## Tag Archives: polynomial ring

### A module which has rank 1 but is not free

Let $R = \mathbb{Z}[x]$ and let $M = (2,x)$ be considered as an $R$-submodule of $R$. Show that $\{2,x\}$ is not a basis of $M$. Show that $M$ has rank 1, but is not free with free rank 1.

Note that $2 \cdot x + (-x) \cdot 2 = 0$, so that $x$ and $2$ are not $R$-linearly independent. Thus $\{2,x\}$ has no hope of being a basis of $M$.

Now suppose $\alpha, \beta \in M$ are nonzero. Similarly, $\beta \cdot \alpha + (-\alpha) \cdot \beta = 0$, so that $\{\alpha, \beta\}$ is not $R$-linearly independent. In particular, the rank of $M$ as an $R$-module is at most 1. On the other hand, since $R$ is an integral domain, every singleton set is linearly independent. So the rank of $M$ over $R$ is exactly 1.

Suppose now that $M$ is free with free rank 1. That is, $M = R\alpha$ for some $\alpha$. Now $2 = r\alpha$ for some $r \in R$, so that $\alpha$ has degree 0. That is, $\alpha \in \mathbb{Z}$, and we have $\alpha \in \{\pm 1, \pm 2\}$. If $\alpha = \pm 1$, then $M = R$. This is a contradiction since (for example) $1 \notin (2,x)$. So, without loss of generality, $\alpha = 2$. But then $M = 2\mathbb{Z}[x]$, and so $x \notin M$, a contradiction. Thus $M$ is not free of free rank 1 as an $R$-module.

### Exhibit a nontrivial alternating bilinear map from a tensor square

Let $R = \mathbb{Z}[x,y]$ and $I = (x,y)$.

1. Prove that if $ax + by = a^\prime x + b^\prime y$ in $R$, then there exists $h \in R$ such that $a^\prime = a + yh$ and $b^\prime = b - xh$.
2. Prove that the map $\varphi(ax+by,cx+dy) = ad-bc \mod (x,y)$ is a well-defined, alternating, bilinear map $I \times I \rightarrow R/I$.

Note that $(a-a^\prime)x = (b^\prime - b)y$. Since $x$ and $y$ are irreducible (hence prime) in the UFD $R$, we have that $y$ divides $a - a^\prime$ and $x$ divides $b^\prime - b$. Say $a - a^\prime = yh$ and $b^\prime - b = xh^\prime$. Since $R$ is a domain, we see that $h = h^\prime$, and thus $a^\prime = a - yh$ and $b^\prime = b + xh$.

Suppose now that $a_1x + b_1y = a_2x + b_2y$ and $c_1x + d_1y = c_2x + d_2y$. Then there exist $h,g \in R$ such that $a_2 = a_1 - yh$, $b_2 = b_1 + xh$, $c_2 = c_1 - yg$, and $d_2 = d_1 + xg$. Evidently, $a_2d_2 - b_2c_2 \equiv a_1d_1 - b_1c_1 \mod (x,y)$. Thus $\varphi$ is well defined. Note that $\varphi((a_1+a_2)x + (b_1+b_2)y, cx+dy) = (a_1+a_2)d - (b_1+b_2)c$ $= (a_1d - b_1c) + (a_2d - b_2c)$ $= \varphi(a_1x+b_1y,cx+dy) + \varphi(a_2x+b_2y,c_x+d_y)$, so that $\varphi$ is linear in the first argument; similarly it is linear in the second argument. Moreover, $\varphi((ax+by)r,cx+dy) = \varphi(arx+bry,cx+dy)$ $= ard-brc$ $= \varphi(ax+by,r(cx+dy))$. Thus $\varphi$ is $R$-bilinear. Since $\varphi(ax+by,ax+by) = ab-ab = 0$, $\varphi$ is alternating. Finally, note that $\varphi(1,0) = 1$. Thus there exists a nontrivial alternating $R$-bilinear map on $I \times I$.

### Polynomial rings over commutative rings are flat modules over their coefficient rings

Let $R$ be a commutative ring. Show that as a left unital $R$-module, $R[x]$ is flat.

We claim that $R[x]$ is in fact a free $R$-module with basis $B = \{x^t \ |\ t \in \mathbb{N} \}$, using the convention that $x^0 = 1$. Certainly every element of $R[x]$ is an $R$-linear combination of elements of $B$, and if $\sum r_ix^i = 0$ then $r_i = 0$ for all $i$.

Free modules are flat, so $R[x]$ is flat.

### Characterize the ideals in F[x] principally generated by monic irreducibles

Let $F$ be a field, let $E$ be an extension of $F$, and let $\alpha \in E$ be algebraic over $F$ with minimal polynomial $p(x)$. Prove that the ideal $(p(x))$ in $F[x]$ is equal to $\{ q(x) \in F[x] \ |\ q(\alpha) = 0 \}$.

If $t(x) \in (p(x))$, then $t(x) = p(x)s(x)$ for some $s(x) \in F[x]$. Thus $t(\alpha) = p(\alpha)s(\alpha) = 0$.

