# Relationship between subrings and ideals

I just want to make sure that I understand these two objects correctly.

Let $R$ be a ring. A subset $R'\subseteq R$ is a subring provided $R'$ is an additive subgroup that is closed with respect to multiplication.

A subset $I\subseteq R$ is an ideal provided $I$ is an additive subgroup such that $rI\subseteq I$ and $Ir\subseteq I$ for all $r\in R$ (where $rI$ and $Ir$ are defined in the "obvious" way).

Now, in my notes, for the definition of a subring, I didn't specify what I meant by "with respect to multiplication." Looking back now, I assume I meant left multiplication.

So then, if this is correct, an ideal is a subring that is closed with respect to both left and right multiplication.

Is this correct?

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Do your rings have identities? The usual definition of a subring requires that it also have a multiplicative identity, although this isn't universal. – Qiaochu Yuan Nov 22 '10 at 19:17
We have not specified that a ring or subring need a multiplicative identity – Bey Nov 22 '10 at 19:44
@Qiaochu: It really depends on whether you are talking about "subrings-with-1" or "subrings" (i.e., subrings in the category of rings with identity, and subrings in the category of rings). Think of the identity as a unary operation in the former. – Arturo Magidin Nov 22 '10 at 19:53

## 4 Answers

A left-ideal is an abelian subgroup (under addition) of the ring which is closed under left-multiplication by elements of the ring, and not just elements in the ideal. In the commutative case we identify left and right multiplication and then an ideal is a subring of the ring. However, that is not enough. Since the subring requires closure under multiplication of elements contained in itself while a left or right ideal requires closure under corresponding multiplication by elements of the ring.

For example, consider the polynomial ring $R=k[x]$ in one variable over a field $k$. Then $R$ is a commutative ring with identity. $k$ is a subring of $R$ being a field. However, any ideal that contains any element of $k$ is actually the whole ring (since if it contains an element of $k$, it contains $1$ by multiplying it by it's inverse and hence contain every element of $R$). So being an ideal is a stronger requirement.

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Since a ring R is also an Abelian group, is it necessary to specify that subrings or ideals are Abelian subgroups? Certainly we have to stipulate they're subgroups under +, but isn't automatic that they're Abelian? – Bey Nov 22 '10 at 19:43
@Bey: You're right, that's superfluous. You just need that they are subgroups under addition. – Timothy Wagner Nov 22 '10 at 19:47

Remember from groups and vector spaces the notions of "sub-group" and of "sub-space": you must have a subset which is also "closed" with respect to the operations. That is, if you take elements from the subset, and you subject them to the operations you have, you should end up with results that are also in your subset.

With groups, that means that if a subset $S$ of a group $G$ is going to be a group, then it must be the case that whenever $s,t\in S$, then $st\in S$ ("closed under multiplication"; if you take two things in $S$ and multiply them, you don't need to go "outside of $S$" to find the answer, it is all already inside $S$; no need to go outside); and if $s\in S$, then you need $s^{-1}\in S$ ("closed under inverses"); and you need $e\in S$ (closed under the operation that gives the identity element).

With vector spaces, if a subset $W$ of a vector space $V$ is going to be a subspace, you need that $\mathbf{0}\in W$ (closed under the operation that gives the identity element); that if $w_1,w_1\in W$, then $w_1+w_2\in W$ (closed under vector addition); and that if $w\in W$ and $\alpha\in F$, then $\alpha w\in W$ (closed under scalar multiplication).

The same is true for rings: if you have a ring $R$, then in order for a subset $S$ to be a subring you need it to be (i) closed under addition; (ii) closed under additive inverses; (iii) contain the identity element of the sum; and (iv) closed under multiplication (if $a,b\in S$, then $ab\in S$). Saying "is a subgroup" is the same as saying (i), (ii), and (iii).

Ideals, however, are a bit more than simply subrings. They play exactly the analogous role to rings as normal subgroups play to groups. You may recall that a normal subgroup $N$ of a group $G$ is a "subgroup-with-something-extra": not only is it "closed" under the three usual operations of the group (identity element, addition, and inverses), it must also be "closed" under a bunch of other operations (conjugation).

Similarly with ideals: not only must they be "closed" under the four usual operations of the ring (additive identity element, addition, additive inverses, and multiplication), it must also be "closed" under a bunch of other operations (that are like the scalar multiplications in the vector space case): left multiplication by any element of $R$, and right multiplication by any element of $R$. Just like "normal subgroup" is "subgroup with something extra", likewise ideals are "subrings with something extra."

The problem with saying "an ideal is a subring that is closed under left and right multiplication" is that saying "closed under left and right multiplication" only says that if $a,b\in I$, then $ab\in I$; you really want to add the coda of "...left and right multiplication by elements of the ring."

Examples abound. Take $R=\mathbb{R}[x]$, the polynomials with coefficients in the real numbers, and consider $\mathbb{Z}[x]$, the polynomials with integer coefficients. The latter is a subggroup of $R$, and if $p(x),q(x)\in\mathbb{Z}[x]$, then the product $p(x)q(x)$ is also in $\mathbb{Z}[x]$. So $\mathbb{Z}[x]$ is a subring of $R$. It is not, however, an ideal of $R$, because $x\in\mathbb{Z}[x]$, and $\pi\in R$, but their product $\pi x\notin\mathbb{Z}[x]$.

Or take $S=\mathbb{R}[x^2]$, the polynomials with real coefficients in which only even powers of $x$ occur. This is a subring of $R$ but not an ideal, because $x^2 \in S$, $x\in R$, but $x(x^2)\notin S$.

On the other hand, $S=\{p(x)\in R\mid p(0)=0\}$, the collection of all polynomials with constant term equal to $0$, is both a subring and an ideal: if $p(x)\in S$ and $q(x)\in R$, then $q(x)p(x)\in S$ since $q(0)p(0)=0$ (and $p(x)q(x)=q(x)p(x)$). That is, $q(x)S\subseteq S$ and $Sq(x)\subseteq S$.

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In the case of the multiplication operation in a (sub)ring the phrase 'with respect to multiplication' doesn't refer to any 'left multiplication': In a ring there is just one multiplication which may be or not be commutative. Being closed with respect to multiplication means that if $a$ and $b$ belong to $R'$, then also $ab$ (and $ba$ by exchanging the roles of $a$ and $b$) belong to $R'$.

In the non-commutative situation we often distinguish the products $ab$ and $ba$ of two elements $a$ and $b$ by saying that we 'multiply $b$ on left/right by $a$'.

Ideal is more than a subring in the sense that we require that products stay in the ideal even when multiplying by elements that don't belong to the ideal.

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"...we require that products stay in the ideal even when multiplying by elements that don't belong to the ideal." This is what I was forgetting that caused some confusion for me. Thanks =) – Bey Nov 22 '10 at 19:48

In noncommutative rings, the terms "left ideal" and "right ideal" are used indicate if you are multiplying from the left or right. The term "ideal" usualy means that it is a left and right ideal at the same time. In commutative ring theory, which is probably what you are studying, this notion of left and right is meaningless since ab and ba are always the same.

Simply put, an ideal is a very special type of subring, with the added property that if a is in the ideal and r is in the ring, r times a is in the ideal (even if r is not in the ideal). Subrings don't have this extra property in general.

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