Ideals

Ideals are the backbone of algebraic geometry, where the Nullstellensatz sets up a dictionary between ideals in polynomial rings and geometric varieties, so that studying a shape becomes studying an ideal. In algebraic number theory, ideals in rings of integers restore unique factorization exactly as Dedekind intended, underlying modern computations of class groups and the arithmetic of number fields. In coding theory and cryptography, cyclic error-correcting codes are literally ideals in quotient rings $\mathbb{F}_q[x]/(x^n-1)$, and ideal lattices underlie lattice-based post-quantum cryptosystems. In commutative algebra and computer algebra systems, Gröbner basis algorithms operate directly on ideals of polynomial rings to solve systems of equations and automate geometric reasoning. In noncommutative algebra and representation theory, one-sided ideals of group rings and matrix algebras correspond to submodules and organize the classification of simple algebras via the Artin–Wedderburn theorem.

Ask the tutor: How does the kernel of a ring homomorphism turn out to be exactly a two-sided ideal, and why is that the deep reason ideals were invented in the first place?

Think of a ring $R$ as a universe of objects you can add, subtract, and multiply. A subring is a self-contained sub-universe: closed under its own operations, but otherwise indifferent to the rest of $R$. An ideal is different in spirit: it is a subset that cannot escape being multiplied into by outside elements. If $a$ is in the ideal and $r$ is anywhere in $R$, then $ra$ (and, for two-sided ideals, $ar$) is forced right back into the ideal. This is a kind of algebraic gravity — the ideal is a "black hole" for multiplication. The picture to keep in mind is $\mathbb{Z}$ and its ideal $6\mathbb{Z} = \{\dots, -12,-6,0,6,12,\dots\}$. Multiply $6$ (or any multiple of $6$) by any integer, and you land back in $6\mathbb{Z}$. Contrast this with the subring $\mathbb{Z} \subseteq \mathbb{Q}$: it is closed under addition and multiplication, but it does not absorb multiplication by arbitrary rationals ($\tfrac12 \cdot 1 = \tfrac12 \notin \mathbb{Z}$), so it is a subring but not an ideal of $\mathbb{Q}$. The absorption property is what lets you build the quotient ring $R/I$, because it guarantees that multiplying cosets $(r+I)(s+I)$ gives a well-defined coset $rs + I$ regardless of which representatives $r,s$ you pick.

Example 1: A non-principal ideal in $\mathbb{Z}[x]$, computed explicitly. Let $R = \mathbb{Z}[x]$, the ring of polynomials with integer coefficients, and let $S = \{2, x\}$. We compute $(S)$ using Part (B) of the theorem (valid since $\mathbb{Z}[x]$ is commutative with identity):

Concretely, this ideal consists of all polynomials in $\mathbb{Z}[x]$ whose constant term is even. Let us verify both inclusions. ($\subseteq$): If $h(x) = 2f(x) + xg(x)$, the constant term of $2f(x)$ is $2f(0)$, an even integer, and the constant term of $xg(x)$ is $0$ (since $x$ divides every term). So the constant term of $h$ is $2f(0)$, which is even. ($\supseteq$): Suppose $h(x) = a_0 + a_1 x + \cdots + a_n x^n \in \mathbb{Z}[x]$ has $a_0$ even, say $a_0 = 2c$. Write

which exhibits $h$ in the form $2f(x) + xg(x)$ with $f(x) = c$ and $g(x) = a_1 + a_2 x + \cdots + a_n x^{n-1}$. So indeed $(2,x) = \{h \in \mathbb{Z}[x] : h(0) \text{ is even}\}$. This ideal is not principal: there is no single polynomial $p(x) \in \mathbb{Z}[x]$ with $(p) = (2,x)$. Briefly: if such $p$ existed, then $2 = p \cdot f$ for some $f$, forcing $p$ to be a constant ($\pm1$ or $\pm2$) by degree considerations; $p = \pm 1$ would give $(p) = \mathbb{Z}[x] \ni 1$, but $1 \notin (2,x)$ since $1$ has odd constant term — contradiction; $p = \pm 2$ would need $x \in (2)$, i.e. $x = 2g(x)$ for some $g \in \mathbb{Z}[x]$, impossible since $2$ does not divide the coefficient $1$ of $x$ in $\mathbb{Z}$. This example shows that even in a very concrete, familiar ring, ideal generation with two generators genuinely needs two generators — a phenomenon central to the later theory of Principal Ideals and Generation and unique factorization.

Example 2: Left vs. right ideals in $M_2(\mathbb{R})$, a genuinely noncommutative computation. Let $R = M_2(\mathbb{R})$, the ring of $2\times 2$ real matrices, which is noncommutative. Consider

We check $L$ is a left ideal but not a right ideal, using the ideal test (Part A). Additive closure: clear, since $L$ is closed under entrywise subtraction (it is a $2$-dimensional real subspace of $M_2(\mathbb{R})$). Left absorption: let $r = \begin{pmatrix} p & q \\ s & t \end{pmatrix} \in R$ and $\ell = \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} \in L$. Then

Wait — we must check this lands in $L$, i.e., that the bottom row is zero. Recompute carefully: row 2 of the product is (row 2 of $r$) times (columns of $\ell$): entry $(2,1) = s\cdot a + t \cdot 0 = sa$, and entry $(2,2) = s \cdot b + t \cdot 0 = sb$. These need not vanish for arbitrary $s,t,a,b$ — for instance take $r = \begin{pmatrix}0&0\\1&0\end{pmatrix}$, $\ell = \begin{pmatrix}1&0\\0&0\end{pmatrix}$: then $r\ell = \begin{pmatrix}0&0\\1&0\end{pmatrix} \notin L$. This shows the subset $L$ as defined is not a left ideal after all — an instructive miscalculation to catch. Let us instead use the correct set that is a left ideal: rows are absorbed from the left when we fix a column, not a row. Define

(matrices whose second column is zero). Check left absorption: for $r = \begin{pmatrix} p&q\\s&t\end{pmatrix}$ and $\ell = \begin{pmatrix} a&0\\b&0\end{pmatrix}$,

since the second column is indeed zero, matching the form of $L'$. Additive closure is immediate as before, so by the ideal test, $L'$ is a left ideal. Now check $L'$ is not a right ideal: take $\ell = \begin{pmatrix}1&0\\0&0\end{pmatrix} \in L'$ and $r = \begin{pmatrix}0&1\\0&0\end{pmatrix} \in R$. Then

since its second column is nonzero. So $\ell r \notin L'$, confirming right absorption fails. This example (corrected after catching the initial error) shows concretely how "column-zero" matrices form a left ideal (they absorb from the left because left multiplication acts on rows, mixing columns internally without touching the zero column), while the symmetric "row-zero" matrices $\{\begin{pmatrix}a&b\\0&0\end{pmatrix}\}$ form a right ideal by an entirely parallel computation: for $\rho = \begin{pmatrix}a&b\\0&0\end{pmatrix}$ and $r = \begin{pmatrix}p&q\\s&t\end{pmatrix}$,

which has zero second row, so lies in the row-zero set — this is exactly right absorption. Neither $L'$ nor the row-zero set is two-sided, and indeed $M_2(\mathbb{R})$ is a simple ring: it has no proper nonzero two-sided ideals at all, only these asymmetric one-sided ideals.