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Let i be the imaginary unit of $\Bbb{C}$ and $\mathbb{Z}[i]:= \{a + ib, ab \in\mathbb{Z} \} $

So the task is to prove that this a ring regarding multiplication and addition in $\mathbb{C}$ which is commutative, free of zero divisors and posseses a 1 element. I already found out that it is a ring and that the 1 Element is (1,0) and 0 Element is (0,0), so in both cases $i= 0$. So I guess this is like proving a trivial Ring. I am just confused because I am not sure about the $i$. How would I complete the proof? I am stuck because I don't know how to prove that it is free of zero divisors and not a field. Thank you for the help.

  • If you proved the 1 element is the same as the 0 element, then you did something incorrect. They are not. – Arturo Magidin Dec 14 '20 at 13:38
  • I think Bruno meant that $0_{\mathbb Z}=0_{\mathbb Z[i]}$, and the same for $1$, not that $0=1$. Anyway, as a hint for why it's not a field: what's the multiplicative inverse of $1+\mathrm i$? Also, it is free of zero divisors since it's a subring of a field, and fields are free of zero divisors. – Vercassivelaunos Dec 14 '20 at 13:39
  • @Vercassivelaunos Wouldn't $2^{-1}\notin \Bbb Z[i]$ suffice? – Divide1918 Dec 14 '20 at 13:41
  • @Surb: Of course it makes sense to say that $i$ is the imaginary unit in $\mathbb C$. $\mathbb Z$ is naturally a subring of $\mathbb C$, so we can freely adjoin elements of $\mathbb C$ to $\mathbb Z$. – Vercassivelaunos Dec 14 '20 at 13:41
  • @Divide1918: Right, that's even easier. – Vercassivelaunos Dec 14 '20 at 13:42
  • @Vercassivelaunos: We indeed understand what it mean (so in this sens it make sense). I just feel a bit strange to use $\mathbb C$ (being a more complex and general structure than $\mathbb Z$) instead of definding $i$ in a easier way (i.e. as being a roof of $X^2+1\in \mathbb Z[X]$) which would make more sense for me). But at the end, it's not so important, and I agree that my comment wasn't really adapted to the question. – Surb Dec 14 '20 at 13:47
  • Perhaps you mean "$\mathbb{Z}[i]:= {a + ib;:; a,b \in\mathbb{Z} }$" and then later "so in both cases $b=0$". – GEdgar Dec 14 '20 at 15:29

6 Answers6

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The map $\Bbb Z[i]\to \Bbb Z/2\Bbb Z$, $a+bi\mapsto a+b+2\Bbb Z$ is a non-trivial ring homomorphism with non-trivial kernel. Such a thing does not exist for fields.

Hagen von Eitzen
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There are no non-zero zero-divisors because $\mathbb{Z}[i]$ is a subring of $\mathbb{C}$, which is a field. Or, if you prefer: $$ (a+bi)(x+yi)=(ax-by)+(bx+ay)i.$$ So the product is equal to zero if and only if $ax-by=bx+ay=0$. Try to deduce that either $a=b=0$ or $x=y=0$.

If $\mathbb{Z}[i]$ were a field, then every non-zero element would have a multiplicative inverse. Let $\alpha$ be an inverse of $2$. But $\mathbb{Z}[i]$ is a subring of $\mathbb{C}$, so $\alpha \in \mathbb{C}$, and inverses are unique, so $\alpha = 1/2$. However, $1/2 \notin \mathbb{Z}[i]$, contradiction.

57Jimmy
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  • Thank you for your help. The idea with the inverse really helped. Than you so much. –  Dec 14 '20 at 14:33
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Consider the norm $|a + bi| = \sqrt{a^2 + b^2} \geq 1$ for $a + bi \in \mathbb{Z} \setminus \{0\}$. This is multiplicative (it is complex norm).

Hence, the element $1 + 2i$ with $|1 + 2i| = \sqrt{5}$ can't be invertible. If it had an inverse $x$, then $|x| \sqrt{5} = | 1 | = 1$, but norms in the non-zero Gaussian integers are at least $1$.

Sofía Marlasca Aparicio
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The quotient field of $\Bbb Z[i]$ obviously is $\Bbb Q(i)$, which is different from $\Bbb Z[i]$. Hence $\Bbb Z[i]$ cannot be a field, because the quotient field of an integral domain $R$ also is the smallest field containing $R$.

References with many details:

Quotient field of gaussian integers

Prove the fractional field of an integral domain is the smallest field containing the integral domain

Dietrich Burde
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Though there are simple direct proofs, it is worth emphasis that this is an immediate consequence of a fundamental property of integral extensions, i.e. extensions obtained by adjoining algebraic integers (roots of polynomials that are monic, i.e. lead coef $= 1).\,$ Namely, integral extension don't alter unit properties of the base ring, so a nonunit (i.e. noninvertible) element remains nonunit in the extension. In particular if the base ring is not a field then ditto for the integral extension. Let's prove this very simply for your special case.

Write $\,\Bbb I := \Bbb Z[i]\cong \Bbb Z[x]/(x^2+1)\,$ for the Gaussian integers. Note every $\,w\in\Bbb I\,$ is integral over $\,\Bbb Z\,$ being a root of a monic polynomial $\,f(x)=(x\!-\!w)(x\!-\!\bar w) = x^2+bx+c \in\Bbb Z[x].\,$ If an integer $n\in\Bbb Z$ is invertible in $\Bbb I\,$ then $\,w = n^{-1}\,$ is a root of such a quadratic so $\,n^{-2}+b\,n^{-1}+c = 0\,$ $\overset{\times\ n}\Rightarrow\, n^{-1} + b + c\,n = 0,\,$ so $\,n^{-1} = -b - c\,n \in \Bbb Z,\,$ i.e. $\,n\,$ was already invertible in $\,\Bbb Z.\,$ Thus nonunits in $\,\Bbb Z\,$ remain nonunits in $\,\Bbb I,\,$ e.g. $\,2\,$ is not invertible in $\,\Bbb I,\,$ so $\,\Bbb I\,$ is not a field.

Exactly the same simple proof generalizes to any integral extension, e.g. see here. Generally this method easily shows that

$\qquad$ If $\, E < D\,$ is an integral extension of domains then $\,E\,$ is a field $\!\iff\! D\,$ is a field.

This preservation of units in integral extensions mean that they faithfully preserve divisibility properties of the base ring, so e.g. we can deduce divisibility properties of integers from divisibility properties of Gaussian integers and other rings of algebraic integers (where proofs my be simpler because we may be able to reduce nonlinear problems to linear problems via factorizations, e.g. classical proofs of Fermat's Last theorem for small exponents). This would not be true if some nonunit changed to a unit in the extension, e.g. if $2$ became a unit in the extension then proofs by parity would break down in the extension, so it would be weaker from a divisibility standpoint.

Bill Dubuque
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You can model $\mathbb Z[i]$ by using the matrix: $$J=\begin{bmatrix}0&-1\\1&0\end{bmatrix} $$ Observe that $J^2=-I$, and observe that the map $$a+bi \mapsto aI + bJ$$ is an embedding of $\mathbb Z[i]$ into $M_2(\mathbb Z)$. Now, we can observe that, for example: $$det (I+J) = det\begin{bmatrix}1&-1\\1&1\end{bmatrix}=2 $$ and thus $I+J$ is not an invertible matrix (in $M_2(\mathbb Z)$, since $2$ is not a unit in $\mathbb Z$), and therefore $\mathbb Z[i]$ is not a field.

Justin Young
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