having trouble seeing a polynomial ring mod some ideal.

Example $1$: Evaluation hom. at $0$: $\ \ \ \varphi: \mathbb Z[x] \to \mathbb Z$ where $\varphi(f(x))=f(0)$. This hom. is surjective and its kernel is the principal ideal generated by $x$, that is $(x)$. So, $\mathbb Z[x]/(x) \cong \mathbb Z$.

Example $2$: Evaluation hom. at $\sqrt 2$: $\ \ \ \psi: \mathbb Q[x] \to \mathbb Q[\sqrt 2]$. We know $\mathbb Q[x]/(x^2-2) \cong \mathbb Q[\sqrt 2]$.

What is $\mathbb Z[x]/(x)$? What do the cosets look like?

What is $\mathbb Q[x]/(x^2-2)$? What do the cosets look like?

Intuitively, what is this telling me? Any other concrete examples for intuition of a quotient ring?

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    $\begingroup$ Maybe this is too facile an explanation, but to me, the statement $\mathbb{Z}[x]/(x) \cong \mathbb{Z}$ says that when think of polynomials with coefficients in $\mathbb{Z}$ as being represented by their constant term, we get something that looks like $\mathbb{Z}$. Does this make sense? $\endgroup$ – Alex Wertheim Apr 26 '15 at 19:42
  • $\begingroup$ Makes sense. But what is $\mathbb Z[x]/(x)$? $\endgroup$ – Al Jebr Apr 26 '15 at 19:44
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    $\begingroup$ $\mathbb{Z}[x]/(x)$ is, formally, equivalence classes of polynomials in $\mathbb{Z}[x]$ up to a $\mathbb{Z}[x]$-multiple of $x$. For example, $3$ and $3+x+5x^2$ are "the same" (i.e., belong to the same equivalence class) in $\mathbb{Z}[x]/(x)$, because $3+x+5x^2 - 3 = x+5x^2$, which is a multiple of $x$ (i.e., lives in the ideal $(x)$). In this way, you can see how all polynomials in $\mathbb{Z}[x]/(x)$ can be identified with their constant term. $\endgroup$ – Alex Wertheim Apr 26 '15 at 19:48
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A general element of $F[X]/(a_nX^n + \dotsb + a_0X^0)$ is of the form $c_{n-1}X^{n-1} + \dotsb + c_0$ where $X^n = \frac{a_{n-1}X^{n-1} + \dotsb + a_0X^0}{a_n}$. Multiplication is performed by multiplying two polynomials in this "general form" and then simplifying the result according to the rule $X^n = \frac{a_{n-1}X^{n-1} + \dotsb + a_0X^0}{a_n}$, which reduces it to the general form again.

An example of this is $\mathbb{R}[X]/(X^2 + 1)$, where the general form of an element is $c_0 + c_1X$, and $X^2 = -1$. This gives the complex numbers. Another example is $\mathbb{R}[X]/(X^2)$, where every element is of the form $c_0 + c_1X$ where $X^2 = 0$; this is the dual numbers. The general form of an element of $\mathbb{R}[X]/(X)$ is the general form of an element of $\mathbb{R}$ because $X = 0$ in this ring. I recommend studying the examples I just gave: the complex numbers, the dual numbers, the bog-standard real numbers when expressed as $\mathbb R[x]/(X)$. Also look at the split-complex numbers, which I haven't shown.

A general element of $\mathbb{Q}[X]/(X^2 - 2)$ is of the form $c_0 + c_1X$ where $X^2 = 2$.

