1
$\begingroup$

Let $A$, $B$ be fields. I showed that $A\times B$ is ring which is not field. I need to show that every ideal in $A\times B$ is principal.

Let $I$ ideal of $A$ and $J$ ideal of $B$. If $I\times J=<(0,0)>$, ideal $I\times J$ is principal. Let $(a,b)\in I\times J$. $<(a,b) >$ is contained in $I\times J$. I need to show $I\times J \subset <(a,b) >$.

I don't know how.

$\endgroup$
  • $\begingroup$ First of all, you cannot show that $I \times J \subseteq \langle(a,b)\rangle$, as shown in quid's answer. Even more importantly, your plan has another flaw: you only consider ideals in $A\times B$ that can be decomposed into a direct product $I \times J$, where $I \subseteq A$ and $J \subseteq B$, but you actually need to consider arbitrary ideals in $A \times B$. $\endgroup$ – Dan Shved Feb 26 '15 at 17:12
1
$\begingroup$

You cannot show this for whatever $(a,b)$. For example note what happens if you happen to choose $(a,b)=(0,0)$. So you need to make sure to choose a "good" element.

Consider the case:

  • $I$ contains an element $(a,b)$ with both $a,b$ nonzero.

  • $I$ does not contain such an element. (Show that then $I$ either contains only elemennts of the form $(a,0)$ or $I$ contains only elemennts of the form $(0,b)$.

$\endgroup$
1
$\begingroup$

All my rings are commutative, the extension of the lemma to the non-commutative case being quite obvious.

Lemma. Let $(A_i)_{i=1,\ldots,n}$ be a (finite) family of rings with $1$. Then ideals of $A = \prod_{i=1}^n A_i$ are of the form $I_1 \times \ldots\times I_n$, where $I_i$ is an ideal of $A_i$ for each $i\in\{1,\ldots,n\}$.

Proof. By induction on $n$, the case $n=0$ beeing trivial, one sees that it suffices to prove the assertion for $n=2$. Let $K$ be an ideal of a product $A\times B$ of two rings with $1$, and note $p : A\times B\to A$ and $q : A\times B\to B$ the two canonical projections. Then $I = p^{-1}(K)$ (resp. $J = q^{-1}(K)$) is an ideal of $A$ (resp. of $B$.) (This is general, the inverse image of an ideal by a ring morphism is always an ideal.) Obviously $K\subseteq I\times J$. To show the inverse inclusion, let $(a,b)\in I\times J$. Then $(a,b') \in K$ and $(a',b)\in K$ for some $(a',b')\in A\times B$. Then $(a,b) = (1,0)(a,b') + (0,1)(a',b) \in K$. $\square$

Corollary. With same notations as in the lemma, if all $A_i$'s are principal ideal rings, so is $A$.

Remark. This is particularly the case if all $A_i$'s are PID's. But beware that in this case $A$ is not a PID if $n>1$, as it is not even a domain.

Apply the corollary to your case, as fields are particulare cases of principal ideal rings.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.