Conversely, if $t(\alpha) = 0$, then the minimal polynomial $p(x)$ divides $t(x)$. So $t(x) \in (p(x))$.

### Establish an isomorphism between two R-algebras

Let $S$ be a commutative ring with 1 and let $R \subseteq S$ be a subring containing 1. Let $I \subseteq R[x_1,\ldots,x_n]$ be an ideal. Prove that as $S$-algebras, $S \otimes_R R[x_1,\ldots,x_n]/I$ and $S[x_1,\ldots,x_n]/IS[x_1,\ldots,x_n]$ are isomorphic.

Define $\varphi : S \times R[x_1,\ldots,x_n] \rightarrow S[x_1,\ldots,x_n]$ by $\varphi(sr) = sr$. This mapping is clearly $R$-bilinear, and so induces an abelian group homomorphism $\Phi : S \otimes_R R[x_1,\ldots,x_n] \rightarrow S[x_1,\ldots,x_n]$. Since $S$ is commutative, $\Phi$ is also a ring homomorphism. Moreover, $\Phi(1 \otimes 1) = 1$ and $\Phi(a(s \otimes r)) = \Phi(as \otimes r)$ $= asr$ $= a \Phi(s \otimes r)$, so that $\Phi$ is an $S$-algebra homomorphism. Now every simple tensor (hence every element) of $S \otimes_R R[x_1,\ldots,x_n]$ can be written in the form $\sum s_i \otimes m_i$ where the $m_i$ are monic monomials in $R[x_1,\ldots,x_n]$. Note that $\Phi(s \otimes m) = sm$ for any monomial $m$ and $s \in S$, so that $\Phi$ is surjective. Moreover, if $\Phi(\sum s_i \otimes m_i) = \sum s_im_i = 0$ where the $m_i$ are distinct, then $s_i = 0$ for all $i$, and thus $\sum s_i \otimes m_i = 0$. So $\mathsf{ker}\ \Phi = 0$, and thus $\Phi$ is injective. So $\Phi$ is an $S$-algebra isomorphism.

Let $\pi : R[x_1,\ldots,x_n] \rightarrow R[x_1,\ldots,x_n]/I$ denote the natural projection, and let $\Psi = (1 \otimes \pi) \circ \Phi^{-1}$. Then $\Psi : S[x_1,\ldots,x_n] \rightarrow S \otimes_R R[x_1,\ldots,x_n]/I$ is a surjective $S$-algebra homomorphism. Now let $r \in I$ and $s = \sum s_im_i \in S[x_1,\ldots,x_n]$. Evidently, $\Psi(rs) = \Psi(\sum s_im_ir)$ $= (1 \otimes \pi)(\sum s_i \otimes m_ir)$ $= 0$. In particular, $IS[x_1, \ldots, x_n] \subseteq \mathsf{ker}\ \Psi$. By the (generalized) first isomorphism theorem, the mapping $\overline{\Psi} : S[x_1,\ldots,x_n]/IS[x_1,\ldots,x_n] \rightarrow S \otimes_R R[x_1,\ldots,x_n]$ given by $\overline{\Psi}(\overline{s}) = \Psi(s)$ is well-defined and a surjective $S$-algebra homomorphism.

Now define $\theta : S \times R[x_1,\ldots,x_n]/I \rightarrow S[x_1,\ldots,x_n]/IS[x_1,\ldots,x_n]$ by $\theta(s,r+I) = sr+IS[x_1,\ldots,x_n]$. Note that if $p-q \in I$, then $sp-sq \in IS[x_1,\ldots,x_n]$, so that $\theta(s,p+I) = \theta(s,q+I)$. That is, $\theta$ is well-defined. Clearly $\theta$ is $R$-bilinear, so that we have an $S$-algebra homomorphism $\Theta : S \otimes_R R[x_1,\ldots,x_n]/I \rightarrow S[x_1,\ldots,x_n]/IS[x_1,\ldots,x_n]$ such that $\Theta(s otimes \overline{r}) = \overline{sr}$.

We claim that $\overline{\Psi}$ and $\Theta$ are mutual inverses. To see this, note that for all $s \in S$ and all monic monomial $m$, we have $(\Theta \circ \overline{\Psi})(\overline{sm}) = \Theta(\overline{\Psi}(\overline{sm}))$ $= \Theta(\Psi(sm))$ $= \Theta((1 \otimes \pi)(\Phi^{-1}(sm)))$ $= \Theta((1 \otimes \pi)(s \otimes m))$ $= \Theta(s \otimes m+I)$ $= \overline{sm}$. Thus $\Theta \circ \overline{\Psi} = \mathsf{id}$. Similarly, $(\overline{\Psi} \circ \Theta)(s \otimes m+I) = \overline{\Psi}(\Theta(s \otimes m+I))$ $= \overline{\Psi}(\overline{sm})$ $= \Psi(sm)$ $= (1 \otimes \pi)(\Phi^{-1}(sm))$ $= (1 \otimes \pi)(s \otimes m)$ $= s \otimes m+I$. Thus $\overline{\Psi} \circ \Theta = \mathsf{id}$.

