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Let $p$ a prime number. $\mathbb{Z}_p \times \mathbb{Z}_p$ with sum given by $(a,b)+(c,d) = (a+c,b+d)$ and multiplication given by $(a,b)*(c,d)=(ac-bd,ad+bc)$ is a field for some prime greater than two and $C_2$ is not a field.

What I did: $C_2$ is not a field, because $(1,0)$ is the neutral, and $(1,1)*(c,d) = (1,0)$ gives $c-d=1, d+c=0$, which have no solution in $\mathbb{Z}_2$, then it is not a field. But I don't know how to do for $p>2$, because all I know is check case by case, and I think it us not supposed to be that much of work.

Thanks.

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The idea is that this multiplication is the same multiplication for complex numbers:

$$ (a + bi)(c + di) = (ac - bd) + (ad + bc)i. $$

Therefore, for $p > 2$, $C_p$ is a field if and only if $i \notin \mathbb Z_p$ meaning no element of $\mathbb Z_p$ has square $-1$. (Why is this true?)

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  • $\begingroup$ $i=(0,1)$ in Z_2, right? How can I see $i$ in the other Z_p? $\endgroup$ – User1618 Sep 17 '18 at 16:28
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    $\begingroup$ @Alnitak $\mathbb Z_2$ is a special case. But for example in $\mathbb Z_5$ we have $2^2 = 4 = -1$ and accordingly $(2 + i)(2 - i) = 2^2 + 1^2 = 0$ which in $C_p$ implies that $(2,1)(2,-1) = (0,0)$. This creates an issue in $C_5$ because $2 = \sqrt{-1}$ so $(2,0) = i$ but we also want $(0,1) = i$. If $i \notin \mathbb Z_p$ then we can write $(a, b) = a + bi$ and perform multiplication as if we had adjoined a root of $-1$ to our ring. $\endgroup$ – Trevor Gunn Sep 17 '18 at 19:01
  • $\begingroup$ Do you know which is the least prime that makes C_p field? I checked 3,7,11 and 13 and got not field. Maybe I'm not doing it right... $\endgroup$ – User1618 Sep 17 '18 at 23:00
  • $\begingroup$ @Alnitak $p = 3$ should be a field. $\endgroup$ – Trevor Gunn Sep 18 '18 at 0:09
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I find this question to be quite intrigueing because it addresses the issue of extending the underlying structure of the field of complex numbers $\Bbb C$ to arbitrary fields, in particular to arbitrary finite fields; in a sense we might say it is about the process of complexification, with which we are familiar from $\Bbb R \subset \Bbb C$, to general fields $F$.

In the case $F = \Bbb R$, which is somewhat prototypical, we may begin by considering the possibility of products on $\Bbb R^2 = \Bbb R \times \Bbb R$ which, given that addition is defined component-wise,

$(a, b) + (c, d) = (a + c, b + d), \; a, b, c, d \in \Bbb R, \tag 1$

are consistent with the field axioms. For example, we may immediately rule out component-wise multiplication,

$(a, b)(c, d) = (ac, bd), \tag 2$

the identity element of which is $(1, 1)$, because it has zero-divisors, e.g., assuming $a \ne 0 \ne d$,

$(a, 0)(0, d) = (a \cdot 0, 0 \cdot d) = (0, 0), \tag 3$

even though

$(a, 0) \ne (0, 0) \ne (0, d); \tag 4$

thus $\Bbb R^2$ with multiplication defined by (2) cannot be a field; if, however we take

$(a, b)(c, d) = (ac - bd, ad + bc), \tag 5$

to be our product, then if

$(a, b)(c, d) = (0, 0), \tag 6$

noting that

$(a, -b)(a, b) = (a, b)(a, -b) = (a^2 + b^2, 0) \tag 7$

we find

$((a^2 + b^2)c, (a^2 + b^2)d) = (a^2 + b^2, 0)(c, d) = (a, -b)(a, b)(c, d) = (a, -b)(0, 0) = (0, 0), \tag 8$

whence

$(a^2 + b^2)c = (a^2 + b^2)d = 0, \tag 9$

which, in the event that $a^2 + b^2 \ne 0$, forces

$c = d = 0, \tag{10}$

and thus we see there are no zero divisors in $\Bbb R^2$ with respect to the product rule (5); to put it another way, $\Bbb R^2$ is an integral domain with the operations (1), (5); in fact, we may show it to be a field under such circumstances, for then we have

$(a(a^2 + b^2)^{-1}, -b(a^2 + b^2)^{-1})(a, b) = (a^2(a^2 + b^2)^{-1} + b^2(a^2 + b^2)^{-1}), 0)$ $= ((a^2 + b^2)(a^2 + b^2)^{-1}, 0) = (1, 0), \tag{11}$

which shows that every $(a, b) \ne (0, 0)$ has an inverse in $\Bbb R^2$, and thus the field axioms are satisfied.

