# Show that the polynomial $f = x^2 + 1$ is irreducible in $\mathbb{Z}_3[x]$. How many elements does $\mathbb{Z}_3[x]/(f)$ have?

Show that the polynomial $$f = x^2 + 1$$ is irreducible in $$\mathbb{Z}_3[x]$$. How many elements does $$\mathbb{Z}_3[x]/(f)$$ have? Write out all of the elements of this field, and find the inverse of each nonzero element.

I am trying to show that the polynomial $$f = x^2 + 1$$ is irreducible in $$\mathbb{Z}_3[x]$$. To do so, I am using the following theorem: (Let $$F$$ be a field, $$f \in F[x]$$ of degree 2 or 3) If $$f$$ has no roots, then $$f$$ is irreducible. So, since $$f(0), f(1)$$ and $$f(2)$$ are all not $$\equiv$$ 0(mod 3), $$f$$ has no roots and thus $$f$$ is irreducible. Now how would I calculate how many elements $$\mathbb{Z}_3[x]/(f)$$ has? Would I need to find all of the polynomials in $$\mathbb{Z}_3[x]$$ that don't have roots? I would appreciate any feedback!

• Each class of $\mathbb{Z}_3[x]/(f)$ has a representative that is the remainder that its elements leave in the division by $f$. Since your $f$ has degree $2$, the remainder has degree at most $1$. So, it is determined by two coefficients that are in $\mathbb{Z}_3$. – topeik Dec 10 '19 at 5:29
• @topeik okay! so would the elements of $\mathbb{Z}_3[x]/(f)$ be all polynomials with degree 1 (i.e. $x + 1$, $x - 3$, $2x + 4$)? – yagayeet Dec 10 '19 at 5:37
• Or degree $0$, like $1,2$, or degree $-\infty$, like $0$. – topeik Dec 10 '19 at 5:39
• The roots of $x^2+1$ are $\pm i$ which certainly are not elements of $\mathbb Z/3\mathbb Z$. – Math1000 Dec 10 '19 at 5:51
• $\mathbb Z_3[x]/(f)$ is a field with $9$ elements – J. W. Tanner Dec 10 '19 at 5:58

The elements of $$\mathbb Z_3[x]/(f)$$ are $$a+bx$$ with $$a,b\in\mathbb Z_3$$.

There are $$3$$ possibilities for $$a$$ and $$3$$ for $$b$$, so $$9$$ elements altogether.

Note that in the quotient ring $$x^2\equiv-1$$.

The multiplicative inverse of $$x$$ is $$2x$$. The inverse of $$1+x$$ is $$2+x$$.

I will leave the rest as an exercise for the reader.

• For the 9 elements I got (with a = 1, 2, 3 and b = 1, 2, 3), $x + 1, 2x + 1, 3x + 1, 3x + 2, 2x + 2, x + 2, x + 3, 2x + 3$ and $3x + 3$. How does the quotient ring $x^2 \equiv -1$ relate to finding the inverse of each nonzero element in $\mathbb{Z}_3[x]/(f)$? – yagayeet Dec 10 '19 at 5:48
• Those $9$ elements are correct, though $x+3\equiv x$, $3x+3\equiv0$, etc. – J. W. Tanner Dec 10 '19 at 5:55
• $(x)(2x)=2x^2\equiv2(-1)=-2\equiv1$; $(1+x)(2+x)=2+3x+x^2\equiv2+0-1=1$ – J. W. Tanner Dec 10 '19 at 5:56
• I am a little confused why you are multiplying $(x)(2x)$? – yagayeet Dec 10 '19 at 6:03
• to demonstrate that inverse of $x$ is $2x$, show their product is $1$ – J. W. Tanner Dec 10 '19 at 6:05