Understanding quotients of $\mathbb{Q}[x]$

We describe the quotient rings in a concrete, computational way.

Let us review the definition, say for $\mathbb{Q}[x]/(x^2-2x)$. Two polynomials $A(x)$ and $B(x)$ are called equivalent modulo $^2-2x$ if their difference is divisible by $x^2-2x$. Then $\mathbb{Q}[x]/(x^2-2x)$ officially consists of the equivalence classes, with the "natural" addition and multiplication modulo $x^2-2x$.

For any polynomial $A(x)$, there exist unique polynomials $Q(x)$ and $R(x)$ such that $R(x)$ has degree $\lt 2$ and $$A(x)=(x^2-2x)Q(x)+R(x).$$ It follows that $A(x)$ is equivalent to $R(x)$. It is easy to see that two polynomials of degree $\lt 2$ are equivalent iff they are equal.

Thus the equivalence classes can be identified with the polynomials of degree $\lt 2$. Addition is the obvious one. Multiplication is a bit trickier. Let us see how to compute the polynomial of degree $\lt 2$ which is equivalent to $(ax+b)(cx+d)$.

Multiply in the natural way, to obtain $ac x^2+(ac+bd)x + bd$. Now recall that $x^2-2x$ is equivalent to $0$, so replace $x^2$ by $2x$. We obtain $(2ac +ac+bd)x +bd$. Now we have an explicit formula for the product.

An analogy: The ring $\mathbb{Z}/(12)$ is officially made up of equivalence classes, where $a$ and $b$ are called equivalent if their difference is divisible by $12$. But it is very useful to think of $\mathbb{Z}/(12)$ as made up of the objects $0,1,2, \dots, 11$, with new addition and multiplication. (For example, $5+9=2$, $5\times 9=9$.)

Similarly, the quotient rings we are looking at can be viewed either abstractly, or concretely as polynomials of degree $\lt 2$, with an unusual multiplication.

Since we have a concrete picture of what is going on, we should be able to answer some questions.

First we show that $\mathbb{Q}[x]/(x^2)$ is not isomorphic to $\mathbb{Q}[x]/(x^2-1)$.

Let's view $\mathbb{Q}[x]/(x^2)$ as the polynomials of degree $\lt 2$, with the natural multiplication except that $x^2$ is always replaced by $0$. Then, in the quotient ring, $(x)(x)=0$. (Strictly speaking, the equivalence class of $x$, times itself, is equal to the equivalence class of $0$.) So $\mathbb{Q}[x]/(x^2)$ has a non-zero element whose square is $0$.

We show that $\mathbb{Q}[x]/(x^2-1)$ has no such element. Suppose to the contrary that in $\mathbb{Q}[x]/(x^2-1)$, the square of (the equivalence class of) $ax+b$ is $0$, that is, $(ax+b)^2$ is equivalent to $0$ modulo $x^2-1$.

Do the squaring. First we get $a^2x^2+2abx+b^2$. Then, since $x^2-1$ is equivalent to $0$, we replace $x^2$ by $1$, and obtain $2abx+b^2+1$. Could this be the $0$ polynomial? No, because the constant term $b^2+1$ cannot be $0$.

So $\mathbb{Q}[x]/(x^2)$ and $\mathbb{Q}[x]/(x^2-1)$ differ in a structural property: The first has a non-zero object whose square is $0$, and the second does not. But any isomorphism $\phi$ from $\mathbb{Q}[x]/(x^2)$ to $\mathbb{Q}[x]/(x^2-1)$ must preserve such structural properties. For completeness we do the details.

Suppose that $w\in \mathbb{Q}[x]/(x^2)$ is non-zero and $\phi(w^2)=0$, then $\phi(w)\ne 0$ and $0=\phi(w^2)=(\phi(w))^2$. So the square of $\phi(w)$ is $0$. This contradicts our earlier calculation, which showed that the square of a non-zero element of $\mathbb{Q}[x]/(x^2-1)$ cannot be zero.

Next we show that $\mathbb{Q}[x]/(x^2-1)$ is isomorphic to $\mathbb{Q}[x]/(x^2-2x)$.

Since $x^2-2x=(x-1)^2-1$, the equivalence class of $x$ in $\mathbb{Q}[x]/(x^2-1)$ should behave like the equivalence class of $x-1$ in $\mathbb{Q}[x]/(x^2-2x)$.

Since we are dealing with equivalence classes modulo two different polynomials, let us change to a more precise notation. Denote the equivalence class of $P(x)$ modulo $x^2-1$ by $P(x)/(x^2-1)$, and modulo $x^2-2x$ by $P(x)/(x^2-2x)$. (We should have used this more precise notation from the beginning, but avoided it for the sake of greater concreteness. But in what follows, don't let the $/(??)$ parts worry you, and maybe even omit them.)

So what should $(ax+b)/(x^2-1)$ be sent to by our isomorphism $\phi$? The natural choice is $(a(x-1) +b)/(x^2-2x)$.
It is clear that (equivalence classes modulo $x^2-1$ of) polynomials of degree $\lt 2$ are sent bijectively by $\phi$ to (equivalence classes modulo $x^2-2x$ of) polynomials of degree $\lt 2$. We must also check that $\phi$ preserves addition and multiplication.