We wish to investigate when the choice of multiplication operation (5) makes $F^2 = F \times F$ into a field; i.e, when $\Bbb R$ is replaced by an arbitrary field $F$.

We accept that $F \times F$ with "addition" given by (1) and multiplication by (5) forms a commutative ring with unit given by $(1, 0)$; these assertions are most easily checked; our concern then is with the existence of multiplicative inverses.

If we trace through the calculations presented in (1) through (12) above we see that many of the essential points carry over directly from the case $\Bbb R^2 = \Bbb R \times \Bbb R$ to $F^2 = F \times F$; for instance, it is clearly shown above that

$a^2 + b^2 \ne 0 \Longrightarrow \exists (a, b)^{-1}, \tag{12}$

and we may simply prove the converse

$\exists (a, b)^{-1} \Longrightarrow a^2 + b^2 \ne 0: \tag{13}$

if $(a, b)$ has an inverse, then there is $(c, d)$ such that

$(a, b)(c, d) = (1, 0); \tag{14}$

then

$(a, -b)(a, b)(c, d) = (a, -b), \tag{15}$

or

$(a^2 + b^2, 0)(c, d) = (a, -b), \tag{16}$

or

$(a^2 + b^2)c = a, \; (a^2 + b^2)d = -b; \tag{17}$

now $(a, b) \ne (0, 0)$ lest (15) fail, and thus $a^2 + b^2 \ne 0$ as well. It follows then that $F \times F$ is a field with multiplication (5) provided

$\forall (0, 0) \ne (a, b) \in F \times F, \; 0 \ne a^2 + b^2 \in F; \tag{18}$

so then $F \times F$ will obey the field axioms if and only if the function $a^2 + b^2$ has no nontrivial zeroes. For finite $F$, this may be checked manually, one pair $(a, b)$ at a time; if $\vert F \vert$ is not too large, this is a feasible calculation. The amount of effort may be reduced a little bit by noting that

$a^2 + b^2 = 0 \Longrightarrow [a, b \ne 0 ] \vee [a = b = 0]; \tag{19}$

that is, either $a$ and $b$ are both zero, or neither is zero; this is easy to see: for example, if $a = 0$, then $a^2 + b^2 = 0$ reduces to

$b^2 = 0 \Longrightarrow b = 0; \tag{20}$

thus in fact we only need evaluate $a^2 + b^2$ on those pairs $a \ne 0 \ne b$.

Some examples: in $\Bbb Z_2$, have $1^2 + 1^2 = 0$, so $\Bbb Z_2 \times \Bbb Z_2$ is not a field with multiplication (5); for $\Bbb Z_3$, however, we have

$1^2 + 1^2 = 1 + 1 = 2; \; 1^2 + 2^2 = 1 + 1 = 2; \; 2^2 + 2^2 = 1 + 1 = 2, \tag{21}$

so $\Bbb Z_3 \times \Bbb Z_3$ is a field; in $\Bbb Z_5$, however,

$1^2 + 2^2 = 1 + 4 = 0, \tag{22}$

so we rule it out; next, in $\Bbb Z_7$,

$1^2 + 1^2 = 2; \; 1^2 + 2^2 = 5; \; 1^2 + 3^2 = 3; \; 1^2 + 4^2 = 3; \; 1^2 + 5^2 = 5; \; 1^2 + 6^2 = 2; \tag{23}$

$2^2 + 2^2 = 1; \; 2^2 + 3^2 = 6; \; 2^2 + 4^2 = 6; \; 2^2 + 5^2 = 1; 2^2 + 6^2 = 5; \tag{24}$

