Find $F = \mathbb Z_3[x] /f(x)$ elements. [closed]

Solution 1:

I'm assuming that your question is in the title and by $g(x)$ you are actually referring to $f(x)$.

As per the comments, I believe that you have shown that $f(x)$ does not have any roots in $\mathbb{Z_3}$, i.e. $f(x)$ is irreducible in $\mathbb{Z_3[x]}$. As TheSilverDoe pointed out, this "trick" of checking irreducibility by checking if the roots lie in the underlying field is typically applicable only for lower degree polynomials(in general, for degree $3$ or less). An irreducible polynomial by definition does not have any proper divisors, i.e. does not factor into two lower degree non-constant polynomials. For polynomials of degrees $2$ or $3$, one can easily verify that a polynomial is irreducible iff it has no root in the underlying field(i.e has a linear polynomial as a proper factor). This cannot be applied in general, for polynomials of degrees greater than $4$. For example $(x^2+1)^2$ has no root in $\mathbb{R[x]}$ but is reducible as it has $(x^2+1)$ as a proper factor.

Note that $\mathbb{Z_3[x]}$ is a PID since $\mathbb{Z_3}$ is a field(polynomial ring in one variable of a field is a PID). So, all non-zero irreducible polynomials in $\mathbb{Z_3[x]}$ in fact generate maximal ideals. Now again as per the comments, I believe that you have shown the result that $I$ is a maximal ideal of a ring R iff $R/I$ is a field. So we have that $F=\mathbb{Z_3[x]}/(f(x))$ is a field.

In case you are familiar with the notion of field extensions, you'll immediately realize that $F=\mathbb{Z_3[x]}/(f(x)) \cong \mathbb{Z_3}(\alpha)$, where $\alpha=\bar{x}=$Image of $x$ in the homomorphism $\mathbb{Z_3[x]} \to \mathbb{Z_3[x]}/(f(x))= x+(x^2+x-1)$ is a root of $f(x)$ in $F$, i.e. $\alpha^2+\alpha-1=0$

The elements of $\mathbb{Z_3}(\alpha)$, the field extension of $\mathbb{Z_3}$ with $\alpha$ are $\{0,1,2,\alpha,\alpha+1,\alpha+2,2\alpha,2\alpha+1,2\alpha+2\}$.This is the smallest field containing $\mathbb{Z_3}$ and $\alpha$. You can verify that this satisfies all the properties of a field, contains $\mathbb{Z_3}$ and $\alpha$ and has no "unnecessary" elements(i.e. the smallest such field). You would have to simplify some of the terms, keeping in mind that we are in $\mathbb{Z_3}$ and the relation $\alpha^2+\alpha-1=0$). This shows that $F \cong \mathbb{Z_3}(\alpha)$ has $9$ elements.