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.
Best Answer
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 $\Bbb Z_5$.)