$3^2 + 3^2 = 4; \; 3^2 + 4^2 = 4; \; 3^2 + 5^2 = 6; \; 3^2 + 6^2 = 3; \tag{25}$

$4^2 + 4^2 = 4; \; 4^2 + 5^2 = 6; \; 4^2 + 6^2 = 3; \tag{26}$

$5^2 + 5^2 = 1; \; 5^2 + 6^2 = 5; \tag{27}$

$6^2 + 6^2 = 2; \tag{28}$

to check $\Bbb Z_{11}$ in this manner, that is, attempting to verify $a^2 + b^2 \ne 0$ for each pair $a, b \in \Bbb Z_{11}$, is really starting to look like a lot of work; however, there is an expedient which can save us hella lot of arithmetic base $11$ or in any other field, for that matter. We note that, for $0 \ne a, b \in F$, the equation

$a^2 + b^2 = 0 \tag{29}$

may be written

$\left ( \dfrac{a}{b} \right )^2 + 1= 0; \tag{30}$

setting

$x = \dfrac{a}{b}, \tag{31}$

we see that (29) has a legitimate solution only if there is $x \in F$ such that

$x^2 + 1 = 0; \tag{32}$

likewise, if (32) is satisfied for some $x \in F$, then we may set $a = bx$ for any $0 \ne b \in F$ and

$a^2 + b^2 = x^2 b^2 + b^2 = (x^2 + 1) b^2 = 0; \tag{33}$

so we see that $F \times F$ is a field with multiplication (5) whenever (32) has no solutions in $F$, for then every $(a, b) \ne (0, 0)$ has an inverse. In the light of (32), we may check $\Bbb Z_{11}$ with relative ease:

$1^2 = 1, \; 2^2 = 4, \; 3^2 = 9, \; 4^2 = 5, \; 6^2 = 3, \; 7^2 = 5, \; 8^2 = 9, \; 9^2 = 4, \; 10^2 = 1; \tag{34}$

we conclude that $\Bbb Z_{11}$ is a field with the product law (5).

To finish off our list, we check "Lucky" $\Bbb Z_{13}$: without presenting the list of all its squares, we simply note that

$5^2 = 25 = -1 \mod{13}, \tag{35}$

so $\Bbb Z_{13}$ may not be "complexified" according to our scheme.

The upshot of all this is that our "complexifcation" of $F \times F$ according to the multiplication (5) goes through precisely for those $F$ in which there is no $x$ for which $x^2 + 1 = 0$, that is, in which a so-called "imaginary unit" does not already exist; when such an $x$ does exist, there will be non-invertible elements with respect to (5) in $F \times F$.

One may in fact inquire further into finite fields $F$ which are not of the form $\Bbb Z_p$; for example, we may look into the field $\Bbb Z_3[x] / (x^2 + x + 2)$, which has $3^2 = 9$ elements all of which may be written as $ay + b$ with $a,b \in \Bbb Z_3$ and $y^2 = -y - 2$; can

$a^2 y^2 + 2aby + b^2 + 1 = (ay + b)^2 + 1 = 0? \tag{36}$

Well, (36) yields

$(2ab - a^2)y + b^2 - 2a^2 + 1$ $= a^2 (-y - 2) + 2aby + b^2 + 1 = a^2 y^2 + 2ab y + b^2 + 1 = 0, \tag{37}$

so (36) is equivalent to

$2ab - a^2 = 0, \; 2a^2 - b^2 = -1; \tag{38}$

if $a \ne 0$, then

$a = 2b, \; 2a^2 - b^2 = 8b^2 - b^2 = b^2 = -1 \Longrightarrow b^2 + 1 = 0, \tag{39}$

which we have already ruled out in $\Bbb Z_3$ (cf. ca. (21)); the case $a = 0$ has already been addressed since then $ay + b \in \Bbb Z_3$.

One could of course continue and attempt similar calculations for many other finite fields of the form $\Bbb Z_p / (p(x))$, or for other cases such as $\Bbb Q(\omega)$, where $\omega$ is a root of unity, but having illustrated these things, I'm ready to stop here.

Nota Bene: In writing up this answer, I did a little research and discovered that in the general theory of quadratic residues it is proved that $-1$ is a square in those $\Bbb Z_p$ with $p \equiv 1 \mod 4$ and is not a square if $p \equiv 3 \mod 4$, a fact corroborated by the cases considered here. End of Note.

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Hint: $(2,1)\times(2,-1)=(0,0)$ in $C_5$

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  • $\begingroup$ You ought to show this is a field for some $p$, not that it isn't. $\endgroup$ – Wojowu Sep 17 '18 at 16:28
  • $\begingroup$ Well, it dealt with half the possible $p$. $\endgroup$ – Empy2 Sep 17 '18 at 16:40

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