I want to show that $\mathbb Z[i]/(3) \cong \mathbb F_9$, and I know that $\mathbb Z[i]/(3) \cong \mathbb Z[x]/(3, x^2 + 1) \cong \mathbb F_3[x]/(x^2 + 1)$. How do we know that $\mathbb F_3[x]/(x^2 + 1) \cong \mathbb F_9$? I know that $x^2 + 1$ is irreducible, is this important?
Why is $\mathbb F_3[x]/(x^2 + 1) \cong \mathbb F_9$
1
$\begingroup$
abstract-algebra
polynomials
ring-theory
2 Answers
0
Since the polynomial x$^{2}$+1 is irreducible, $\mathbb{F}_{3}$[x]/(x)$^{2}$+1) is definitely a field. It is like dividing a ring by its prime element.
Now the mod group is basically remainder mod(x$^{2}$+1), which is basically a linear polynomial (ax+b) where a,b$\in\mathbb{F}_{3}$.
Since, $\mathbb{F}_{3}$ has 3 elements, we have 3 choices for a and b each. Therefore the mod group has 9 elements.
-
0Thanks, could you expand on why $\mathbb F_3[x]/(x^2 + 1)$ is a field? I know that $R/I$ is a field iff $I$ is maximal, so in this case why do we know that $(x^2 + 1)$ is maximal in $\mathbb F_3[x]$? – 2017-02-15
-
0yes, (x$^{2}$+1) is maximal because $\mathbb{F}_{3}$[x] is a PID and x$^{2}$+1 is an irreducible polynomial. – 2017-02-15
-
0Ah yes, because $ R[x]$ is a PID iff $R$ is a field. Thanks! – 2017-02-15
0
$F_3[X]/(x^2+1)$ is an extension of degree 2 of $F_3$ so its cardinal is $3^2$. Two finite field which have the same order are isomorphic.
-
0Sorry, can you explain what an extension of degree $2$ means? And where does the irreducibility of $x^2 + 1$ come into play? – 2017-02-15
-
0Its dimension as $F_3$ vector space is 2. – 2017-02-15