a) How many elements does the quotient ring$\displaystyle \frac{\mathbb Z_5[x]}{\langle x^2+1\rangle} $ have?

Is it an integral domain?

I can see that the polynomial, $\displaystyle p(x)= x^2+1=(x-2)(x-3)$ is reducible over the field of integers modulo $5$ but can't proceed further.

b) What if we had $\displaystyle \frac{\mathbb Z_{11}[x]}{\langle x^2+1\rangle}$.

Where the polynomial was irreducible over the field of integers modulo $11$.

I had a look at some solutions which say that the elements in this quotient ring will be of type $ax+b$ and then we have $11$ choices for each of the two and consequently $121$ elements.

I could not follow why the elements will be of $ax+b$ form. Please explain.

  • $\begingroup$ you remember Chinese remainder analogue in ring theory?? $\endgroup$ – user87543 Jan 27 '14 at 10:35
  • $\begingroup$ No. I just know tha basics of ring theory. Isomorphisms, Ideals etc. $\endgroup$ – prat Jan 27 '14 at 10:37

By the division algorithm you can write any polynomial uniquely as $$q(X^2+1)+r$$ with $\deg r<2$, or $r=0$ (Careful! The pair $(q,r)$ is unique, but of course two different polynomials might yield the same remainder $r$, and this will happpen!) Since any polynomial of degree at most $1$ can be written as $aX+b$, letting $a,b$ range over all elts of $\Bbb Z_5$ gives you $5\times 5=25$ elements. Note however the elements will be of the form $aX+b+(X^2+1)$, rather.

Note also that by your work $(X-2)+I$ and $(X-3)+I$ are zero divisors in the quotient, hence your quotient is not a domain (precisely because your polynomial is reducible over $\Bbb Z_5$.)

  • $\begingroup$ So it shouldn't make any difference whether the polynomial is reducible or irreducible over the respective field, right? I mean, with respect to number of elements. $\endgroup$ – prat Jan 27 '14 at 10:41
  • $\begingroup$ @prat Some things matter, some things don't. Be specific? For example, computing the number of elements or proving the power basis does not require any ir/reducibility hypothesis, whereas showing the quotient is a domain does require a hypothesis. $\endgroup$ – anon Jan 27 '14 at 10:43
  • $\begingroup$ @prat You mean over the respective ring? It does! If the polynomial is irreducible, the quotient is a field, if it is reducible, it is not even a domain! $\endgroup$ – Pedro Tamaroff Jan 27 '14 at 10:43
  • $\begingroup$ @prat I recommend you read anon's answer with care. It is a good checklist of things you should really know! =) $\endgroup$ – Pedro Tamaroff Jan 27 '14 at 10:48

If $F$ is a field and $f(x)\in F[x]$ is a polynomial then $\{1,x,\cdots,x^{n-1}\}$ is a basis for $F[x]/(f(x))$, where $n=\deg f(x)$. This is a fact you should be aware of. To prove it, note that it suffices to see that any power $\ge n$ can be reduced to smaller powers. Then, note $x^m=x^{m-n}x^n=x^{m-n}(\cdots)$ (can you figure out how to write $x^n$ in terms of smaller powers of $x$ using the relation $f(x)=0$?). One can make this argument even more swift; any polynomial in $F[x]$ has a unique remainder under division by $f(x)$ which has degree $\le$ that of $f(x)$. (The "division algorithm.")

You can use this to show that $(\Bbb Z/p\Bbb Z)[x]/f(x)$ has dimension $\deg f(x)$ as a vector space over $\Bbb Z/p\Bbb Z$ and hence has $|\Bbb Z/p\Bbb Z|^{\deg f(x)}=p^n$ elements. What is this in your case?

For $(\Bbb Z/5\Bbb Z)[x]/(x^2+1)$ to be an integral domain, we must have $ab=0\Rightarrow a=0$ or $b=0$. This is equivalent to saying that $(x^2+1)\mid a(x)b(x)\Rightarrow (x^2+1)\mid a(x)$ or $(x^2+1)\mid b(x)$. Can you think of a counterexample to this? Hint: you've already written down the $a(x)$ and $b(x)$ in your question.

Have you seen the proof that $F[x]$ is a PID? Have you seen the proof that if $R$ is a PID and $a\in R$ is an irreducible element, that $a$ must be a prime element and $(a)$ is a prime ideal? Have you seen the proof that the quotient of a ring by a prime ideal is a domain?


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