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Prove the direct product of rings never form a ring.

And be the direct product $A\times B$:

$+$: $(x,y) + (z,t) = (x+z,y+t)$

$*$: $(x,y) * (z,t) = (x*z, y*t)$

$x,z\in A$.

If we define $A$ and $B$ both commutative rings, with units and inverses itself (So $A$ and $B$ are fields).

Now, $A\times B$ is commutative, it is a ring, there is a unity element.

And since $A$ and $B$ have inverses, the inverse of each element in $A\times B$ $(x,y)$ is just $(x^{-1},y^{-1})$, being $x^{-1}$ the inverse of $x$ in $A$, and so to $y$ in $B$.

So $$(x,y) * (x^{-1},y^{-1}) = (e_{a},e_{b})$$

Defining the unity of $A\times B$ as $(e_{a},e_{b})$

I can't see the error in my writings.

Ottavio
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Gabriela Da Silva
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1 Answers1

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The direct product of fields is never a field; because we have zero divisors, $(1,0)$ and $(0,1)$, for instance, it is not even an integral domain.

But the direct product of rings is a ring.

  • Ops i wanted to say field instead of rings... Anyway, what is the problem in have zero divisors? The definition i saw of fields is:

    Commutative, Unity, and nonzero invertible.

    If it has no zero divisors, so it is an integral domain.

    Is this definition wrong?

     – Gabriela Da Silva Dec 18 '20 at 23:13
  • A field is a commutative ring in which all the nonzero elements are units. A zero divisor is nonzero but can't be a unit. –  Dec 18 '20 at 23:14
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    @GabrielaDaSilva Zero divisors are never invertible under multiplication. Prove this. – CyclotomicField Dec 18 '20 at 23:15
  • Except in the zero ring! – Qiaochu Yuan Dec 18 '20 at 23:57