Hence, as $S$-algebras, we have $S \otimes_R R[x_1,\ldots,x_n]/I \cong S[x_1,\ldots,x_n]/IS[x_1,\ldots,x_n]$.

### Establish an isomorphism between two R-algebras

Let $S$ be a commutative ring with 1 and let $R \subseteq S$ be a unital subring. Prove that as $S$-algebras, $S[x]$ and $S \otimes_R R[x]$ are isomorphic.

Define $\varphi : S \times R[x] \rightarrow S[x]$ by $\varphi(s,r) = sr$. This mapping is certainly $R$-bilinear, and so (since $S$ is commutative) induces an additive group homomorphism $\Phi : S \otimes_R R[x] \rightarrow S[x]$ such that $s \otimes r \mapsto sr$. Note that $\Phi((s_1 \otimes r_1)(s_2 \otimes r_2)) = \Phi(s_1s_2 \otimes r_1r_2)$ $= s_1s_2r_1r_2$ $= s_1r_1s_2r_2$ $= \Phi(s_1 \otimes r_1) \Phi(s_2 \otimes r_2)$, so that $\Phi$ is a ring homomorphism. Moreover, we have $\Phi(1 \otimes 1) = 1$ and $\Phi(a (s \otimes r)) = \Phi(as \otimes r)$ $= asr$ $= a\Phi(s \otimes r)$, so that $\Phi$ is an $S$-algebra homomorphism.

Since $\Phi(s \otimes x^k) = sx^k$, $\Phi$ is surjective. Note that every simple tensor (hence every element) of $S \otimes_R R[x]$ can be written in the form $\sum s_i \otimes x^i$. If $\Phi(\sum s_i \otimes x^i) = \sum s_ix^i = 0$, then we have $s_i = 0$ for all $i$, and thus $\sum s_i \otimes x^i = 0$. In particular, $\mathsf{ker}\ \Phi = 0$, so that $\Phi$ is injective.

Thus $S[x]$ and $S \otimes_R R[x]$ are isomorphic as $S$-algebras.

### Relatively prime polynomials over a field have no common roots

Let $F$ be a field and let $p,q \in F[x]$ be relatively prime. Prove that $p$ and $q$ can have no roots in common.

We noted previously that $F[x]$ is a Bezout domain. In particular, if $p$ and $q$ are relatively prime then there exist $a,b \in F[x]$ such that $ap + bq = 1$. If $\zeta$ is a root of both $p$ and $q$, then we have $1 = 0$, a contradiction.

### In a polynomial ring over a field, greatest common divisors exist and are unique up to units and Bezout’s identity holds

Let $F$ be a field and let $a(x),b(x) \in F[x]$. Then a greatest common divisor $d(x)$ of $a(x)$ and $b(x)$ exists and is unique up to multiplication by a unit. Moreover, there exist $s(x)$ and $t(x)$ such that $d = as+bt$.

This follows because $F[x]$ is a Euclidean domain.

### Divisors of irreducible polynomials over a field

Let $F$ be a field and suppose $p(x) \in F[x]$ is irreducible. If $q(x)|p(x)$, what can we say about $q(x)$?

Note that the units in $F[x]$ are precisely the nonzero constant polynomials. If $q(x)|p(x)$, then we have $p(x) = q(x)t(x)$ for some $t(x) \in F[x]$. Since $p(x)$ is irreducible, either $q(x) \in F^\times$ is a unit, or $q$ is a constant multiple of $p(x)$.

### Hilbert’s Basis Theorem for polynomial rings: Every ideal in a polynomial ring over a field is finitely generated

Let $F$ be a field and let $R = F[x_1, \ldots, x_t]$. Let $I \subseteq R$ be a nonzero ideal. Fix a monomial order. Prove that $I$ has a Gröbner basis with respect to the monomial order and conclude that $I$ is finitely generated.

Let $I \subseteq R$ be a nonzero ideal, and consider the monomial ideal $LT(I)$. By Dickson’s lemma, $LT(I)$ is finitely generated by monomials; say $LT(I) = (a_1, \ldots, a_k)$. Without loss of generality we may assume that no $a_i$ divides another. Recall that $LT(I) = (LT(g_i) \ |\ g_i \in I)$. In particular, we have that for each $i \in [1,k]$, there exist $g_i \in I$ and $j \in [1,k]$ such that $a_j|LT(g_i)|a_i$. In fact, $a_j = a_i$, so that $a_i$ and $LT(g_i)$ are associates. That is, we have $LT(I) = (LT(g_1), \ldots, LT(g_k))$ for some $g_i \in I$. By Proposition 24 on page 322 of D&F, $G = \{g_1, \ldots, g_k\}$ is a Gröbner basis for $I$. Then $I = (G)$ is finitely generated.