# The ideals of $A \times B$ for fields $A,B$.are principal

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.

• 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 $A \times B$. – Dan Shved Feb 26 '15 at 17:12

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 $(0,b)$.

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.