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  • $\begingroup$ Ok, so I am understanding the examples where the ideal is $(x^n)$. But can you elaborate on quotient ring modded by $(x^2-2)$? Your last line. $\endgroup$ – Al Jebr Apr 26 '15 at 20:52
  • $\begingroup$ If you mod by a polynomial $f(X)$ of degree $n$, then every element of the quotient ring will be a polynomial of degree $n-1$. The multiplication rule is found by starting with the equation $f(X) = 0$ and rearranging to find $X^n$. Also see the split-complex numbers. The only shortcoming of these examples is that they are quotients by quadratics. If you want a quotient by a cubic, I can give you that. $\endgroup$ – ogogmad Apr 26 '15 at 20:54
  • $\begingroup$ A general element of $\mathbb{Q}[X]/(X^2 - 2)$ is of the form $c_0 + c_1X$ where $X^2 = 2$. I can't wrap my head around this. $\endgroup$ – Al Jebr Apr 26 '15 at 21:02
  • $\begingroup$ What do you mean "where $X^2=2$"? $\endgroup$ – Al Jebr Apr 26 '15 at 21:44
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    $\begingroup$ @Pacman $\, R[x]/(x^2-2) \cong R[x]\pmod{x^2-2},\,$ where $\,x^2\equiv 2,\,$ which can be used as a rewrite rule $\,x^2 \to 2\,$ to reduce all powers $\,x^n,\ n > 2,\,$ till one gets a linear poly $\equiv f(x)\,$ (a rewritng form of the Polynomial Division Algorithm when we need only the Remainder). See also this answer. $\endgroup$ – Bill Dubuque Apr 26 '15 at 22:29
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In the polynomial rings you are exploring, there is a division algorithm. Suppose $p(x)$ is a monic polynomial and $q(x)$ is any polynomial, then over $\mathbb Z$ or $\mathbb Q$ we have $$q(x)=p(x)s(x)+r(x)$$ for some polynomial $s(x)$ is less than where the degree of $r(x)$ is less than the degree of $p(x)$. The elements of the quotient ring of $\mathbb Q[x]$ or $\mathbb Z[x]$ by the ideal generated by $p(x)$ can be thought of as the remainders $r(x)$. The construction is equivalent to setting all the multiples of $p(x)$ equal to zero.

If two remainders $r(x)$ and $r'(x)$ are in the same coset of the ideal then $r(x)-r'(x)$ is divisible by $p(x)$ - but $p(x)$ has a higher degree, so the difference must be zero and $r(x)=r'(x)$. The remainders can be added (no problem) and multiplied if the result is reduced to the remainder on division by $p(x)$ - we often say reduced mod $p(x)$.

This answer is just a little introduction to a big and important subject, and can't cover everything. You might like to think for yourself what happens over $\mathbb Z$ when $p(x)$ is not monic (say $p(x)=2x+1$). This doesn't make a difference over $\mathbb Q$ because $\frac 12\in \mathbb Q$. But $2$ is not invertible over $\mathbb Z$.


In your second example with $x^2-2$, the remainders are all linear of the form $ax+b$. Since $x^2-2$ has zero remainder, we can say $x^2=2$ and identify $x$ with one of the square roots of $2$ (note there are two possibilities which are algebraically equivalent - this is a significant fact - we cannot distinguish them algebraically). The elements can be identified with $a\sqrt 2+b$.

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  • $\begingroup$ What do you mean $x^2-2$ has zero remainder? $\endgroup$ – Al Jebr Apr 26 '15 at 21:20
  • $\begingroup$ For $\mathbb Z/(x)$, I see that $a_0+a_1x+\dots + a_nx^n + (x)$ can be written as $a_0 + x(a_1+a_2x+ \dots + a_nx^{n-1}) +(x)$. But this is $a_0 +(x)$ since $(x)$ is an ideal. This can be rewritten as $a_0 +xf(x)$. I'm having trouble doing this with $\mathbb Q[x]/(x^2-2)$. How would we use this with $\mathbb Q[x]/(x^2-2)$? $\endgroup$ – Al Jebr Apr 26 '15 at 21:24
  • $\begingroup$ @Pacman $x^2-2=1\cdot(x^2-2)+0$ so the polynomial $x^2-2$ is in the coset which contains zero, and is therefore equivalent to zero in the quotient space. To convert an arbitrary polynomial into its remainder in this particular case, replace $x$ with $\sqrt 2$ throughout so that $x^{2n}$ becomes $2^n$ and $x^{2n+1}=2^n\sqrt 2$. This is equivalent to using the division algorithm, as you may be able to see fro the identity $x^r=x^{r-2}(x^2-2)+2x^{r-2}\equiv 2x^{r-2}$ which you can use successively to reduce the power. The division algorithm always works - short cuts depend on circumstances. $\endgroup$ – Mark Bennet Apr 27 '15 at 6:09
  • $\begingroup$ @Pacman I used the division algorithm and remainders to show that a coset has a representative of low degree and to give you an idea what those representatives are. A polynomial $q(x)$ is contained in the coset $r(x)+I$ where $I$ is the set of multiples of $p(x)$. It is an expression in powers of $x$, where $p(x)\equiv 0$ so $x$ can be regarded as a root of $p(x)$. $\endgroup$ – Mark Bennet Apr 27 '15 at 6:13
  • $\begingroup$ math.stackexchange.com/questions/3777658/… $\endgroup$ – Consider Non-Trivial Cases Aug 2 '20 at 15:21

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