# Surjective ring homomorphism - cardinality of the domain

I've been doing some exam preparation examples, and I wanted to ask if my idea to this example makes sense - at least if I am on the right path.

A field $P \cong \mathbb{Z}_7$ and a polynomial ring $P[x]$ are given. Also, a polynomial $f(x)=x^3 +2$ is given, and this polynomial induces an ideal called $I_f$. The first step was to prove that $f(x)$ irreducible in $P[x]$, which I managed to do.

As a next part a surjective ring homomorphism $\varphi _f:P[x] \rightarrow R_f$ is given, with only information about $R_f$ is that it is a ring and that the kernel of $\varphi_f$ is the ideal $I_f$. The question was to determine the cardinality of the domain $R_f$.

Since this ideal is a kernel of this surjective homomorphism, by the first isomorphism theorem, $R_f \cong P[x]/_{I_f}$. So the cardinality of $R_f$ should be the number of all the polynomials that are left after the factorisation: polynomials of degree 2 and 1, and constant polynomials, without the zero - polynomial since it is already in the ideal. After counting the possibilities, and since $\mathbb Z_7$ has only 6 elements that are not zero, I got to $6^2$ for degree $2$, $6^2$ for degree $1$ and $6$ for degree $0$. That, with the zero element of the domain, sums op to $259$.

I am not sure if I can really count the elements like that and if I applied the first isomorphism theorem correctly in this case. I would also like to have some way to formalise this proof.

I would appreciate any thoughts on this solution.

• At this level you should start making the distinction between a polynomial and a polynomial function. Elements of $P[x]$ are not functions. – Arnaud Mortier Jun 10 '18 at 10:56
• Thanks, @ArnaudMortier, I edited it. – Milena Jun 10 '18 at 10:58
• There is no reason to include polynomials of degree $2$ in the quotient. $x^2=-2$. Also no reason to exclude $0$ as a coefficient. – Arnaud Mortier Jun 10 '18 at 11:06
• Can you explain why? – Milena Jun 10 '18 at 11:39
• See the answer I gave here (how to think of a quotient without thinking of cosets). – Arnaud Mortier Jun 10 '18 at 11:44

Each $g(x)\in P[X]$ can be written in a unique way as $g(x)=q(x)f(x)+r(x)$ with $\deg r<\deg f$. Two poylnomials are equivalent modulo $f$ iff they lead to the same $r$. Clearly, there are $7\cdot 7$ choices for the coefficients of $r$.