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Let $p$ be a prime Let $n$ be an element in $Z_{p}^*$ where $n\not=\pm1$. Define a ring structure on $F = Z_p \times Z_p$. We define the addition by $$ (a_1,b_1) + (a_2,b_2) = (a_1 + a_2, b_1 + b_2). $$

And define multiplication by:

$$ (a_1,b_1) * (a_2,b_2) = (a_1 a_2 + b_1 b_2 n, a_1 b_2 + a_2 b_1). $$

Prove that $F$ is a field.

Okay, so our goal is to show that that both $<R,+>$ is an abelian group as well as $<R,*>$ is an abelian group with identity element not equal to 0. We also want the distributive laws to hold.

$Z_n\times Z_n$ is a cyclic group and is abelian under addition. So that is obvious. And the way the addition is done, our addition fulfills all requirements of a ring. I have on paper also proved communitivity for our multiplication and the distributive property holding.

What I have trouble with proving the inverse element existing for multiplication and it being unique. I know that unity is (1,0) and the additive identity is (0,0).

mvw
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user121615
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2 Answers2

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The map $(a,b) \mapsto \overline{a+bX}$ gives an isomorphism of $F$ with $(\mathbb{Z}/p) [X] / (X^2 - n)$.

If $n$ is not a square modulo $p$, then this is a field of order $p^2$ because $X^2-n$ is irreducible.

But if $n$ is a square modulo $p$, then there are zero divisors: if $k^2 \equiv n\pmod{p}$, then $(k,1)\ast(-k,1) = (0,0)$, so $F$ is not a field.

One elementary way to see that $F$ is a field (and, more generally, which elements have inverses):

By symmetry, $(a,b)$ is invertible if and only if $(a,-b)$ is invertible. But $(a,b)\ast (a,-b) = (a^2 - b^2 n,0)$, which is invertible if and only if $a^2 - b^2 n$ is invertible in $\mathbb{Z}/p$.

Andrew Dudzik
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  • I was told that F is indeed a field. – user121615 Apr 28 '15 at 16:54
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    @user121615 It is not always a field, regardless of what you were told. Most likely, whoever wrote this problem overlooked the correct condition on $n$—it should not be that $n\neq \pm 1$, but rather that $n$ is a quadratic nonresidue modulo $p$. – Andrew Dudzik Apr 28 '15 at 16:58
  • But if this is indeed a proof, it is using very complicated mathematics that I have not encountered yet. I am about 20 chapters into Freiligh's abstract algebra book. Is there a proof on that level anywhere? – user121615 Apr 28 '15 at 16:58
  • Oh okay. I'm not second guessing you or anything. This isn't the first time my professor has given incorrect info. It's just that I have never seen a field like this and I have trouble understanding exactly what this field is. – user121615 Apr 28 '15 at 17:00
  • @user121615 mvw's answer gives the correct inverse, and a more elementary way to think about $F$ as a ring of matrices. My answer should make more sense to you by Chapter 27. – Andrew Dudzik Apr 28 '15 at 17:02
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Regarding the multiplicative inverse: $$ (1,0) = (a_1, b_1) * (a_2, b_2) = \left( \begin{matrix} a_1 & b_1 n \\ b_1 & a_1 \end{matrix} \right) \left( \begin{matrix} a_2 \\ b_2 \end{matrix} \right) = \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) \iff \\ \left( \begin{matrix} a_2 \\ b_2 \end{matrix} \right) = \left( \begin{matrix} a_1 & b_1 n \\ b_1 & a_1 \end{matrix} \right)^{-1} \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \frac{1}{a_1^2 - b_1^2 n} \left( \begin{matrix} a_1 & -b_1 n \\ -b_1 & a_1 \end{matrix} \right) \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) = \frac{1}{a_1^2 - b_1^2 n} \left( \begin{matrix} a_1 \\ -b_1 \end{matrix} \right) $$ This gives $$ (a_1, b_1)^{-1} = \left(\frac{a_1}{a_1^2 -b_1^2 n}, -\frac{b_1}{a_1^2 -b_1^2 n} \right) $$ for $a_1^2 \ne b_1^2 n$. So depending on the choice of $n$ and $p$ there might be elements of $F = \mathbb{Z}_p \times\mathbb{Z}_p$ that get no inverse this way.

If $n < 0$ then $a_1^2 - b_1^2 n = a_1^2 + b_1^2 (-n)$ only vanishes for $(a_1,b_1)=(0,0)$, which will not need one.

If $n > 0$ then $a_1^2 = b_1^2 n \iff a_1 = \pm b_1 \sqrt{n}$. So for non-integer $\sqrt{n}$ everything ist fine.

mvw
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