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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!

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  • $\begingroup$ 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 $\mathbb{Z}_3$. $\endgroup$ – topeik Dec 10 '19 at 5:29
  • $\begingroup$ @topeik okay! so would the elements of $\mathbb{Z}_3[x]/(f)$ be all polynomials with degree 1 (i.e. $x + 1$, $x - 3$, $2x + 4$)? $\endgroup$ – yagayeet Dec 10 '19 at 5:37
  • $\begingroup$ Or degree $0$, like $1,2$, or degree $-\infty$, like $0$. $\endgroup$ – topeik Dec 10 '19 at 5:39
  • $\begingroup$ The roots of $x^2+1$ are $\pm i$ which certainly are not elements of $\mathbb Z/3\mathbb Z$. $\endgroup$ – Math1000 Dec 10 '19 at 5:51
  • $\begingroup$ $\mathbb Z_3[x]/(f)$ is a field with $9$ elements $\endgroup$ – J. W. Tanner Dec 10 '19 at 5:58
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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.

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

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