Checking the addition is very easy. Let's deal with the multiplication. Look at $(ax+b)(cx+d)$. Modulo $x^2-1$, this is (equivalent to) $(ad+bc)x + ac+bd$.

Note that $\phi$ maps $(ax+b)/(x^2-1)$ to $(a(x-1)+b)/(x^2-2x)$ and maps $(cx+d)/(x^2-1)$ to $(c(x-1)+d)/(x^2-2x)$. Also, $\phi$ maps
$((ad+bc)x + ac+bd)/(x^2-1)$ to $((ac+bd)(x-1)+ ac+bd)/(x^2-2x)$. So we need to verify verify that $$(a(x-1)+b)(c(x-1)+d) \quad\text{is equivalent to}\quad (ac+bd)(x-1)+ ac+bd$$ modulo $x^2-2x$.

Multiply out the left-hand side, using the fact that $(x-1)^2-1$ is equivalent to $0$. We get $(ac+bd)(x-1)+ac+bd$, exactly what is wanted.

Comment: Look for example at $\mathbb{Q}[x]/(x^2-2x)$. The fact that $x^2-2x$ factors nicely means that we can express $\mathbb{Q}[x]/(x^2-1)$ as a direct product of simpler structures. This important structural information is easiest to approach through a more abstract approach. However, a concrete view of things is always useful, both for the understanding and for computational algebra.

Every ideal of $\mathbb{Q}[x]$ is principal. Given $f(x)$ and $g(x)$, we have that $(f(x))\subseteq (g(x))$ if and only if $g(x)$ divides $f(x)$. Moreover, $(f(x))$ is maximal if and only if $f(x)$ is irreducible, if and only if $(f(x))$ is a nonzero prime ideal.

You can always do division: every element of $\mathbb{Q}[x]/(f)$ is "represented" by a residue class of the form $g(x) + (f)$, where $g(x)$ is a polynomial of degree strictly less than $f$: just take an arbitrary $a(x)$, divide by $f$, and let $g(x)$ be the remainder. Then $a(x)+(f) = g(x)+(f)$. This is in complete analogy to the quotients of $\mathbb{Z}$.

Also, if $f(x)$ and $g(x)$ are coprime, then we know that $(f(x))+(g(x)) = \mathbb{Q}[x]$, and that $((f(x))\cap((g(x)) = (f(x))(g(x)) = (f(x)g(x))$.

From the Chinese Remainder Theorem, it follows that if $f_1,\ldots,f_n$ are pairwise relatively prime polynomials in $\mathbb{Q}[x]$, then $$\frac{\mathbb{Q}[x]}{(f_1\cdots f_n)} \cong \frac{\mathbb{Q}[x]}{(f_1)}\times \cdots \times \frac{\mathbb{Q}[x]}{(f_n)}.$$

Since $x^2-x = x(x-1)$, you have that $$\frac{\mathbb{Q}[x]}{(x^2-x)} \cong \frac{\mathbb{Q}[x]}{(x)}\times\frac{\mathbb{Q}[x]}{(x-1)}.$$ Now, $\mathbb{Q}[x]/(x)$ is easy; what about $\mathbb{Q}[x]/(x-1)$? Consider the map $\mathbb{Q}[x]\to\mathbb{Q}$ induced by evaluating at $x=1$.

(More generally, if $f(x) = ax+b$ with $a\neq 0$, then what is $\mathbb{Q}[x]/(ax+b)$? Consider the evaluation map from $\mathbb{Q}[x]$ to $\mathbb{Q}$ given by evaluating at $x=-\frac{b}{a}$).

You should now be able to see that the first quotient is indeed isomorphic to $\mathbb{Q}[x]/(x^2-1)$ as well, by considering the factorization of $x^2-1$.

(More generally, if $f(x) = p_1^{a_1}\cdots p_m^{a_m}$ is a factorization of $f(x)$ into irreducibles, the Chinese Remainder Theorem tells you that $\mathbb{Q}[x]/(f)$ is isomorphic to the product of the quotients $\mathbb{Q}[x]/(p_i^{a_i})$).

As for $\mathbb{Q}[x]/(x^2)$, that's a bit more difficult. But notice that the element $x+(x^2)$ is not zero, and yet its square is zero. Is there any such element in $\mathbb{Q}[x]/(x^2-x)$?

Hint $\,\ (1)\, \ $ What shifty automorphism of $\,\mathbb Q[x]\,$ maps $\,x(x-2)\,$ to $\,(x+1)(x-1)\,$ ?

$(2)\, \ $ $ f\,$ squarefree $\,\Rightarrow \mathbb Q[x]/f\,$ has no nilpotents $\ne 0\ $ since $\ f\ |\ g^{\,n} \Rightarrow f\mid g\,$ for squarefree $\,f.\,$ Therefore $\,\mathbb Q[x]/(x^n)\,\not\cong\, \mathbb Q[x]/((x\!-\!a_1)\,\cdots\,(x\!-\!a_k))\ $ for $\ n>1,\,$ distinct $\,a_i\in\mathbb Q$.