In general, I struggle working with quotients of polynomial rings, so I was hoping someone might be able to help with the following exercise:

Show that $\mathbb{Q}[x]/(x^2-2x)$ is isomorphic to $\mathbb{Q}[x]/(x^2-1)$, but that the latter is not isomorphic to $\mathbb{Q}[x]/(x^2)$.

Mostly I want to understand what these quotients all "look like." We did few examples in class and few homework problems with such quotients, and the ones we do work with are much simpler (typically involving $\mathbb{Z}_2$ or $\mathbb{Z}_3$ or $\mathbb{R}$ instead of $\mathbb{Q}$).

Any help is appreciated, thank you!

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    $\begingroup$ Do you know what $\mathbb{Q}[x]/(x - a)$ looks like? Are you familiar with the Chinese Remainder Theorem in this context? $\endgroup$ – Qiaochu Yuan Aug 16 '11 at 4:12
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    $\begingroup$ For first part, $x^2-2x=(x-1)^2-1$ suggests the mapping to use. $\endgroup$ – André Nicolas Aug 16 '11 at 4:12
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    $\begingroup$ You may view taking a quotient in a ring as "adding a relation" in the ring. For example $\mathbb{Q}[x]/(x^2-2x)$ is the ring $\mathbb{Q}$ to which you add a "formal element" $x$ satisfying the relation $x^2 - 2x = x(x-2) = 0$ (roughly $x$ can be either $0$ or $2$ indifferently). I suggest you start with this even simpler exercise : $\mathbb{Q}[x]/(x-1)$ is isomorphic to $\mathbb{Q}$. $\endgroup$ – Joel Cohen Aug 16 '11 at 4:22
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    $\begingroup$ These examples do not depend on $\mathbb{Q}$. For that, consider the difference between $\mathbb{Q}[x]/(x^2 - 2)$ and $\mathbb{R}[x]/(x^2 - 2)$. $\endgroup$ – lhf Aug 16 '11 at 10:54

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.

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  • $\begingroup$ Thank you for this very thorough answer. This is tremendously helpful =) $\endgroup$ – Bey Aug 16 '11 at 18:18
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    $\begingroup$ @Bey: You are welcome. One of the problems with learning "abstract" algebra is that it is easy to lose sight of the fact that it deals with real objects. Once you have a sufficiently large collection of concrete examples, testing possible proof ideas for reasonableness becomes much easier. Algebra without concrete examples is like geometry without pictures. $\endgroup$ – André Nicolas Aug 16 '11 at 18:25

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 $\mathbb{Q}[x]/(x^2-x)$?

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  • $\begingroup$ I had a hard time choosing between your answer, Arturo, and Andre's as the best. Both answered my question clearly and thoroughly, but the constructive nature of Andre's solution appealed to me. I have studied the CRT, and I will certainly keep it in my toolbox from now on, but have done very few exercises with it. I will attempt to verify some assertions you made that seem clear but I haven't yet proved for myself (e.g., that (f(x))(g(x))=(f(x)g(x)) for f(x),g(x) relatively prime). Thank you for your solution =) $\endgroup$ – Bey Aug 16 '11 at 18:23
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    $\begingroup$ @Bey: No worries and no need to explain. You accept the answer that you found most helpful; that's by its very nature a subjective thing, and there's no need to apologize for your choice (so long as you don't make a habit of not accepting answers, which you certainly are not) $\endgroup$ – Arturo Magidin Aug 16 '11 at 19:02

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

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

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